Shapes of Molecules
Ionic bonds formed by electrostatic attraction between ions are non-directional in space. When covalent bonds are formed between atoms they occur by the overlapping of atomic orbitals. Some of these orbitals point in particular directions in space (p, d and f). It therefore follows that the bonds formed will have preferred directions in space and the molecules will have definite shapes.
Explanation in terms of electron pair repulsion theory for the shapes of molecules containing up to four pairs of electrons around the central atom such as BeCl2, BF3, CH4, NH3, H2O, and CO2. (Questions will not be set on hybridisation of orbitals). The departure of the bond angles in NH3 and H2O from the predicted tetrahedral, explained in terms of the increasing repulsion between bonding pair-bonding pair, lone pair-bonding pair and lone pair-lone pair electrons.
When you have finished this section you should be able to:
· · Apply the electron-pair-repulsion theory to molecules involving two, three and four pairs of bonding electrons.
· · Extend the electron-pair-repulsion theory to include non-bonding pairs (lone pairs) of electrons.
· · Use the following terms correctly to describe the shapes of molecules - linear, bent or V-shaped, trigonal planar, tetrahedral, trigonal pyramidal, trigonal bipyramidal, octahedral.
The shapes of covalent molecules
A covalent molecule will have a shape which is determined by the angles between the bonds joining the atoms together. A simple theory to account for the shapes of molecules was put forward by Sidgwick and Powell in 1940. It is called the VSEPR (Valence Shell Electron Pair Repulsion) theory. Note that the term valence shell is often used for what we call the outer-shell electrons.
The basic idea of the theory is that charge clouds (i.e. electron pairs) in the outer –shell of an atom in a molecule will arrange themselves to stay as far away from each other as possible. This will minimise the total energy of electrostatic repulsion between them.
With the help of this theory it is possible to predict
(i) The general geometric shape of a molecule.
(ii) The departure of interbond angles from the values for regular
geometric shapes.
Linear molecules
Exercise 1
(a) (a) How many pairs of bonding electrons are there in a molecule of beryllium hydride, BeH2?
(b) If the electron pairs repel each other so that they occupy regions of space as far apart as possible, what shape must the molecules have?
Molecules formed from the bonding of two atoms such as HCl, Cl2, HI and Br2 must be linear as it will always be possible to connect the nuclei of the two atoms by a straight line.
Molecules of the type
A B A
having just two electron pairs around the central atom (an electron deficient compound) will also be linear. A linear arrangement of atoms ( with a bond angle of 180o ) puts the two electron clouds as far apart as possible. e.g. Cl-Be-Cl
You can apply the same principle to molecules containing multiple bonds. Electron pairs in multiple bonds are assumed to occupy the position of one electron pair in a single bond.
Exercise 2
What shape would you expect for the following molecules?
Explain your answer.
(a) CO2 (b) HCN
Trigonal (triangular) planar molecules
Exercise 3
(a) How many pairs of bonding electrons are there in a molecule of boron trifluoride, BF3?
(b) By considering repulsion of electron pairs, what shape do you expect for BF3?
When there are three pairs of electrons around the central
Atom (or their equivalent), the bonds lie in the same plane at an angle of 120o. The arrangement is described as trigonal planar.
A
B A
A
Exercise 4
The molecule of methanal (formaldehyde) is trigonal planar, and the molecule of ethene is based on a trigonal planar arrangement around each carbon atom, as shown below.
H H H
120o C=O C=C
H H H
methanal ethene
Explain why the bonds around each carbon atom are in a trigonal planar arrangement.
Tetrahedral molecules
Both the shapes we have considered so far are planar and are therefore easy to draw on paper. For this reason, dot-and-cross diagrams for linear and trigonal planar molecules can correspond also to their actual shapes. This is not the case for tetrahedral molecules, which are three-dimensional, as you can see in the next exercise.
Exercise 5
(a) How many pairs of bonding electrons are there in a molecule of methane, CH4?
(b) (b) By considering repulsion of electron pairs, what shape do you expect for the CH4 molecule?
(c) By considering bond angles, show that electron pair repulsion would not give the square planar shape of the dot-and-cross diagram below.
(d) State whether you expect the following molecules and ions to have an identical shape or a very similar shape to the CH4 molecule. Give reasons.
(i) NH4+ (ii) SiCl4
When there is a complete octet in the valence shell with four electron pairs around the central atom the molecule adopts a tetrahedral structure with bond angles of 109.5o.
The 3-dimensional tetrahedral shape can be represented as shown :
B
109½ o
A
B
B
B
It is most important for you to be able to visualise tetrahedral bonding and to be able to draw it adequately, because it forms the basis of so many molecular shapes, especially in organic chemistry.
Molecules with lone pairs
So far we have considered only molecules where all the outer shell electrons are used in bonding. However, many molecules have some electrons , called non-bonding pairs or lone pairs, which are not shared between two atoms. These electrons also affect the shape of the molecule by electron pair repulsion.
Exercise 6
(a) (a) Draw a dot-and-cross diagram of the ammonia molecule, and count the number of pairs of electrons surrounding the nitrogen atom.
(b) (b) In which directions (from the N atom) would you expect to find the electron pairs (or regions of greatest charge density)?
(c) (c) What shape is outlined by the four atoms in the molecule?
The tetrahedron is not quite regular, because the four electron pairs are not identical. The lone pair, since it is not shared, remains closer to the N atom than the bonding pairs. The repulsion between non-bonding pairs and bonding pairs is greater, which pushes the hydrogen atoms closer together. This gives a slightly smaller H-N-H bond angle than the tetrahedral angle of 109½ o.
The shape of the ammonia molecule is described as trigonal pyramidal.
When describing the shape only the positions of the bonded atoms are considered (even thought the non-bonded pairs help to determine the overall shape).
The strength of the repulsion between electron pairs decreases in the order:
lone-pair / lone-pair > bonded-pair / lone-pair > bonded-pair / bonded-pair
strongest strong least strong
Two non-bonding pairs
Exercise 7
(a) (a) Draw a dot-and-cross diagram of the water molecule.
(b) (b) Draw the shape of a water molecule, showing how it fits into a tetrahedron.
(c) (c) Use your answer to the previous exercise to predict approximately the H-O-H bond angle.
The shape of the water molecule is described as bent or V-shaped.
Thursday, October 22, 2009
Isomerism in Organic Compounds
Isomerism in Organic Compounds
Structural isomerism for aliphatic compounds containing up to six carbon atoms, to include branched structures and position of the C=C double bonds in alkenes. (Cyclic compounds excluded).
Isomerism (greek isos, equal; meros, parts) occurs when there are two or more compounds (called isomers) with the same molecular formula but different arrangements of their atoms. Isomers have different physical and chemical properties but the differences may be great or small depending on the type of isomerism.
There are two main classes of isomerism;
(i) (i) structural isomerism
(ii) (ii) stereoisomerism
which are themselves sub-divided.
ISOMERISM
structural isomerism stereoisomerism
chain positional functional group geometric optical
isomerism isomerism isomerism isomerism isomerism
Structural isomerism
Structural isomers are molecules with the same molecular formula but with different structural arrangements of the atoms.
· · Chain isomers
This occurs where isomers have different arrangements of the carbon chain. There is only one alkane corresponding to each of the formulae CH4, C2H6 and C3H8.
For butane C4H10 two arrangements are possible.
H H H H H H H
│ │ │ │ │ │ │
H—C—C—C—C—H --- H—C C C—H
│ │ │ │ │ │ │
H H H H H H—C—H H
│
H
butane methylpropane
b.pt. 273K b.pt. 261K
As the number of carbon atoms increases the number of possible isomers increases rapidly.
Number of structural isomers
C5H12
3
C6H14
5
C7H16
9
C10H22
75
C20H42
366,319
C30H62
Over 4000 million
Positional isomers
This occurs when isomers have the same carbon skeleton but the functional group is in different positions in the molecule.
H H H H H H
│ │ │ │ │ │
H—C—C—C—OH-- H—C C C—H
│ │ │ │ │ │
H H H H OH H
propan-1-ol propan-2-ol
Exercise 1
(a) Draw and name the five structural isomers of C6H14.
(b) (b) Write the structural formulae of all the alcohols of molecular formula C4H10O. Name the isomers.
Stereoisomerism : cis-trans isomers for compounds containing one C=C bond, the energy barrier to rotation in these compounds.
Stereoisomerism
Stereoisomers have identical molecular formulae, and the atoms are linked together in the same order, but have different relative positions in space.
The two types of stereoisomerism are
(i) (i) Geometric (or cis-trans ) isomerism
(ii) (ii) Optical isomerism [See Module 4.5]
·Exercise 2
1. Draw and name all the structural isomers, including geometric isomers, of C4H8.
Structural isomerism for aliphatic compounds containing up to six carbon atoms, to include branched structures and position of the C=C double bonds in alkenes. (Cyclic compounds excluded).
Isomerism (greek isos, equal; meros, parts) occurs when there are two or more compounds (called isomers) with the same molecular formula but different arrangements of their atoms. Isomers have different physical and chemical properties but the differences may be great or small depending on the type of isomerism.
There are two main classes of isomerism;
(i) (i) structural isomerism
(ii) (ii) stereoisomerism
which are themselves sub-divided.
ISOMERISM
structural isomerism stereoisomerism
chain positional functional group geometric optical
isomerism isomerism isomerism isomerism isomerism
Structural isomerism
Structural isomers are molecules with the same molecular formula but with different structural arrangements of the atoms.
· · Chain isomers
This occurs where isomers have different arrangements of the carbon chain. There is only one alkane corresponding to each of the formulae CH4, C2H6 and C3H8.
For butane C4H10 two arrangements are possible.
H H H H H H H
│ │ │ │ │ │ │
H—C—C—C—C—H --- H—C C C—H
│ │ │ │ │ │ │
H H H H H H—C—H H
│
H
butane methylpropane
b.pt. 273K b.pt. 261K
As the number of carbon atoms increases the number of possible isomers increases rapidly.
Number of structural isomers
C5H12
3
C6H14
5
C7H16
9
C10H22
75
C20H42
366,319
C30H62
Over 4000 million
Positional isomers
This occurs when isomers have the same carbon skeleton but the functional group is in different positions in the molecule.
H H H H H H
│ │ │ │ │ │
H—C—C—C—OH-- H—C C C—H
│ │ │ │ │ │
H H H H OH H
propan-1-ol propan-2-ol
Exercise 1
(a) Draw and name the five structural isomers of C6H14.
(b) (b) Write the structural formulae of all the alcohols of molecular formula C4H10O. Name the isomers.
Stereoisomerism : cis-trans isomers for compounds containing one C=C bond, the energy barrier to rotation in these compounds.
Stereoisomerism
Stereoisomers have identical molecular formulae, and the atoms are linked together in the same order, but have different relative positions in space.
The two types of stereoisomerism are
(i) (i) Geometric (or cis-trans ) isomerism
(ii) (ii) Optical isomerism [See Module 4.5]
·Exercise 2
1. Draw and name all the structural isomers, including geometric isomers, of C4H8.
Bonding
Whenever atoms, ions or molecules approach each other, there are electrostatic forces acting between them. When the net forces are forces of attraction, and they are strong enough to bind the particles together, we refer to them as chemical bonds. When particles are bound together by chemical bonds, the resulting arrangement is known as the structure of the substance concerned.
All the noble gases except He have completely full inner shells and an outer octet of electrons. As these gases are monatomic and rarely enter into chemical combination it is assumed that the outer octet of electrons is a very stable arrangement and therefore when atoms combine they will try to obtain the noble gas configuration.
The three main types of bonds are:
1. 1. Ionic or electrovalent
2. 2. Covalent
3. 3. Metallic
Ionic bonding restricted to elements in groups I, II, VI and VII, the ions of which have a noble gas structure. Dot and cross diagrams. Characteristic properties of ionic compounds. Sodium chloride as a typical ionic crystal.
When you have finished this section you should be able to:
· · Understand that ionic bonding involves attraction between oppositely charged ions formed by electron transfer
· · Describe the various electron configurations of simple stable ions.
· · List properties typical of ionic compounds and indicate how useful each one is in deciding whether a substance is ionic.
Ionic Bonding
Atoms can achieve noble gas electronic configuration by loss or gain of electrons to form ions. Metals (with low electronegativities) lose electrons to form positive ions and non-metals gain electrons to form negative ions.
Sodium can attain the stable electron configuration of neon by losing one electron
Na (1s22s22p63s1) Na+ (1s22s22p6) + e-
With ten electrons and eleven protons he species formed has a positive charge.
Fluorine is one electron short of the neon electronic configuration. If it obtains one electron (from sodium) it can achieve a full outer shell of eight electrons.
F (1s22s22p5) + e- F- (1s22s22p6)
The species formed has ten electrons and nine protons; it is a negatively charged fluoride ion.
You will soon be aware of the limitations of these diagrams, but they are nevertheless quite useful even in A-level work for describing electron transfer and for checking that all electrons are accounted for. However, electron shell diagrams are rather tedious to draw, and give more information than we usually need when only the outer shell electrons are involved in reactions. We can simplify the above diagram to draw what we call a 'dot-and-cross' diagram showing only the outer shell electrons:
Electrons shown by dots and crosses are of course indistinguishable - we use different symbols only to show where they come from. Note also the use of brackets to separate the charge symbol from the symbols for electrons.
In the first exercise you draw some similar dot-and-cross diagrams yourself
Exercise 1
Draw dot-and-cross diagrams for the formation of:
(a) (a) sodium chloride from sodium atoms and chlorine atoms;
(b) (b) magnesium oxide from magnesium atoms and oxygen atoms;
(c ) calcium fluoride from calcium atoms and fluorine atoms.
Properties of Ionic Compounds
1. 1. Crystalline solids : Ionic compounds are crystalline solids. Ionic crystals are quite hard due to the strong electrostatic forces between the ions
2. 2. High melting and boiling points : All ionic compounds have high melting and boiling points as a large amount of heat energy is required to break the strong electrostatic forces of attraction between the ions.
3. 3. Soluble in water : Many ionic compounds are soluble in water. The water molecules attract the ions and pull the structure apart. They also stabalise the ions once in solution
4. 4. Conduct electricity when molten or dissolved in water : All ionic compounds conduct electricity when molten and are decomposed in the process, which is called electrolysis. Most ionic compounds dissolve in water, which frees the ions so that they can move and carry electrical charge.
Structure of sodium chloride
An ion may be regarded as an electrically charged sphere. A charged sphere is surrounded by a uniform electric field and therefore attracts oppositely charged spheres in all directions.
No particular orientation is favoured, so we often say that ionic bonding is non-directional. What we really mean by this is that the forces of attraction are non-directional.
However, when large numbers of oppositely-charged spherical ions are attracted to each other, the repulsion of the like-charges also comes into play. Mutual repulsion of similarly-charged ions limits the number which can come into contact with an oppositely-charged ion, and effectively fixes their relative positions. Ions, therefore, tend to cling together in large clusters known as ionic lattices in which attractive and repulsive forces are balanced. The particular arrangement of ions depends on their relative charges and sizes.
The ionic lattice of sodium chloride consists of a regular arrangement of alternating sodium and chloride ions extending in three dimensions:
Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl-
Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na+
Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl-
Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na
The lattice can be imagined as consisting of two inter-penetrating face centred cubic structures of sodium ions and chloride ions.
Na+ Na+ Cl- Cl-
Na+ Cl-
Na+ Na+ Cl- Cl-
Each sodium ion is surrounded by six chloride ions as nearest neighbours in the lattice and each chloride ion by six sodium ions. The lattice is said to have 6:6 coordination.
Stereoscopic picture of sodium chloride. The large circles represent Cl- ions and the small circles Na+ ions.
You can see from the above figure that it is impossible to say that a particular sodium ion ‘belongs’ to a particular chloride ion, as they are an equal distance from six. This means that ‘molecules’ of an ionic compound are not formed. The formula used is the atoms in their combining ratio.
Formation of covalent bonds in terms of the sharing of electron pairs between atoms. Dot and cross diagrams. Multiple bonds exemplified by C2H4, N2 and CO2. The octet rule and its limitations, eg BeCl2, BF3. The coordinate bond as a special case of the covalent bond eg NH4+. Dot and cross diagrams are required to show only the outermost electrons but, where appropriate, charges should be included.
When you have finished this section you should be able to:
· · Describe covalent bonding in terms of shared pairs of electrons.
· · Draw dot-and-cross diagrams for a variety of covalent molecules.
· · Explain the term dative covalent bond.
· · Draw dot-and-cross diagrams and structural formulae including dative bonds.
· · For polyatomic ions, distinguish between the covalent bonding within the ion and the ionic bonding between ions.
· · Draw dot-and-cross diagrams for at least two covalent compounds in which an atom other than hydrogen has fewer than eight outer shell electrons.
· · Draw dot-and-cross diagrams for at least two covalent compounds in which an atom has more than eight outer shell electrons.
1.2 Covalent Bonding
Covalent bonds are formed by the sharing of electrons between atoms. The two atoms have to approach sufficiently close to each other for their atomic orbitals to overlap. The shared pair of electrons constitutes a single covalent bond and occupy the same orbital with opposing spins.
Dot-and cross Diagrams
Dot-and cross diagrams are simplified versions of the diagram for the chlorine molecule on the left below, showing only the outer shell electrons
Note that the use of two (or even more) different symbols for the electrons does not mean that the electrons are different - it simply identifies the ‘parent’ atoms.
You may also encounter dot-and-cross diagrams with bond lines added, and also versions showing only the bonding electrons,
Cleave crystal of calcite to show cleavage planes and hardness.
Show model of NaCl structure from booklet and from model building kit.
Whenever atoms, ions or molecules approach each other, there are electrostatic forces acting between them. When the net forces are forces of attraction, and they are strong enough to bind the particles together, we refer to them as chemical bonds. When particles are bound together by chemical bonds, the resulting arrangement is known as the structure of the substance concerned.
All the noble gases except He have completely full inner shells and an outer octet of electrons. As these gases are monatomic and rarely enter into chemical combination it is assumed that the outer octet of electrons is a very stable arrangement and therefore when atoms combine they will try to obtain the noble gas configuration.
The three main types of bonds are:
1. 1. Ionic or electrovalent
2. 2. Covalent
3. 3. Metallic
Ionic bonding restricted to elements in groups I, II, VI and VII, the ions of which have a noble gas structure. Dot and cross diagrams. Characteristic properties of ionic compounds. Sodium chloride as a typical ionic crystal.
When you have finished this section you should be able to:
· · Understand that ionic bonding involves attraction between oppositely charged ions formed by electron transfer
· · Describe the various electron configurations of simple stable ions.
· · List properties typical of ionic compounds and indicate how useful each one is in deciding whether a substance is ionic.
Ionic Bonding
Atoms can achieve noble gas electronic configuration by loss or gain of electrons to form ions. Metals (with low electronegativities) lose electrons to form positive ions and non-metals gain electrons to form negative ions.
Sodium can attain the stable electron configuration of neon by losing one electron
Na (1s22s22p63s1) Na+ (1s22s22p6) + e-
With ten electrons and eleven protons he species formed has a positive charge.
Fluorine is one electron short of the neon electronic configuration. If it obtains one electron (from sodium) it can achieve a full outer shell of eight electrons.
F (1s22s22p5) + e- F- (1s22s22p6)
The species formed has ten electrons and nine protons; it is a negatively charged fluoride ion.
You will soon be aware of the limitations of these diagrams, but they are nevertheless quite useful even in A-level work for describing electron transfer and for checking that all electrons are accounted for. However, electron shell diagrams are rather tedious to draw, and give more information than we usually need when only the outer shell electrons are involved in reactions. We can simplify the above diagram to draw what we call a 'dot-and-cross' diagram showing only the outer shell electrons:
Electrons shown by dots and crosses are of course indistinguishable - we use different symbols only to show where they come from. Note also the use of brackets to separate the charge symbol from the symbols for electrons.
In the first exercise you draw some similar dot-and-cross diagrams yourself
Exercise 1
Draw dot-and-cross diagrams for the formation of:
(a) (a) sodium chloride from sodium atoms and chlorine atoms;
(b) (b) magnesium oxide from magnesium atoms and oxygen atoms;
(c ) calcium fluoride from calcium atoms and fluorine atoms.
Properties of Ionic Compounds
1. 1. Crystalline solids : Ionic compounds are crystalline solids. Ionic crystals are quite hard due to the strong electrostatic forces between the ions
2. 2. High melting and boiling points : All ionic compounds have high melting and boiling points as a large amount of heat energy is required to break the strong electrostatic forces of attraction between the ions.
3. 3. Soluble in water : Many ionic compounds are soluble in water. The water molecules attract the ions and pull the structure apart. They also stabalise the ions once in solution
4. 4. Conduct electricity when molten or dissolved in water : All ionic compounds conduct electricity when molten and are decomposed in the process, which is called electrolysis. Most ionic compounds dissolve in water, which frees the ions so that they can move and carry electrical charge.
Structure of sodium chloride
An ion may be regarded as an electrically charged sphere. A charged sphere is surrounded by a uniform electric field and therefore attracts oppositely charged spheres in all directions.
No particular orientation is favoured, so we often say that ionic bonding is non-directional. What we really mean by this is that the forces of attraction are non-directional.
However, when large numbers of oppositely-charged spherical ions are attracted to each other, the repulsion of the like-charges also comes into play. Mutual repulsion of similarly-charged ions limits the number which can come into contact with an oppositely-charged ion, and effectively fixes their relative positions. Ions, therefore, tend to cling together in large clusters known as ionic lattices in which attractive and repulsive forces are balanced. The particular arrangement of ions depends on their relative charges and sizes.
The ionic lattice of sodium chloride consists of a regular arrangement of alternating sodium and chloride ions extending in three dimensions:
Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl-
Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na+
Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl-
Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na
The lattice can be imagined as consisting of two inter-penetrating face centred cubic structures of sodium ions and chloride ions.
Na+ Na+ Cl- Cl-
Na+ Cl-
Na+ Na+ Cl- Cl-
Each sodium ion is surrounded by six chloride ions as nearest neighbours in the lattice and each chloride ion by six sodium ions. The lattice is said to have 6:6 coordination.
Stereoscopic picture of sodium chloride. The large circles represent Cl- ions and the small circles Na+ ions.
You can see from the above figure that it is impossible to say that a particular sodium ion ‘belongs’ to a particular chloride ion, as they are an equal distance from six. This means that ‘molecules’ of an ionic compound are not formed. The formula used is the atoms in their combining ratio.
Formation of covalent bonds in terms of the sharing of electron pairs between atoms. Dot and cross diagrams. Multiple bonds exemplified by C2H4, N2 and CO2. The octet rule and its limitations, eg BeCl2, BF3. The coordinate bond as a special case of the covalent bond eg NH4+. Dot and cross diagrams are required to show only the outermost electrons but, where appropriate, charges should be included.
When you have finished this section you should be able to:
· · Describe covalent bonding in terms of shared pairs of electrons.
· · Draw dot-and-cross diagrams for a variety of covalent molecules.
· · Explain the term dative covalent bond.
· · Draw dot-and-cross diagrams and structural formulae including dative bonds.
· · For polyatomic ions, distinguish between the covalent bonding within the ion and the ionic bonding between ions.
· · Draw dot-and-cross diagrams for at least two covalent compounds in which an atom other than hydrogen has fewer than eight outer shell electrons.
· · Draw dot-and-cross diagrams for at least two covalent compounds in which an atom has more than eight outer shell electrons.
1.2 Covalent Bonding
Covalent bonds are formed by the sharing of electrons between atoms. The two atoms have to approach sufficiently close to each other for their atomic orbitals to overlap. The shared pair of electrons constitutes a single covalent bond and occupy the same orbital with opposing spins.
Dot-and cross Diagrams
Dot-and cross diagrams are simplified versions of the diagram for the chlorine molecule on the left below, showing only the outer shell electrons
Note that the use of two (or even more) different symbols for the electrons does not mean that the electrons are different - it simply identifies the ‘parent’ atoms.
You may also encounter dot-and-cross diagrams with bond lines added, and also versions showing only the bonding electrons,
Cleave crystal of calcite to show cleavage planes and hardness.
Show model of NaCl structure from booklet and from model building kit.
(Halogens : restricted to chlorine, bromine and iodine)
Valid deductions may be expected about other elements in the group.
Physical properties of halogens, limited to colour and physical state at room temperature.
The halogens are a group of reactive non-metals, which are essentially similar to each other with only gradual changes as the atomic number increases
They are all p-block elements with a simple molecular structure consisting of covalently bonded diatomic molecules, X2.
o o o o
o o o
o X o X o X X
o o o o
There are only weak Van der Waals forces between the molecules. The strength of the forces increases as the number of electrons (Mr) in the molecule increases.
F2< Cl2< Br2< I2
In the case of iodine, the forces are sufficiently strong to bind the iodine molecules together in a 3-D crystal lattice. The X-X bond strength decreases down the group
Cl2> Br2 > I2 as the atoms get larger and the attraction of the nucleus for the shared electrons decreases (electronegativity decreases).
There is a slight tendency to metallic character with increasing atomic number. The halogens complete their octet by gaining one electron forming a halide ion, X- (see electron affinity values) or by sharing one electron. Fluorine is restricted to an oxidation state of -1 but the remaining elements have empty d orbitals and can promote electrons to give oxidation states of +1, +3, +5 and +7.
They are all oxidising agents and combine readily with metals and hydrogen.
Chlorine is a greenish-yellow gas.
Bromine is a red-brown volatile liquid.
Iodine is a black shiny solid, which sublimes on heating to produce a purple vapour.
Reactions of chlorine gas with elements (see Section 7.5.1), water, alkalis (under different conditions to form ClO- and ClO3-), other halides in aqueous solution, iron(II) ions in solution, hydrocarbons (see Sections 7.3.2, 7.3.3 and 7.3.4).
Reaction of halides with elements
Metals
The halogens combine readily with most metals forming the metal halides.
The vigour of the reaction decreases from chlorine to iodine.
Group I and II halides are ionic.
2Na (s) + Cl2 (g) 2Na+Cl- (s)
Mg (s) + Cl2 (g) Mg2+2Cl- (s)
The halides of Group III are predominantly covalent.
2Al (s) + 3Cl2 (g) 2AlCl3 (s)
Non-metals
The elements react directly with many non-metals the oxidising power decreasing from chlorine to iodine.
The elements combine directly with phosphorus, the oxidation state of the product depending on the oxidising power of the halogen.
2P (s) + 5Cl2 (g) 2PCl5 (s)
2P (s) + 3Br2 (l) 2PBr3 (l)
Solubility of the elements
All three elements are only slightly soluble in water because of the relatively strong hydrogen-bonding between the water molecules, which does not exist between the halogen molecules
i.e. solvent-solvent attractions > solute-solvent attractions > solute-solute
attractions.
Cl2> Br2>I2
solubility decreasing
They are soluble in non-polar organic solvents such as toluene and TCE.
(Why?)
Chlorine reacts slowly with water forming hydrochloric acid and chloric(I) acid. This reaction involves disproportionation:- a change in which one particular molecule, atom or ion is simultaneously both oxidised and reduced.
reduction
Cl2 (g) + H2O (l) HCl (aq) + HClO (aq) (chlorine water)
o.n. 0 -1 +1
oxidation
Exercise 1
Write balanced equations for each of the above reactions.
(Use the symbol X for a general halogen reaction)
The identification of halide ions in solution by the use of silver ions and aqueous ammonia; the [Ag(NH3)2]+ ion
Reaction of the halide ions in solution, X-(aq)
Most metal halides are soluble except lead and silver halide. Therefore solutions of lead and silver ions are used to test for the presence of halide ions in solution.
Bromine and iodine disproportionate in a similar way but to a lesser extent.
Reaction of chlorine with aqueous sodium hydroxide.
Chlorine reacts faster with dilute sodium hydroxide than with water.
When chlorine is added to cold dilute alkali it disproportionates to chloride and chlorate(l).
(i)
reduction
Cl2 (g) + 2NaOH (aq) NaCl (aq) + NaOCl (aq) + H2O
o.n. 0 -1 +1
oxidation
( 2OH- + Cl2 Cl- + OCl- + H2O )
(ii) In hot concentrated alkali, if the solution is warmed to 70oC, the chlorate(I) disproportionates further to chlorate(V).
reduction
3NaOCl (aq) 2NaCl (aq) + NaClO3 (aq)
o.n. +1 -1 +5
oxidation
If chlorine is bubbled directly into hot conc. alkali then
(iii) reduction
3Cl2 (g) + 6NaOH(aq) 5NaCl (aq) + NaClO3 (aq)
o.n. 0 -1 +5
oxidation
( 6OH- + 3Cl2 5Cl- + ClO3- + 3H2O )
For bromine, both reactions (i) and (ii) are fast at 15oC.
For iodine, decomposition of IO- occurs rapidly at 0oC so it is difficult to prepare NaIO free from NaIO3.
NaClO is a mild antiseptic (Milton).
NaClO3 powerful weed killer.
Relative oxidising ability of the halogens linked to redox potentials.
Displacement reactions of the halogens
Since they are very electronegative, all the halogens are oxidising agents. Their standard electrode potentials, Eq, become less positive on descending the group, showing that their oxidising power decreases.
X2 + 2e- 2X-
s.e.p. Eq /volts
Cl2 (g) /2Cl- + 1.36
Br2 (g) / 2Br- + 1.09
I2 (g) / 2I- + 0.54
Therefore chlorine oxidises bromide ions to bromine and iodide ions to iodine.
These are displacement reactions.
Cl2 (g) + 2Br- (aq) ----- Br2 (l)+ 2Cl- (aq)
(colourless) (yellow/orange)
Cl2 (g) + 2I- (aq) -- I2 (s) + 2Cl- (aq)
(colourless) (red/brown)
Bromine oxidises iodide to iodine
Br2 (g) + 2I- (aq) ---- I2 (s) + 2Br- (aq)
Iodine does not oxidise any of the others.
Exercise 2
Write an equation for the reaction of sodium chloride solution with
(a) (a) lead nitrate solution and
(b) silver nitrate solution followed by the addition of ammonia.
The reaction of solid halides with concentrated sulphuric acid to illustrate the relative reducing ability of halide ions and the hydrogen halides. Presence of halide ions in sea water.
Reaction of the metal halides with conc. sulphuric acid H2SO4
When concentrated sulphuric acid is added to a sodium halide the first product is fumes of the hydrogen halide HX, because each of these compounds is more volatile than sulphuric acid.
NaCl (s) + H2SO4 (1) HCl (g) + NaHSO4 (s)
NaBr (s) + H2SO4 (1) HBr(g) + NaHSO4 (s)
However conc. sulphuric acid is also a strong oxidising agent and will oxidise
HBr Br2 and HI I2, but not HF and HCl.
oxidised
2HBr(g) + H2SO4 Br2 + SO2 (g) + 2H2O
reduced
Similarly 2HI(g) + H2SO4 I2 + SO2(g) + 2H2O
Therefore conc. sulphuric acid cannot be used for the preparation of
HBr(g) and HI(g).
However conc. phosphoric(V) acid, H3PO4, can be used for the preparation since it is relatively non volatile and a poor oxidising agent.
NaBr (s) + H3PO4 (1) HBr (g) + NaH2PO4
NaI (s) + H3PO4 (1) HI (g) + NaH2PO4
Exercise 3
1. 1. Research: Look up the presence of halide ions in sea water.
2. 2. Make notes on the uses of the halogens and their compounds.
Valid deductions may be expected about other elements in the group.
Physical properties of halogens, limited to colour and physical state at room temperature.
The halogens are a group of reactive non-metals, which are essentially similar to each other with only gradual changes as the atomic number increases
They are all p-block elements with a simple molecular structure consisting of covalently bonded diatomic molecules, X2.
o o o o
o o o
o X o X o X X
o o o o
There are only weak Van der Waals forces between the molecules. The strength of the forces increases as the number of electrons (Mr) in the molecule increases.
F2< Cl2< Br2< I2
In the case of iodine, the forces are sufficiently strong to bind the iodine molecules together in a 3-D crystal lattice. The X-X bond strength decreases down the group
Cl2> Br2 > I2 as the atoms get larger and the attraction of the nucleus for the shared electrons decreases (electronegativity decreases).
There is a slight tendency to metallic character with increasing atomic number. The halogens complete their octet by gaining one electron forming a halide ion, X- (see electron affinity values) or by sharing one electron. Fluorine is restricted to an oxidation state of -1 but the remaining elements have empty d orbitals and can promote electrons to give oxidation states of +1, +3, +5 and +7.
They are all oxidising agents and combine readily with metals and hydrogen.
Chlorine is a greenish-yellow gas.
Bromine is a red-brown volatile liquid.
Iodine is a black shiny solid, which sublimes on heating to produce a purple vapour.
Reactions of chlorine gas with elements (see Section 7.5.1), water, alkalis (under different conditions to form ClO- and ClO3-), other halides in aqueous solution, iron(II) ions in solution, hydrocarbons (see Sections 7.3.2, 7.3.3 and 7.3.4).
Reaction of halides with elements
Metals
The halogens combine readily with most metals forming the metal halides.
The vigour of the reaction decreases from chlorine to iodine.
Group I and II halides are ionic.
2Na (s) + Cl2 (g) 2Na+Cl- (s)
Mg (s) + Cl2 (g) Mg2+2Cl- (s)
The halides of Group III are predominantly covalent.
2Al (s) + 3Cl2 (g) 2AlCl3 (s)
Non-metals
The elements react directly with many non-metals the oxidising power decreasing from chlorine to iodine.
The elements combine directly with phosphorus, the oxidation state of the product depending on the oxidising power of the halogen.
2P (s) + 5Cl2 (g) 2PCl5 (s)
2P (s) + 3Br2 (l) 2PBr3 (l)
Solubility of the elements
All three elements are only slightly soluble in water because of the relatively strong hydrogen-bonding between the water molecules, which does not exist between the halogen molecules
i.e. solvent-solvent attractions > solute-solvent attractions > solute-solute
attractions.
Cl2> Br2>I2
solubility decreasing
They are soluble in non-polar organic solvents such as toluene and TCE.
(Why?)
Chlorine reacts slowly with water forming hydrochloric acid and chloric(I) acid. This reaction involves disproportionation:- a change in which one particular molecule, atom or ion is simultaneously both oxidised and reduced.
reduction
Cl2 (g) + H2O (l) HCl (aq) + HClO (aq) (chlorine water)
o.n. 0 -1 +1
oxidation
Exercise 1
Write balanced equations for each of the above reactions.
(Use the symbol X for a general halogen reaction)
The identification of halide ions in solution by the use of silver ions and aqueous ammonia; the [Ag(NH3)2]+ ion
Reaction of the halide ions in solution, X-(aq)
Most metal halides are soluble except lead and silver halide. Therefore solutions of lead and silver ions are used to test for the presence of halide ions in solution.
Bromine and iodine disproportionate in a similar way but to a lesser extent.
Reaction of chlorine with aqueous sodium hydroxide.
Chlorine reacts faster with dilute sodium hydroxide than with water.
When chlorine is added to cold dilute alkali it disproportionates to chloride and chlorate(l).
(i)
reduction
Cl2 (g) + 2NaOH (aq) NaCl (aq) + NaOCl (aq) + H2O
o.n. 0 -1 +1
oxidation
( 2OH- + Cl2 Cl- + OCl- + H2O )
(ii) In hot concentrated alkali, if the solution is warmed to 70oC, the chlorate(I) disproportionates further to chlorate(V).
reduction
3NaOCl (aq) 2NaCl (aq) + NaClO3 (aq)
o.n. +1 -1 +5
oxidation
If chlorine is bubbled directly into hot conc. alkali then
(iii) reduction
3Cl2 (g) + 6NaOH(aq) 5NaCl (aq) + NaClO3 (aq)
o.n. 0 -1 +5
oxidation
( 6OH- + 3Cl2 5Cl- + ClO3- + 3H2O )
For bromine, both reactions (i) and (ii) are fast at 15oC.
For iodine, decomposition of IO- occurs rapidly at 0oC so it is difficult to prepare NaIO free from NaIO3.
NaClO is a mild antiseptic (Milton).
NaClO3 powerful weed killer.
Relative oxidising ability of the halogens linked to redox potentials.
Displacement reactions of the halogens
Since they are very electronegative, all the halogens are oxidising agents. Their standard electrode potentials, Eq, become less positive on descending the group, showing that their oxidising power decreases.
X2 + 2e- 2X-
s.e.p. Eq /volts
Cl2 (g) /2Cl- + 1.36
Br2 (g) / 2Br- + 1.09
I2 (g) / 2I- + 0.54
Therefore chlorine oxidises bromide ions to bromine and iodide ions to iodine.
These are displacement reactions.
Cl2 (g) + 2Br- (aq) ----- Br2 (l)+ 2Cl- (aq)
(colourless) (yellow/orange)
Cl2 (g) + 2I- (aq) -- I2 (s) + 2Cl- (aq)
(colourless) (red/brown)
Bromine oxidises iodide to iodine
Br2 (g) + 2I- (aq) ---- I2 (s) + 2Br- (aq)
Iodine does not oxidise any of the others.
Exercise 2
Write an equation for the reaction of sodium chloride solution with
(a) (a) lead nitrate solution and
(b) silver nitrate solution followed by the addition of ammonia.
The reaction of solid halides with concentrated sulphuric acid to illustrate the relative reducing ability of halide ions and the hydrogen halides. Presence of halide ions in sea water.
Reaction of the metal halides with conc. sulphuric acid H2SO4
When concentrated sulphuric acid is added to a sodium halide the first product is fumes of the hydrogen halide HX, because each of these compounds is more volatile than sulphuric acid.
NaCl (s) + H2SO4 (1) HCl (g) + NaHSO4 (s)
NaBr (s) + H2SO4 (1) HBr(g) + NaHSO4 (s)
However conc. sulphuric acid is also a strong oxidising agent and will oxidise
HBr Br2 and HI I2, but not HF and HCl.
oxidised
2HBr(g) + H2SO4 Br2 + SO2 (g) + 2H2O
reduced
Similarly 2HI(g) + H2SO4 I2 + SO2(g) + 2H2O
Therefore conc. sulphuric acid cannot be used for the preparation of
HBr(g) and HI(g).
However conc. phosphoric(V) acid, H3PO4, can be used for the preparation since it is relatively non volatile and a poor oxidising agent.
NaBr (s) + H3PO4 (1) HBr (g) + NaH2PO4
NaI (s) + H3PO4 (1) HI (g) + NaH2PO4
Exercise 3
1. 1. Research: Look up the presence of halide ions in sea water.
2. 2. Make notes on the uses of the halogens and their compounds.
The Periodic Table- work sheet
The Periodic Table
Section A
For each of the following questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the response sheet.
1.B Which one of the following equations does not represent a reaction of chlorine under suitable conditions?
A Cl2 + H2O HCl + HOCl
B 2Cl2 + CH4 CCl4 + 2H2
C 3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O
D Cl2 + 2NaOH NaCl + NaOCl + H2O
2.A When the hydrogen halides dissolve in water acidic solutions are formed. Which has the highest pH (assuming all are equal concentrations)?
A Hydrogen fluoride
B Hydrogen chloride
C Hydrogen iodide
D Hydrogen bromide
3. In which one of the following changes has chlorine been oxidised?
A 3Cl2 + 2Fe 2FeCl3
B Cl2 + I2 2IC1
C Cl2 + 2KBr Br2 + 2KCl
D Cl2 + 2KOH H2O + KClO + KCl
4. 4. In which one of the following sequences are the oxides of the elements classified as basic, amphoteric and acidic respectively?
A Na, Mg, Al
B Na, K, S
C K, Al, P
D S, P, Mg
5. Which one of the following is true of the halogens as the Group is descended from chlorine to iodine?
A The atomic radius decreases.
B The colour of the element lightens.
C The melting point of the element increases.
D The oxidising power of the element increases.
5. 5. Which one of the following equations represents a reaction of chlorine?
A Cl2 + NaOH NaOCl + HCl
B Cl2 + Fe FeCl2
C Cl2 + H2O HCl + HOCl
D Cl2 + CH4 CH2Cl2 + H2
Section A
For each of the following questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the response sheet.
1.B Which one of the following equations does not represent a reaction of chlorine under suitable conditions?
A Cl2 + H2O HCl + HOCl
B 2Cl2 + CH4 CCl4 + 2H2
C 3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O
D Cl2 + 2NaOH NaCl + NaOCl + H2O
2.A When the hydrogen halides dissolve in water acidic solutions are formed. Which has the highest pH (assuming all are equal concentrations)?
A Hydrogen fluoride
B Hydrogen chloride
C Hydrogen iodide
D Hydrogen bromide
3. In which one of the following changes has chlorine been oxidised?
A 3Cl2 + 2Fe 2FeCl3
B Cl2 + I2 2IC1
C Cl2 + 2KBr Br2 + 2KCl
D Cl2 + 2KOH H2O + KClO + KCl
4. 4. In which one of the following sequences are the oxides of the elements classified as basic, amphoteric and acidic respectively?
A Na, Mg, Al
B Na, K, S
C K, Al, P
D S, P, Mg
5. Which one of the following is true of the halogens as the Group is descended from chlorine to iodine?
A The atomic radius decreases.
B The colour of the element lightens.
C The melting point of the element increases.
D The oxidising power of the element increases.
5. 5. Which one of the following equations represents a reaction of chlorine?
A Cl2 + NaOH NaOCl + HCl
B Cl2 + Fe FeCl2
C Cl2 + H2O HCl + HOCl
D Cl2 + CH4 CH2Cl2 + H2
Tuesday, April 21, 2009
Determining the enthalpy change of a reaction
Determining the enthalpy change of a reaction
Aims
The purpose of this experiment is to determine the enthalpy change for the displacement reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
By adding an excess of zinc powder to a known amount of copper(II) sulphate solution, and measuring the temperature change over a period of time, you can calculate the enthalpy change for the reaction.
Apparatus
Goggles
Bench mat
25cm3 pipette
Pipette filler
Polystyrene cup with lid
Weighing bottle
Spatula
Balance
Thermometer
Stop clock
Zinc powder
1.0M copper(II) sulphate solution
Methods
1. Pipette 25.0cm3 of the copper(II) sulphate solution into the polystyrene cup.
2. Weigh about 6g of zinc powder in the weighing bottle – as this is an excess, there is no need to be accurate.
3. Put the thermometer through the hole in the lid, stir, and record the temperature every half minute for 2½ minutes in the table below.
4. At precisely 3 minutes, add the zinc powder to the cup.
5. Continue stirring, and record the temperature for an additional 6 minutes in the table below.
makaeResults table with time AND temperature
Plot the temperature (vertical axis) against time (horizontal axis).
2. Extrapolate the curve back to 3.0 minutes to establish the maximum temperature rise as shown in the example below: ΔT time (minutes)temperature (°C) 0 246810
3. Calculate the enthalpy change, ΔH, for the quantities used, using the formula:
ΔH = m c ΔT where m = mass of solution (g)
c = specific heat capacity of water = 4.18 J g–1 K–1
ΔT = rise in temperature (K)
4. Calculate the enthalpy change for one mole of Zn and CuSO4(aq).
5. Calculate the maximum error for each piece of apparatus, and then the total overall apparatus error.
6. The accepted value for this reaction is –217 kJ mol–1.
Compare your result with this value by calculating the percentage error in your answer:
error = experimental value - accepted valueaccepted value x 100%
Evaluation
1. Compare your total apparatus error with your answer to part 6 above – is the apparatus error enough to account for any difference between the accepted value and your experimental value?
2. List some possible reasons for any difference between your value and the accepted value (these should not be the apparatus errors mentioned above).
3. Why do you think the temperature increases for a few readings after adding the zinc?
(Hint: the temperature does not go even higher if more zinc is used, or if the powder is finely divided).
Aims
The purpose of this experiment is to determine the enthalpy change for the displacement reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
By adding an excess of zinc powder to a known amount of copper(II) sulphate solution, and measuring the temperature change over a period of time, you can calculate the enthalpy change for the reaction.
Apparatus
Goggles
Bench mat
25cm3 pipette
Pipette filler
Polystyrene cup with lid
Weighing bottle
Spatula
Balance
Thermometer
Stop clock
Zinc powder
1.0M copper(II) sulphate solution
Methods
1. Pipette 25.0cm3 of the copper(II) sulphate solution into the polystyrene cup.
2. Weigh about 6g of zinc powder in the weighing bottle – as this is an excess, there is no need to be accurate.
3. Put the thermometer through the hole in the lid, stir, and record the temperature every half minute for 2½ minutes in the table below.
4. At precisely 3 minutes, add the zinc powder to the cup.
5. Continue stirring, and record the temperature for an additional 6 minutes in the table below.
makaeResults table with time AND temperature
Plot the temperature (vertical axis) against time (horizontal axis).
2. Extrapolate the curve back to 3.0 minutes to establish the maximum temperature rise as shown in the example below: ΔT time (minutes)temperature (°C) 0 246810
3. Calculate the enthalpy change, ΔH, for the quantities used, using the formula:
ΔH = m c ΔT where m = mass of solution (g)
c = specific heat capacity of water = 4.18 J g–1 K–1
ΔT = rise in temperature (K)
4. Calculate the enthalpy change for one mole of Zn and CuSO4(aq).
5. Calculate the maximum error for each piece of apparatus, and then the total overall apparatus error.
6. The accepted value for this reaction is –217 kJ mol–1.
Compare your result with this value by calculating the percentage error in your answer:
error = experimental value - accepted valueaccepted value x 100%
Evaluation
1. Compare your total apparatus error with your answer to part 6 above – is the apparatus error enough to account for any difference between the accepted value and your experimental value?
2. List some possible reasons for any difference between your value and the accepted value (these should not be the apparatus errors mentioned above).
3. Why do you think the temperature increases for a few readings after adding the zinc?
(Hint: the temperature does not go even higher if more zinc is used, or if the powder is finely divided).
Simple calorimetry to find the enthalpy of combustion of alcohols
Simple calorimetry to find the enthalpy of combustion of alcohols
Aims
You will use simple calorimetry estimate the enthalpy of combustion of an alcohol.
Apparatus
Goggles
Bench mat
Stand, boss, clamp
Thermometer
100cm3 measuring cylinder
Steel can
Digital balance
Access to spirit burners containing:
methanol, ethanol,
propanol or butanol
Method
1. Draw up a suitable table or tables to record your results.
2. Measure 100cm3 of water in the measuring cylinder.
Pour the water into the steel can and record its temperature.
3. Choose a spirit burner.
Record the name of the fuel, and the mass of the whole burner (including the lid and fuel inside).
4. Clamp the steel can, and set it up so that the spirit burner will fit comfortably under it.
5. Light the wick of the spirit burner, and put it under the steel can.
6. Stir the water gently with the thermometer, and watch the temperature.
When it has increased by 20°C, put the lid on the spirit burner to put the flame out.
Record the new mass of the whole burner (including the lid and fuel inside).
7. Using fresh water each time, repeat the experiment at least twice with the same fuel.
Analysis
1. Calculate the energy transferred to the water using the equation q = mcΔT
Assume that 1cm3 of water has a mass of 1g and c = 4.18 J g–1 K–1.
2. For each replicate experiment, perform the calculations described below:
a) Calculate the mass of fuel burnt.
b) Calculate the Mr of the fuel used. Use your answer to part a) to work out the amount of fuel burnt.
c) Work out the energy transferred to the water in kJ mol–1, and so the enthalpy of combustion.
3. Estimate the maximum errors in the using each piece of apparatus, and the total apparatus error.
Aims
You will use simple calorimetry estimate the enthalpy of combustion of an alcohol.
Apparatus
Goggles
Bench mat
Stand, boss, clamp
Thermometer
100cm3 measuring cylinder
Steel can
Digital balance
Access to spirit burners containing:
methanol, ethanol,
propanol or butanol
Method
1. Draw up a suitable table or tables to record your results.
2. Measure 100cm3 of water in the measuring cylinder.
Pour the water into the steel can and record its temperature.
3. Choose a spirit burner.
Record the name of the fuel, and the mass of the whole burner (including the lid and fuel inside).
4. Clamp the steel can, and set it up so that the spirit burner will fit comfortably under it.
5. Light the wick of the spirit burner, and put it under the steel can.
6. Stir the water gently with the thermometer, and watch the temperature.
When it has increased by 20°C, put the lid on the spirit burner to put the flame out.
Record the new mass of the whole burner (including the lid and fuel inside).
7. Using fresh water each time, repeat the experiment at least twice with the same fuel.
Analysis
1. Calculate the energy transferred to the water using the equation q = mcΔT
Assume that 1cm3 of water has a mass of 1g and c = 4.18 J g–1 K–1.
2. For each replicate experiment, perform the calculations described below:
a) Calculate the mass of fuel burnt.
b) Calculate the Mr of the fuel used. Use your answer to part a) to work out the amount of fuel burnt.
c) Work out the energy transferred to the water in kJ mol–1, and so the enthalpy of combustion.
3. Estimate the maximum errors in the using each piece of apparatus, and the total apparatus error.
Thermometric titration
Thermometric titration
Your task
You will use thermometric titration to determine the concentration of hydrochloric acid. Neutralisation is an exothermic reaction and the maximum temperature is reached at the end-point.
To gain full marks, you should:
• complete the experiment without guidance
(but remember – always ask for help if you need it);
• work carefully and safely, making accurate measurements and detailed observations; and
• record all your results clearly in an appropriate way.
Use the following reagents in your experiment:
• 1.000M sodium hydroxide solution
• approx. 2M hydrochloric acid
Method
1. Read through the Methods, then construct suitable blank tables for your Results.
2. Transfer 50cm3 of sodium hydroxide solution to a polystyrene cup.
Allow it to stand for a few minutes, then record the temperature of the solution.
3. Add 5.0cm3 of hydrochloric acid from a burette to the cup.
Immediately stir the mixture with the thermometer and record its temperature.
Repeat until you have added a total of 50.0cm3 of acid.
Analysis
1. Plot a graph of temperature (vertical axis) against total volume of acid added (horizontal axis). Total volume of acid added (cm3)Temperature (OC)
Draw straight lines of best fit and extend them until they cross (see diagram right).
The point at which the two lines meet corresponds to the volume of acid needed for neutralisation and to the maximum temperature.
2. Use information from your graph to calculate the concentration of the acid.
Evaluation
Comment on your results, their accuracy, and the likely sources of error in the experiment. Consider the limitations of the experiment, and possible improvements to it.
Thermometric titration
Technician’s Notes
Per pupil:
1 x polystyrene cup
1 x burette and stand
1 x 25cm3 pipette
1 x pipette filler
1 x plastic filter funnel
1 x thermometer
Per class:
sodium hydroxide solution: 1.000M (allow about 200cm3 per student)
hydrochloric acid: 2M approx. (allow about 150cm3 per student)
ethanoic acid: 2M approx. (allow about 150cm3 per student)
*Health and Safety Notes
Hydrochloric acid -Corrosive.
Refer to Hazcards for correct method to prepare the 2M (approx.) solution.
Ethanoic acid
Corrosive.
Harmful vapour.
Refer to Hazcards for correct method to prepare the 2M (approx.) solution.
Sodium hydroxide solid and solutions
Sodium hydroxide is very caustic and forms strongly alkaline solutions.
Exercise care in handling - wear gloves and eye protection.
If spilt, wash with a lot of water.
Your task
You will use thermometric titration to determine the concentration of hydrochloric acid. Neutralisation is an exothermic reaction and the maximum temperature is reached at the end-point.
To gain full marks, you should:
• complete the experiment without guidance
(but remember – always ask for help if you need it);
• work carefully and safely, making accurate measurements and detailed observations; and
• record all your results clearly in an appropriate way.
Use the following reagents in your experiment:
• 1.000M sodium hydroxide solution
• approx. 2M hydrochloric acid
Method
1. Read through the Methods, then construct suitable blank tables for your Results.
2. Transfer 50cm3 of sodium hydroxide solution to a polystyrene cup.
Allow it to stand for a few minutes, then record the temperature of the solution.
3. Add 5.0cm3 of hydrochloric acid from a burette to the cup.
Immediately stir the mixture with the thermometer and record its temperature.
Repeat until you have added a total of 50.0cm3 of acid.
Analysis
1. Plot a graph of temperature (vertical axis) against total volume of acid added (horizontal axis). Total volume of acid added (cm3)Temperature (OC)
Draw straight lines of best fit and extend them until they cross (see diagram right).
The point at which the two lines meet corresponds to the volume of acid needed for neutralisation and to the maximum temperature.
2. Use information from your graph to calculate the concentration of the acid.
Evaluation
Comment on your results, their accuracy, and the likely sources of error in the experiment. Consider the limitations of the experiment, and possible improvements to it.
Thermometric titration
Technician’s Notes
Per pupil:
1 x polystyrene cup
1 x burette and stand
1 x 25cm3 pipette
1 x pipette filler
1 x plastic filter funnel
1 x thermometer
Per class:
sodium hydroxide solution: 1.000M (allow about 200cm3 per student)
hydrochloric acid: 2M approx. (allow about 150cm3 per student)
ethanoic acid: 2M approx. (allow about 150cm3 per student)
*Health and Safety Notes
Hydrochloric acid -Corrosive.
Refer to Hazcards for correct method to prepare the 2M (approx.) solution.
Ethanoic acid
Corrosive.
Harmful vapour.
Refer to Hazcards for correct method to prepare the 2M (approx.) solution.
Sodium hydroxide solid and solutions
Sodium hydroxide is very caustic and forms strongly alkaline solutions.
Exercise care in handling - wear gloves and eye protection.
If spilt, wash with a lot of water.
Volumetric Analysis 3 To determine the relative molecular mass of a soluble base
Volumetric Analysis 3
To determine the relative molecular mass of a soluble base
Introduction
In Volumetric Analysis 1 & 2 you prepared a standard solution of sodium carbonate and used it to standardise an unknown concentration of dilute hydrochloric acid. In this practical you will use your new-found skills to find out the relative molecular mass of an unknown group 1 carbonate – the mysterious "Substance Z". Group 1 carbonates are soluble in water (although Li2CO3 is only sparingly soluble) and will react with dilute hydrochloric acid according to the overall equation below:
X2CO3(aq) + 2HCl(aq) → 2XCl(aq) + CO2(g) + H2O(l)
(X represents a group 1 element)
If you know the amount of hydrochloric acid that will react with a known amount of Substance Z, you should be able to determine the Mr of Substance Z and so identify the group 1 element in it.
You will need to make careful notes about your experiment as you go along today.
Apparatus
Consult your notes from Volumetric Analysis 1 & 2 to decide upon the apparatus you need.
Make sure that your practical write-up includes the apparatus you use today.
Method
Consult your notes from Volumetric Analysis 1 & 2 and The Burette to remind yourself of the procedures needed for safe and accurate working.
Make sure that your practical write-up includes the methods you use today.
1. Weigh out accurately between 1.3g and 1.7g of Substance Z.
Record your weighings in a suitable form. Dissolve your weighed Substance Z in de-ionised water, and make up the solution to 250cm3 in a volumetric flask.
2. Clean your burette with de-ionised water and then with the standard 0.100M hydrochloric acid to be used for the titration.
3. Pipette 25cm3 of the Substance Z solution into a clean conical flask.
Using methyl orange indicator, titrate with the standard hydrochloric acid.
4. Repeat step 3 until concordant results are obtained.
Record your results as in Volumetric Analysis 2.
After cleaning and clearing away, determine the identity of Substance Z as described overleaf.
Copyright © 2003 Nigel Saunders N-ch1-37 Copyright © 2003 Nigel Saunders N-ch1-37
Analysis
As in Volumetric Analysis 2, 1 mole of X2CO3 will react with 2 moles of HCl (see equation below):
X2CO3(aq) + 2HCl(aq) → 2XCl(aq) + CO2(g) + H2O(l)
(X represents a group 1 element)
1. Calculate the number of moles of HCl there were in your mean titre.
2. Calculate the number of moles of HCl that would react with the entire 250cm3 of Substance Z solution.
3. Work out the number of moles of X2CO3 were there in the 250cm3 of Substance Z solution.
You now know:
•the mass of X2CO3 in your Substance Z solution; and
•the number of moles of X2CO3 in your Substance Z solution.
4. Calculate the mass of one mole of X2CO3.
5. What is Substance Z, and why? Copyright © 2003 Nigel Saunders N-ch1-37
Volumetric Analysis 3
To determine the relative molecular mass of a soluble base
Technician's Notes
Prior to practical
Sodium carbonate*
Heat required amount of sodium carbonate (Na2CO3) to drive off water of crystallisation.
Either: heat in an evaporating dish over a Bunsen burner for 30 minutes approx., or
heat in a drying oven at about 110oC for 1 hour.
Agitate the solid periodically with a clean glass rod.
Transfer to a desiccator after heating, and label it "Substance Z - Harmful".
Care: Use tongs and eye protection.
Beware of hot solid and apparatus.
Sodium carbonate forms caustic alkaline solutions with water; if spilt on skin wash with plenty of water.
Analytical balances
Please check cleanliness and correct functioning of analytical balances.
De-ionised water
Please check 6th Form wash bottles are clean and filled with de-ionised water.
Make sure that additional de-ionised water is available in the aspirator.
Burettes
Please check the cleanliness and correct functioning of the burettes.
Per class
Sodium carbonate solid (see above). Allow approx. 2.5g per student.
Analytical balances (see above).
Top pan digital balances (minimum of two if possible).
De-ionised water (see above).
0.100M hydrochloric acid* ( a good home-made solution should suffice for this practical).
Allow 200cm3 per student.
Methyl orange indicator solution (the more bottles the better).
Per student
(Normally found in lab anyway)
1 x pair of safety goggles
1 x bench mat
2 x 100cm3 beaker
2 x 250cm3 beaker
1 x 250cm3 conical flask
1 x glass funnel (check that it will enter the neck of the volumetric flask easily) Copyright © 2003 Nigel Saunders N-ch1-37
Per student
(Additional apparatus to put out)
1 x glass rod (long)
1 x 250cm3 volumetric flask with stopper to fit
1 x 25cm3 bulb pipette
1 x pipette filler (check correct functioning)
1 x burette (see overleaf)
1 x burette stand
1 x weighing bottle with lid
1 x 6th Form wash bottle containing de-ionised water
1 x small spatula
1 x white tile
1 x small plastic filter funnel
1 x copy of N-ch1-37 (student guide to practical)
*Health and Safety Notes
Hydrochloric acid
Corrosive.
Use pre-prepared standard solution, or refer to Hazcards for correct method to prepare an accurate 0.100M solution.
Sodium carbonate (solutions and solid)
Sodium carbonate solutions are alkaline and therefore caustic.
Exercise care in handling - wear eye protection and, if spilt, wash with a lot of water.
To determine the relative molecular mass of a soluble base
Introduction
In Volumetric Analysis 1 & 2 you prepared a standard solution of sodium carbonate and used it to standardise an unknown concentration of dilute hydrochloric acid. In this practical you will use your new-found skills to find out the relative molecular mass of an unknown group 1 carbonate – the mysterious "Substance Z". Group 1 carbonates are soluble in water (although Li2CO3 is only sparingly soluble) and will react with dilute hydrochloric acid according to the overall equation below:
X2CO3(aq) + 2HCl(aq) → 2XCl(aq) + CO2(g) + H2O(l)
(X represents a group 1 element)
If you know the amount of hydrochloric acid that will react with a known amount of Substance Z, you should be able to determine the Mr of Substance Z and so identify the group 1 element in it.
You will need to make careful notes about your experiment as you go along today.
Apparatus
Consult your notes from Volumetric Analysis 1 & 2 to decide upon the apparatus you need.
Make sure that your practical write-up includes the apparatus you use today.
Method
Consult your notes from Volumetric Analysis 1 & 2 and The Burette to remind yourself of the procedures needed for safe and accurate working.
Make sure that your practical write-up includes the methods you use today.
1. Weigh out accurately between 1.3g and 1.7g of Substance Z.
Record your weighings in a suitable form. Dissolve your weighed Substance Z in de-ionised water, and make up the solution to 250cm3 in a volumetric flask.
2. Clean your burette with de-ionised water and then with the standard 0.100M hydrochloric acid to be used for the titration.
3. Pipette 25cm3 of the Substance Z solution into a clean conical flask.
Using methyl orange indicator, titrate with the standard hydrochloric acid.
4. Repeat step 3 until concordant results are obtained.
Record your results as in Volumetric Analysis 2.
After cleaning and clearing away, determine the identity of Substance Z as described overleaf.
Copyright © 2003 Nigel Saunders N-ch1-37 Copyright © 2003 Nigel Saunders N-ch1-37
Analysis
As in Volumetric Analysis 2, 1 mole of X2CO3 will react with 2 moles of HCl (see equation below):
X2CO3(aq) + 2HCl(aq) → 2XCl(aq) + CO2(g) + H2O(l)
(X represents a group 1 element)
1. Calculate the number of moles of HCl there were in your mean titre.
2. Calculate the number of moles of HCl that would react with the entire 250cm3 of Substance Z solution.
3. Work out the number of moles of X2CO3 were there in the 250cm3 of Substance Z solution.
You now know:
•the mass of X2CO3 in your Substance Z solution; and
•the number of moles of X2CO3 in your Substance Z solution.
4. Calculate the mass of one mole of X2CO3.
5. What is Substance Z, and why? Copyright © 2003 Nigel Saunders N-ch1-37
Volumetric Analysis 3
To determine the relative molecular mass of a soluble base
Technician's Notes
Prior to practical
Sodium carbonate*
Heat required amount of sodium carbonate (Na2CO3) to drive off water of crystallisation.
Either: heat in an evaporating dish over a Bunsen burner for 30 minutes approx., or
heat in a drying oven at about 110oC for 1 hour.
Agitate the solid periodically with a clean glass rod.
Transfer to a desiccator after heating, and label it "Substance Z - Harmful".
Care: Use tongs and eye protection.
Beware of hot solid and apparatus.
Sodium carbonate forms caustic alkaline solutions with water; if spilt on skin wash with plenty of water.
Analytical balances
Please check cleanliness and correct functioning of analytical balances.
De-ionised water
Please check 6th Form wash bottles are clean and filled with de-ionised water.
Make sure that additional de-ionised water is available in the aspirator.
Burettes
Please check the cleanliness and correct functioning of the burettes.
Per class
Sodium carbonate solid (see above). Allow approx. 2.5g per student.
Analytical balances (see above).
Top pan digital balances (minimum of two if possible).
De-ionised water (see above).
0.100M hydrochloric acid* ( a good home-made solution should suffice for this practical).
Allow 200cm3 per student.
Methyl orange indicator solution (the more bottles the better).
Per student
(Normally found in lab anyway)
1 x pair of safety goggles
1 x bench mat
2 x 100cm3 beaker
2 x 250cm3 beaker
1 x 250cm3 conical flask
1 x glass funnel (check that it will enter the neck of the volumetric flask easily) Copyright © 2003 Nigel Saunders N-ch1-37
Per student
(Additional apparatus to put out)
1 x glass rod (long)
1 x 250cm3 volumetric flask with stopper to fit
1 x 25cm3 bulb pipette
1 x pipette filler (check correct functioning)
1 x burette (see overleaf)
1 x burette stand
1 x weighing bottle with lid
1 x 6th Form wash bottle containing de-ionised water
1 x small spatula
1 x white tile
1 x small plastic filter funnel
1 x copy of N-ch1-37 (student guide to practical)
*Health and Safety Notes
Hydrochloric acid
Corrosive.
Use pre-prepared standard solution, or refer to Hazcards for correct method to prepare an accurate 0.100M solution.
Sodium carbonate (solutions and solid)
Sodium carbonate solutions are alkaline and therefore caustic.
Exercise care in handling - wear eye protection and, if spilt, wash with a lot of water.
Volumetric Analysis 2
Volumetric Analysis 2
To standardise hydrochloric acid
Introduction
In the last practical you prepared a standard solution of sodium carbonate.
Today, you will use it to find the concentration of dilute hydrochloric acid by titration.
This process is known as standardising the hydrochloric acid.
The reaction between sodium carbonate and hydrochloric acid takes place in two stages:
Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq) (1)
NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) (2)
Two indicators are needed to cover both stages:
•in stage 1, phenolphthalein is most suitable, and will respond to the pH change associated with the formation of sodium hydrogencarbonate, NaHCO3.
•in stage 2, methyl orange is most suitable, and will respond to the pH change associated with the final formation of sodium chloride, NaCl.
As a result, this practical gives you experience of titration using two different indicators (the phenolphthalein colour change is easy to spot, whereas the methyl orange colour change is quite difficult to judge).
Apparatus
Goggles
Bench mat
100cm3 beaker
250cm3 beaker
250cm3 conical flask
25cm3 bulb pipette
Pipette filler
Burette
Burette stand and holder
Plastic filter funnel
White tile
Teat pipette
Access to:
your standard sodium carbonate solution
dilute hydrochloric acid to standardise
phenolphthalein indicator solution
methyl orange indicator solution
Methods
1. Transfer a 25cm3 aliquot (portion) of your sodium carbonate solution to a 250cm3 capacity conical flask. Add a few drops of phenolphthalein indicator solution.
2. Titrate with the hydrochloric acid. The end-point of the titration is when the solution just changes from pink to colourless. Note the titre, then add a few drops of methyl orange.
3. Titrate with the hydrochloric acid. The end-point of the titration is when the solution just changes from yellow to red. Note the second titre.
4. Repeat steps 1 - 3 until concordance (i.e. until the readings are the same or within 0.1cm3).
Tabulate your titrations as described in The Burette sheet. You will need two sets of tables.
5. After tidying away, do the calculations described overleaf.
Copyright © 2003 Nigel Saunders N-ch1-36 N-ch1-36 (N.S. 2003)
Calculations
1. Calculate the Mr of Na2CO3.
Ar (Na) = 23 Ar (C) = 12 Ar (O) = 16
2. Look back at the accurate mass of sodium carbonate you used in the last practical.
Using your answer to step 1, calculate the number of moles of Na2CO3 that you dissolved in 250cm3 of water during Volumetric Analysis 1.
3. Use your answer to step 2 to calculate the number of moles of Na2CO3 in the 25cm3 transferred to the conical flask.
Stage 1 Phenolphthalein results
4. The equation for the first stage of the reaction between sodium carbonate and hydrochloric acid is shown below again:
Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq)
From the equation, you can see that 1 mole of Na2CO3 will react with 1 mole of HCl.
How many moles of HCl will react with the number of moles of Na2CO3 calculated in step 3?
5. The answer to step 4 tells you how many moles of HCl were in your first mean titre.
Divide this number by the volume of the first mean titre: this is the concentration of HCl in mol dm−3.
Stage 2 Methyl orange results
6. The equation for the second stage of the reaction between sodium carbonate and hydrochloric acid is shown below again:
NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)
From the equation, you can see that 1 mole of NaHCO3 will react with 1 mole of HCl.
The number of moles of NaHCO3 is equal to the number of moles of Na2CO3.
How many moles of HCl will react with the number of moles of NaHCO3 calculated in step 3?
7. The answer to step 6 tells you how many moles of HCl were in your second mean titre.
Divide this number by the volume of the second mean titre: this is also the concentration of HCl.
8. Your answers to steps 5 and 7 should be identical. Comment on your findings. N-ch1-36 (N.S. 2003)
Volumetric Analysis 2
To standardise hydrochloric acid
Name . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Date . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Results
1. Phenolphthalein indicator (first part of each run)
Burette reagent
Approx. 0.075M hydrochloric acid
Conical flask reagent
Standard sodium carbonate solution
Indicator
Phenolphthalein
To standardise hydrochloric acid
Introduction
In the last practical you prepared a standard solution of sodium carbonate.
Today, you will use it to find the concentration of dilute hydrochloric acid by titration.
This process is known as standardising the hydrochloric acid.
The reaction between sodium carbonate and hydrochloric acid takes place in two stages:
Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq) (1)
NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) (2)
Two indicators are needed to cover both stages:
•in stage 1, phenolphthalein is most suitable, and will respond to the pH change associated with the formation of sodium hydrogencarbonate, NaHCO3.
•in stage 2, methyl orange is most suitable, and will respond to the pH change associated with the final formation of sodium chloride, NaCl.
As a result, this practical gives you experience of titration using two different indicators (the phenolphthalein colour change is easy to spot, whereas the methyl orange colour change is quite difficult to judge).
Apparatus
Goggles
Bench mat
100cm3 beaker
250cm3 beaker
250cm3 conical flask
25cm3 bulb pipette
Pipette filler
Burette
Burette stand and holder
Plastic filter funnel
White tile
Teat pipette
Access to:
your standard sodium carbonate solution
dilute hydrochloric acid to standardise
phenolphthalein indicator solution
methyl orange indicator solution
Methods
1. Transfer a 25cm3 aliquot (portion) of your sodium carbonate solution to a 250cm3 capacity conical flask. Add a few drops of phenolphthalein indicator solution.
2. Titrate with the hydrochloric acid. The end-point of the titration is when the solution just changes from pink to colourless. Note the titre, then add a few drops of methyl orange.
3. Titrate with the hydrochloric acid. The end-point of the titration is when the solution just changes from yellow to red. Note the second titre.
4. Repeat steps 1 - 3 until concordance (i.e. until the readings are the same or within 0.1cm3).
Tabulate your titrations as described in The Burette sheet. You will need two sets of tables.
5. After tidying away, do the calculations described overleaf.
Copyright © 2003 Nigel Saunders N-ch1-36 N-ch1-36 (N.S. 2003)
Calculations
1. Calculate the Mr of Na2CO3.
Ar (Na) = 23 Ar (C) = 12 Ar (O) = 16
2. Look back at the accurate mass of sodium carbonate you used in the last practical.
Using your answer to step 1, calculate the number of moles of Na2CO3 that you dissolved in 250cm3 of water during Volumetric Analysis 1.
3. Use your answer to step 2 to calculate the number of moles of Na2CO3 in the 25cm3 transferred to the conical flask.
Stage 1 Phenolphthalein results
4. The equation for the first stage of the reaction between sodium carbonate and hydrochloric acid is shown below again:
Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq)
From the equation, you can see that 1 mole of Na2CO3 will react with 1 mole of HCl.
How many moles of HCl will react with the number of moles of Na2CO3 calculated in step 3?
5. The answer to step 4 tells you how many moles of HCl were in your first mean titre.
Divide this number by the volume of the first mean titre: this is the concentration of HCl in mol dm−3.
Stage 2 Methyl orange results
6. The equation for the second stage of the reaction between sodium carbonate and hydrochloric acid is shown below again:
NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)
From the equation, you can see that 1 mole of NaHCO3 will react with 1 mole of HCl.
The number of moles of NaHCO3 is equal to the number of moles of Na2CO3.
How many moles of HCl will react with the number of moles of NaHCO3 calculated in step 3?
7. The answer to step 6 tells you how many moles of HCl were in your second mean titre.
Divide this number by the volume of the second mean titre: this is also the concentration of HCl.
8. Your answers to steps 5 and 7 should be identical. Comment on your findings. N-ch1-36 (N.S. 2003)
Volumetric Analysis 2
To standardise hydrochloric acid
Name . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Date . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Results
1. Phenolphthalein indicator (first part of each run)
Burette reagent
Approx. 0.075M hydrochloric acid
Conical flask reagent
Standard sodium carbonate solution
Indicator
Phenolphthalein
Volumetric Analysis 1
Volumetric Analysis 1
To make a standard solution of sodium carbonate
Introduction
A standard solution is one whose concentration is known exactly. Standard solutions of liquids, for example acids, are easy to prepare and are usually supplied. Standard solutions of solids can be prepared by weighing a mass of solid, and dissolving it in a known volume of solution in a volumetric flask. Today, you are going to prepare a standard solution of sodium carbonate to use later in another practical.
Apparatus
Goggles
Bench mat
100 cm3 beaker
250 cm3 beaker
250 cm3 volumetric flask with stopper
Filter funnel
Glass rod
Teat pipette
Spatula
Label
De-ionised water
Anhydrous sodium carbonate, Na2CO3(s) Copyright © Nigel Saunders N-ch1-35
Methods
Read through the Methods. Make a suitable blank results table, complete with units in the headings.
1. Using the ± 0.1g balance, weigh approximately between 1.2g and 1.4g of sodium carbonate into the small beaker. Do not record the mass.
2. Using the ± 0.01g balance, weigh the small beaker and its contents accurately.
Record this mass.
3. Transfer the contents of the small beaker into the large beaker.
Weigh the small beaker again using the ± 0.01g balance.
Record this mass.
The difference between the two accurate masses is the mass of sodium carbonate in your beaker.
4. Add de-ionised water cautiously down the side of the large beaker.
Use about 150cm3 of water, and swirl the beaker to mix the contents.
5. Stir using a glass rod to dissolve the solid completely.
6. Transfer the solution into the volumetric flask using the funnel.
Remember: pour down the glass rod;
remove the last drop of solution from the glass rod onto the funnel.
Wash the beaker, rod and funnel several times using de-ionised water from the wash bottle, letting the washings go into the flask.
7. Make up to the mark on the volumetric flask with de-ionised water.
Stopper firmly, and shake the flask thoroughly to mix the contents.
8. Label the flask clearly with your name, the date, and the contents of the flask. Copyright © 2003 N. Saunders N-ch1-35
Volumetric Analysis 1
To make a standard solution of sodium carbonate:
Technician's Notes
Prior to practical
Sodium carbonate
Heat required amount of sodium carbonate (Na2CO3) to drive off water of crystallisation.
Either: heat in an evaporating dish over a Bunsen burner for 30 minutes approx., or
heat in a drying oven at about 110°C for 1 hour.
Agitate the solid periodically with a clean glass rod.
Transfer to a desiccator after heating, and label it "sodium carbonate".
Care: Use tongs and eye protection.
Beware of hot solid and apparatus.
Sodium carbonate forms caustic alkaline solutions with water; if spilt on skin wash with plenty of water.
Analytical balances
Please check cleanliness and correct functioning of analytical balances.
De-ionised water
Please check 6th Form wash bottles are clean and filled with de-ionised water.
Make sure that additional de-ionised water is available in the aspirator.
Requirements per class
Sodium carbonate solid (see above). Minimum of 3g per student approx.
Analytical balances (see above).
Top pan digital balances (minimum of two if possible).
De-ionised water (see above).
Requirements per student
1 x 250cm3 beaker (dry)
1 x 250cm3 volumetric flask with stopper to fit
1 x glass funnel (check that it will enter the neck of the volumetric flask easily)
1 x glass rod
1 x weighing bottle with lid
1 x 6th Form wash bottle containing de-ionised water
1 x teat pipette
1 x small spatula
1 x self-adhesive label
To make a standard solution of sodium carbonate
Introduction
A standard solution is one whose concentration is known exactly. Standard solutions of liquids, for example acids, are easy to prepare and are usually supplied. Standard solutions of solids can be prepared by weighing a mass of solid, and dissolving it in a known volume of solution in a volumetric flask. Today, you are going to prepare a standard solution of sodium carbonate to use later in another practical.
Apparatus
Goggles
Bench mat
100 cm3 beaker
250 cm3 beaker
250 cm3 volumetric flask with stopper
Filter funnel
Glass rod
Teat pipette
Spatula
Label
De-ionised water
Anhydrous sodium carbonate, Na2CO3(s) Copyright © Nigel Saunders N-ch1-35
Methods
Read through the Methods. Make a suitable blank results table, complete with units in the headings.
1. Using the ± 0.1g balance, weigh approximately between 1.2g and 1.4g of sodium carbonate into the small beaker. Do not record the mass.
2. Using the ± 0.01g balance, weigh the small beaker and its contents accurately.
Record this mass.
3. Transfer the contents of the small beaker into the large beaker.
Weigh the small beaker again using the ± 0.01g balance.
Record this mass.
The difference between the two accurate masses is the mass of sodium carbonate in your beaker.
4. Add de-ionised water cautiously down the side of the large beaker.
Use about 150cm3 of water, and swirl the beaker to mix the contents.
5. Stir using a glass rod to dissolve the solid completely.
6. Transfer the solution into the volumetric flask using the funnel.
Remember: pour down the glass rod;
remove the last drop of solution from the glass rod onto the funnel.
Wash the beaker, rod and funnel several times using de-ionised water from the wash bottle, letting the washings go into the flask.
7. Make up to the mark on the volumetric flask with de-ionised water.
Stopper firmly, and shake the flask thoroughly to mix the contents.
8. Label the flask clearly with your name, the date, and the contents of the flask. Copyright © 2003 N. Saunders N-ch1-35
Volumetric Analysis 1
To make a standard solution of sodium carbonate:
Technician's Notes
Prior to practical
Sodium carbonate
Heat required amount of sodium carbonate (Na2CO3) to drive off water of crystallisation.
Either: heat in an evaporating dish over a Bunsen burner for 30 minutes approx., or
heat in a drying oven at about 110°C for 1 hour.
Agitate the solid periodically with a clean glass rod.
Transfer to a desiccator after heating, and label it "sodium carbonate".
Care: Use tongs and eye protection.
Beware of hot solid and apparatus.
Sodium carbonate forms caustic alkaline solutions with water; if spilt on skin wash with plenty of water.
Analytical balances
Please check cleanliness and correct functioning of analytical balances.
De-ionised water
Please check 6th Form wash bottles are clean and filled with de-ionised water.
Make sure that additional de-ionised water is available in the aspirator.
Requirements per class
Sodium carbonate solid (see above). Minimum of 3g per student approx.
Analytical balances (see above).
Top pan digital balances (minimum of two if possible).
De-ionised water (see above).
Requirements per student
1 x 250cm3 beaker (dry)
1 x 250cm3 volumetric flask with stopper to fit
1 x glass funnel (check that it will enter the neck of the volumetric flask easily)
1 x glass rod
1 x weighing bottle with lid
1 x 6th Form wash bottle containing de-ionised water
1 x teat pipette
1 x small spatula
1 x self-adhesive label
Estimating Errors in Chemistry
Copyright © 2003 Nigel Saunders N-ch1-24 Estimating Errors in Chemistry
Introduction
You must be able to estimate the size and importance of errors in practical work to gain full marks in your practical assessments. The table summarises what you need to do about errors in each Skill:
Skill
You must be able to:
Skill P
Planning
Organise the procedure to be followed, selecting appropriate techniques, reagents and apparatus, with due regard to precision of measurement and scale of working.
Skill I
Implementing
Make and record measurements to a degree of precision allowed by the apparatus used.
Skill A
Analysing
Identify sources of error, and recognise the limitations of experimental measurements.
Skill E
Evaluating
Assess the reliability of your data and the conclusions drawn from them, taking into account the errors in the data obtained.
Introduction
You must be able to estimate the size and importance of errors in practical work to gain full marks in your practical assessments. The table summarises what you need to do about errors in each Skill:
Skill
You must be able to:
Skill P
Planning
Organise the procedure to be followed, selecting appropriate techniques, reagents and apparatus, with due regard to precision of measurement and scale of working.
Skill I
Implementing
Make and record measurements to a degree of precision allowed by the apparatus used.
Skill A
Analysing
Identify sources of error, and recognise the limitations of experimental measurements.
Skill E
Evaluating
Assess the reliability of your data and the conclusions drawn from them, taking into account the errors in the data obtained.
Making Standard Solutions
Making Standard Solutions
What is a standard solution?
A standard solution is a solution whose concentration is known accurately. Its concentration is usually given in mol dm–3. When making up a standard solution it is important that the correct mass of substance is accurately measured. It is also important that all of this is successfully transferred to the volumetric flask used to make up the solution. The following procedure will make sure that this happens.
Background calculations
1. Work out the number of moles needed to make up a solution with the required volume and concentration.
Show your working in the space below.
2. Now work out the relative formula mass, Mr, of the chosen substance. Show your working in the space below.
3. Work out the mass of substance needed using your answers from steps 1 and 2. Show your working in the space below.
Making up the solution
• Take a watch glass and place it on the balance. Tare the balance (set it to zero).
Carefully weigh out the required mass of substance.
• Transfer this amount to a beaker. Add water from a wash bottle to dissolve it.
Use some of the water to rinse all the substance off the watch glass. Do this at least twice.
• Stir with a glass rod until all the solid is dissolved, then transfer the solution to the volumetric flask.
Use more water from the wash bottle to rinse out the beaker and the glass rod. Do this at least twice.
• Add water to just below the line on the volumetric flask.
Add the final drops with a teat pipette to ensure that the bottom of the meniscus is on the line.
• Put the lid on the flask and turn the flask over a couple of times to mix the solution.
• Label your solution with your name, the date, and the contents, e.g. 2.0M NaCl. Then tidy up!
moles = concentration x volume
concentration is in mol dm–3 … and …
volume is in dm3, so if the volume is given in cm3, divide it by 1000 to get dm3
To find the Mr of carbon dioxide, CO2:
CO2 has … 1 carbon atom … 1 x 12 = 12
2 oxygen atoms … 2 x 16 = 32
add together … = 44
It helps to remember that:
"mass is mister mole", or
mass = Mr x mole Copyright © 2003 Nigel Saunders & Sally Burch N-ch1-49
What is a standard solution?
A standard solution is a solution whose concentration is known accurately. Its concentration is usually given in mol dm–3. When making up a standard solution it is important that the correct mass of substance is accurately measured. It is also important that all of this is successfully transferred to the volumetric flask used to make up the solution. The following procedure will make sure that this happens.
Background calculations
1. Work out the number of moles needed to make up a solution with the required volume and concentration.
Show your working in the space below.
2. Now work out the relative formula mass, Mr, of the chosen substance. Show your working in the space below.
3. Work out the mass of substance needed using your answers from steps 1 and 2. Show your working in the space below.
Making up the solution
• Take a watch glass and place it on the balance. Tare the balance (set it to zero).
Carefully weigh out the required mass of substance.
• Transfer this amount to a beaker. Add water from a wash bottle to dissolve it.
Use some of the water to rinse all the substance off the watch glass. Do this at least twice.
• Stir with a glass rod until all the solid is dissolved, then transfer the solution to the volumetric flask.
Use more water from the wash bottle to rinse out the beaker and the glass rod. Do this at least twice.
• Add water to just below the line on the volumetric flask.
Add the final drops with a teat pipette to ensure that the bottom of the meniscus is on the line.
• Put the lid on the flask and turn the flask over a couple of times to mix the solution.
• Label your solution with your name, the date, and the contents, e.g. 2.0M NaCl. Then tidy up!
moles = concentration x volume
concentration is in mol dm–3 … and …
volume is in dm3, so if the volume is given in cm3, divide it by 1000 to get dm3
To find the Mr of carbon dioxide, CO2:
CO2 has … 1 carbon atom … 1 x 12 = 12
2 oxygen atoms … 2 x 16 = 32
add together … = 44
It helps to remember that:
"mass is mister mole", or
mass = Mr x mole Copyright © 2003 Nigel Saunders & Sally Burch N-ch1-49
Sunday, April 19, 2009
Anion analysis
Chloride Cl-
AgCl, PbCl2, Hg2Cl2 and CuCl are insoluble in water.
Concentrated sulphuric acid liberates steamy acidic fumes of HCl from solid chlorides:
NaCl(s) + H2SO4(l) " NaHSO4(s) + HCl(g)
Silver nitrate solution added to a solution of a chloride that has been acidified (test with blue litmus paper) with dilute nitric acid gives a white precipitate of silver chloride. The precipitate is readily soluble in dilute ammonia or in sodium thiosulphate solution:
Ag+(aq) + Cl-(aq) " AgCl(s)
AgCl(s) + 2NH3(aq) " [Ag(NH3)2]+(aq) + Cl-(aq)
AgCl(s) + 2S2O32-(aq) " [Ag(S2O3)2]3-(aq) + Cl-(aq)
Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by giving spurious precipitates.
Concentrated solutions of sulphates can give a precipitate of silver sulphate in this test; its appearance is wholly different from AgCl. The latter is truly white; the sulphate is a pearly white, rather like pearlescent nail varnish.
Bromide Br-
AgBr, PbBr2, Hg2Br2 and CuBr are insoluble in water.
Concentrated sulphuric acid gives a mixture of hydrogen bromide, bromine and sulphur dioxide with solid bromides; the HBr produced is oxidised by sulphuric acid. The mixture evolves steamy brownish acidic fumes:
NaBr(s) + H2SO4(l) " NaHSO4(s) + HBr(g)
2HBr + H2SO4 " Br2 + SO2 + 2H2O
Silver nitrate solution added to a solution of a bromide that has been acidified (test with blue litmus paper) with dilute nitric acid gives a cream precipitate of silver bromide. The precipitate is readily soluble in concentrated ammonia:
Ag+(aq) + Br-(aq) " AgBr(s)
AgBr(s) + 2NH3(aq) " [Ag(NH3)2]+(aq) + Br-(aq)
Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by giving spurious precipitates.
Oxidising agents oxidise bromide to bromine, which is yellow or orange in aqueous solution. Bromine can be extracted from the solution by shaking with an immiscible organic solvent, for example hexane, the organic layer then turning orange.
A suitable oxidising agent is sodium chlorate(I); this is added to the test solution, followed by a little dilute hydrochloric acid and a few cm3 of hexane:
OCl-(aq) + 2H+(aq) + 2Br-(aq) " Br2(aq) + Cl-(aq) + H2O(l)
Iodide I-
AgI, PbI2, Hg2I2 and CuI are insoluble in water.
Concentrated sulphuric acid gives a mixture of hydrogen iodide, iodine, hydrogen sulphide, sulphur and sulphur dioxide when added to solid iodides; the HI produced is oxidised by sulphuric acid. The mixture evolves purple acidic fumes, turns to a brown slurry, and is a mess:
NaI(s) + H2SO4(l) " NaHSO4(s) + HI(g)
2HI + H2SO4 " I2 + SO2 + 2H2O
6HI + H2SO4 " 3I2 + S + 4H2O
8HI + H2SO4 " 4I2 + H2S + 4H2O
There are no state symbols in these equations because the mixture is such a mess, and is more sulphuric acid than water.
Silver nitrate solution added to a solution of an iodide that has been acidified (test with blue litmus paper) with dilute nitric acid gives a yellow precipitate of silver iodide. The precipitate is insoluble even in concentrated ammonia:
Ag+(aq) + I-(aq) " AgI(s)
Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by giving spurious precipitates.
Oxidising agents oxidise iodide to iodine, which is yellow or orange in aqueous solution. Iodine can be extracted from the solution by shaking with an immiscible organic solvent, for example hexane, the organic layer then turning purple.
A suitable oxidising agent is sodium chlorate(I); this is added to the test solution, followed by a little dilute hydrochloric acid and a few cm3 of hexane:
OCl-(aq) + 2H+(aq) + 2Ir-(aq) " I2(aq) + Cl-(aq) + H2O(l)
Lead ethanoate or lead nitrate solutions give a bright yellow precipitate of lead(II) iodide with iodides:
Pb2+(aq) + 2I-(aq) " PbI2(s)
The colour of lead(II) iodide comes from interactions in the lattice; dissolving the salt in boiling water gives a colourless solution which deposits glittering yellow plates on cooling.
Solutions of copper(II) salts give a brown mixture containing iodine and copper(I) iodide when added to solutions of iodides. Addition of sodium thiosulphate solution decolourises the iodine and leaves pinkish-cream copper(I) iodide as a precipitate.
2Cu2+(aq) + 4I-(aq) " 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) " 2I-(aq) + S4O62-(aq)
This reaction is the basis for the volumetric estimation of copper; the iodine liberated from a known amount of a copper(II) solution is titrated with standard sodium thiosulphate solution.
Sulphite SO32-
Sulphurous acid is considerably stronger than carbonic acid, so sulphites do not give the effervescence that is characteristic of carbonates when dilute acid is added.
Dilute hydrochloric acid on warming with a sulphite evolves sulphur dioxide; this turns acidified potassium dichromate(VI) solution (or paper) green:
SO32-(aq) + 2H+(aq) " H2O(l) + SO2(g)
Barium chloride solution gives a white precipitate of barium sulphite; addition of dilute hydrochloric acid causes the precipitate to dissolve without effervescence:
SO32-(aq) + Ba2+(aq) " BaSO3(s)
Sulphate SO42-
BaSO4, SrSO4 and PbSO4 are insoluble; CaSO4 is sparingly soluble.
Barium chloride solution added to the test solution acidified with dilute hydrochloric acid gives a white precipitate of barium sulphate:
Ba2+(aq) + SO42-(aq) " BaSO4(s)
HSO4- does the same thing with barium ions; however the original test solution would then be very acidic, so that should be tested for.
The addition of HCl destroys any carbonate or sulphite ions present so prevents the spurious positive result due to the precipitation of these barium salts. Barium nitrate solution can be used instead of barium chloride.
18.2 Lead ethanoate solution gives a precipitate of white lead sulphate:
Pb2+(aq) + SO42-(aq) " PbSO4(s)
Nitrate NO3-
Since all nitrates are water soluble, there is no precipitation reaction for this ion.
Solid nitrates decompose on heating; those of group 1 (except Li) give the nitrite and oxygen;
2NaNO3(s) " 2NaNO2(s) + O2(g)
All others give the metal oxide, nitrogen dioxide, and oxygen. A brown gas is emitted that re-lights a glowing splint:
2Pb(NO3)2(s) " 2PbO(s) + O2(g) + 2NO2(g)
Nitrate ions are reduced to ammonia by boiling with aluminium or with Devarda’s Alloy in sodium hydroxide solution. Devarda’s Alloy contains aluminium, zinc and copper. Since ammonium ions also give ammonia with NaOH, the test solution must be boiled with NaOH and the vapour tested for ammonia; if present heating must continue until all the ammonia has gone. The mixture is then cooled, Devarda’s Alloy (or a piece of aluminium foil) added, and the mixture re-heated. A gas that turns moist red litmus paper blue indicates nitrate in the original solution:
3NO3-(aq) + 8Al(s) + 18H2O(l) + 21 OH-(aq) " 8[Al(OH)6]3-(aq) + 3NH3(g)
Not an equation to be remembered!
Carbonate CO32-
Only the alkali metal and ammonium carbonates are water soluble. Some carbonates (e.g. zinc, copper(II)) are basic carbonates and contain a proportion of the hydroxide in their structure.
Heating decomposes all but the alkali and alkaline earth metal carbonates (at Bunsen temperatures) giving the oxide and carbon dioxide:
CuCO3(s) " CuO(s) + CO2(g)
Dilute hydrochloric acid gives vigorous effervescence with carbonates, evolving carbon dioxide:
CO32-(aq or s) + 2H+(aq) " H2O(l) + CO2(g)
Bicarbonates also give this effervescence. The reaction of carbonates with acid is exothermic; bicarbonates react endothermically.
Hydrogen carbonate (bicarbonate) HCO3-
Only the alkali metal and ammonium bicarbonates are obtainable as solids; group 2 bicarbonates exist only in solution.
Calcium chloride solution on addition to a bicarbonate solution gives no precipitate since calcium bicarbonate is soluble; this distinguishes it from carbonate, which does give a precipitate. On heating the calcium chloride/bicarbonate mixture a white precipitate appears since the bicarbonate decomposes to carbonate:
Ca2+(aq) + 2HCO3-(aq) " CaCO3(s) + CO2(g) + H2O(l)
Ethanoate CH3COO-
Ethanoates on heating with dilute hydrochloric acid give ethanoic acid, recognisable by its vinegary smell.
Neutral iron(III) chloride solution added to neutral solutions of ethanoate ion give a deep red colouration owing to formation of iron(III) ethanoate.
Ethanedioate C2O42-
Concentrated sulphuric acid added to a solid ethanedioate salt gives a mixture of carbon monoxide and carbon dioxide from dehydration of the ethanedioic acid formed:
HOOC-COOH " H2O + CO2 + CO
Potassium manganate(VII) solution acidified with dilute sulphuric acid added to a solution of an ethanedioate causes the purple colour to disappear:
2MnO4-(aq) + 5C2O42-(aq) + 16H+(aq) " 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
Calcium chloride solution added to a solution of an ethanedioate gives a white precipitate of calcium ethanedioate:
Ca2+(aq) + C2O42-(aq) " CaC2O4(s)
The precipitate dissolves readily in dilute hydrochloric acid.
Chromate(VI) CrO42- and dichromate(VI) Cr2O72-
These ions are related through the equilibrium
Cr2O72-(aq) + 2OH-(aq) D 2CrO42-(aq) + H2O(l)
In alkaline solution the yellow chromate(VI) dominates, in acidic solution orange dichromate(VI). All dichromate(VI) salts are soluble; addition of dichromate(VI) ions to solutions of ions of metals which have insoluble chromate(VI) salts leads to the precipitation of chromates. This means that the only dichromates that can exist are those of group 1 metals, ammonium, magnesium, calcium and strontium.
Barium chloride solution added to a chromate(VI) or dichromate(VI) solution precipitates bright yellow barium chromate(VI):
Ba2+(aq) + CrO42-(aq) " BaCrO4(s)
The addition of a heavy metal ion to potassium dichromate solution precipitates the chromate and therefore moves the equilibrium to the right hand side. If the chromate is sparingly soluble (e.g. strontium) the supernatant liquid will remain yellow. Very insoluble chromates, such as lead, remove all the colour from the supernatant liquid.
Dichromate(VI) ion solution in sulphuric acid is an oxidising agent; oxidation is shown by the solution turning from orange to green (Cr(III)). The following can be oxidised:
(a) iron(II) to iron(III):
Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) " 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
(b) iodide to iodine (the solution turns murky greenish-brown):
Cr2O72-(aq) + 14H+(aq) + 6I-(aq) " 2Cr3+(aq) + 7H2O(aq) + 3I2(aq)
(c) sulphite to sulphate:
Cr2O72-(aq) + 8H+(aq) + 3SO32-(aq) " 2Cr3+(aq) + 4H2O(l) + 3SO42-(aq)
(d) nitrite to nitrate:
Cr2O72-(aq) + 8H+(aq) + 3NO2-(aq) " 2Cr3+(aq) + 4H2O(l) + 3NO3-(aq)
(e) hydrogen peroxide reacts with acidified dichromate(VI) solutions to give a blue compound that rapidly turns green and evolves oxygen. The blue compound can be extracted into an organic solvent such as butan-1-ol. The blue compound is CrO5, which contains a peroxy structure. It is covalent, and is stable in organic solvents though not in water.
Cr2O72-(aq) + 8H+(aq) + 3H2O2 (aq) " 2Cr3+(aq) + 7H2O(l) + 3 O2(g)
Alcohols are oxidised by acidified potassium dichromate(VI) solution. Primary alcohols give aldehydes and then acids, secondary alcohols give ketones. Ethanol can be used to test for dichromate(VI) ions, therefore, the solution turning green and the apple smell of ethanal being evident.
AgCl, PbCl2, Hg2Cl2 and CuCl are insoluble in water.
Concentrated sulphuric acid liberates steamy acidic fumes of HCl from solid chlorides:
NaCl(s) + H2SO4(l) " NaHSO4(s) + HCl(g)
Silver nitrate solution added to a solution of a chloride that has been acidified (test with blue litmus paper) with dilute nitric acid gives a white precipitate of silver chloride. The precipitate is readily soluble in dilute ammonia or in sodium thiosulphate solution:
Ag+(aq) + Cl-(aq) " AgCl(s)
AgCl(s) + 2NH3(aq) " [Ag(NH3)2]+(aq) + Cl-(aq)
AgCl(s) + 2S2O32-(aq) " [Ag(S2O3)2]3-(aq) + Cl-(aq)
Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by giving spurious precipitates.
Concentrated solutions of sulphates can give a precipitate of silver sulphate in this test; its appearance is wholly different from AgCl. The latter is truly white; the sulphate is a pearly white, rather like pearlescent nail varnish.
Bromide Br-
AgBr, PbBr2, Hg2Br2 and CuBr are insoluble in water.
Concentrated sulphuric acid gives a mixture of hydrogen bromide, bromine and sulphur dioxide with solid bromides; the HBr produced is oxidised by sulphuric acid. The mixture evolves steamy brownish acidic fumes:
NaBr(s) + H2SO4(l) " NaHSO4(s) + HBr(g)
2HBr + H2SO4 " Br2 + SO2 + 2H2O
Silver nitrate solution added to a solution of a bromide that has been acidified (test with blue litmus paper) with dilute nitric acid gives a cream precipitate of silver bromide. The precipitate is readily soluble in concentrated ammonia:
Ag+(aq) + Br-(aq) " AgBr(s)
AgBr(s) + 2NH3(aq) " [Ag(NH3)2]+(aq) + Br-(aq)
Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by giving spurious precipitates.
Oxidising agents oxidise bromide to bromine, which is yellow or orange in aqueous solution. Bromine can be extracted from the solution by shaking with an immiscible organic solvent, for example hexane, the organic layer then turning orange.
A suitable oxidising agent is sodium chlorate(I); this is added to the test solution, followed by a little dilute hydrochloric acid and a few cm3 of hexane:
OCl-(aq) + 2H+(aq) + 2Br-(aq) " Br2(aq) + Cl-(aq) + H2O(l)
Iodide I-
AgI, PbI2, Hg2I2 and CuI are insoluble in water.
Concentrated sulphuric acid gives a mixture of hydrogen iodide, iodine, hydrogen sulphide, sulphur and sulphur dioxide when added to solid iodides; the HI produced is oxidised by sulphuric acid. The mixture evolves purple acidic fumes, turns to a brown slurry, and is a mess:
NaI(s) + H2SO4(l) " NaHSO4(s) + HI(g)
2HI + H2SO4 " I2 + SO2 + 2H2O
6HI + H2SO4 " 3I2 + S + 4H2O
8HI + H2SO4 " 4I2 + H2S + 4H2O
There are no state symbols in these equations because the mixture is such a mess, and is more sulphuric acid than water.
Silver nitrate solution added to a solution of an iodide that has been acidified (test with blue litmus paper) with dilute nitric acid gives a yellow precipitate of silver iodide. The precipitate is insoluble even in concentrated ammonia:
Ag+(aq) + I-(aq) " AgI(s)
Acidification with nitric acid is necessary to eliminate carbonate or sulphite, both of which interfere with the test by giving spurious precipitates.
Oxidising agents oxidise iodide to iodine, which is yellow or orange in aqueous solution. Iodine can be extracted from the solution by shaking with an immiscible organic solvent, for example hexane, the organic layer then turning purple.
A suitable oxidising agent is sodium chlorate(I); this is added to the test solution, followed by a little dilute hydrochloric acid and a few cm3 of hexane:
OCl-(aq) + 2H+(aq) + 2Ir-(aq) " I2(aq) + Cl-(aq) + H2O(l)
Lead ethanoate or lead nitrate solutions give a bright yellow precipitate of lead(II) iodide with iodides:
Pb2+(aq) + 2I-(aq) " PbI2(s)
The colour of lead(II) iodide comes from interactions in the lattice; dissolving the salt in boiling water gives a colourless solution which deposits glittering yellow plates on cooling.
Solutions of copper(II) salts give a brown mixture containing iodine and copper(I) iodide when added to solutions of iodides. Addition of sodium thiosulphate solution decolourises the iodine and leaves pinkish-cream copper(I) iodide as a precipitate.
2Cu2+(aq) + 4I-(aq) " 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) " 2I-(aq) + S4O62-(aq)
This reaction is the basis for the volumetric estimation of copper; the iodine liberated from a known amount of a copper(II) solution is titrated with standard sodium thiosulphate solution.
Sulphite SO32-
Sulphurous acid is considerably stronger than carbonic acid, so sulphites do not give the effervescence that is characteristic of carbonates when dilute acid is added.
Dilute hydrochloric acid on warming with a sulphite evolves sulphur dioxide; this turns acidified potassium dichromate(VI) solution (or paper) green:
SO32-(aq) + 2H+(aq) " H2O(l) + SO2(g)
Barium chloride solution gives a white precipitate of barium sulphite; addition of dilute hydrochloric acid causes the precipitate to dissolve without effervescence:
SO32-(aq) + Ba2+(aq) " BaSO3(s)
Sulphate SO42-
BaSO4, SrSO4 and PbSO4 are insoluble; CaSO4 is sparingly soluble.
Barium chloride solution added to the test solution acidified with dilute hydrochloric acid gives a white precipitate of barium sulphate:
Ba2+(aq) + SO42-(aq) " BaSO4(s)
HSO4- does the same thing with barium ions; however the original test solution would then be very acidic, so that should be tested for.
The addition of HCl destroys any carbonate or sulphite ions present so prevents the spurious positive result due to the precipitation of these barium salts. Barium nitrate solution can be used instead of barium chloride.
18.2 Lead ethanoate solution gives a precipitate of white lead sulphate:
Pb2+(aq) + SO42-(aq) " PbSO4(s)
Nitrate NO3-
Since all nitrates are water soluble, there is no precipitation reaction for this ion.
Solid nitrates decompose on heating; those of group 1 (except Li) give the nitrite and oxygen;
2NaNO3(s) " 2NaNO2(s) + O2(g)
All others give the metal oxide, nitrogen dioxide, and oxygen. A brown gas is emitted that re-lights a glowing splint:
2Pb(NO3)2(s) " 2PbO(s) + O2(g) + 2NO2(g)
Nitrate ions are reduced to ammonia by boiling with aluminium or with Devarda’s Alloy in sodium hydroxide solution. Devarda’s Alloy contains aluminium, zinc and copper. Since ammonium ions also give ammonia with NaOH, the test solution must be boiled with NaOH and the vapour tested for ammonia; if present heating must continue until all the ammonia has gone. The mixture is then cooled, Devarda’s Alloy (or a piece of aluminium foil) added, and the mixture re-heated. A gas that turns moist red litmus paper blue indicates nitrate in the original solution:
3NO3-(aq) + 8Al(s) + 18H2O(l) + 21 OH-(aq) " 8[Al(OH)6]3-(aq) + 3NH3(g)
Not an equation to be remembered!
Carbonate CO32-
Only the alkali metal and ammonium carbonates are water soluble. Some carbonates (e.g. zinc, copper(II)) are basic carbonates and contain a proportion of the hydroxide in their structure.
Heating decomposes all but the alkali and alkaline earth metal carbonates (at Bunsen temperatures) giving the oxide and carbon dioxide:
CuCO3(s) " CuO(s) + CO2(g)
Dilute hydrochloric acid gives vigorous effervescence with carbonates, evolving carbon dioxide:
CO32-(aq or s) + 2H+(aq) " H2O(l) + CO2(g)
Bicarbonates also give this effervescence. The reaction of carbonates with acid is exothermic; bicarbonates react endothermically.
Hydrogen carbonate (bicarbonate) HCO3-
Only the alkali metal and ammonium bicarbonates are obtainable as solids; group 2 bicarbonates exist only in solution.
Calcium chloride solution on addition to a bicarbonate solution gives no precipitate since calcium bicarbonate is soluble; this distinguishes it from carbonate, which does give a precipitate. On heating the calcium chloride/bicarbonate mixture a white precipitate appears since the bicarbonate decomposes to carbonate:
Ca2+(aq) + 2HCO3-(aq) " CaCO3(s) + CO2(g) + H2O(l)
Ethanoate CH3COO-
Ethanoates on heating with dilute hydrochloric acid give ethanoic acid, recognisable by its vinegary smell.
Neutral iron(III) chloride solution added to neutral solutions of ethanoate ion give a deep red colouration owing to formation of iron(III) ethanoate.
Ethanedioate C2O42-
Concentrated sulphuric acid added to a solid ethanedioate salt gives a mixture of carbon monoxide and carbon dioxide from dehydration of the ethanedioic acid formed:
HOOC-COOH " H2O + CO2 + CO
Potassium manganate(VII) solution acidified with dilute sulphuric acid added to a solution of an ethanedioate causes the purple colour to disappear:
2MnO4-(aq) + 5C2O42-(aq) + 16H+(aq) " 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
Calcium chloride solution added to a solution of an ethanedioate gives a white precipitate of calcium ethanedioate:
Ca2+(aq) + C2O42-(aq) " CaC2O4(s)
The precipitate dissolves readily in dilute hydrochloric acid.
Chromate(VI) CrO42- and dichromate(VI) Cr2O72-
These ions are related through the equilibrium
Cr2O72-(aq) + 2OH-(aq) D 2CrO42-(aq) + H2O(l)
In alkaline solution the yellow chromate(VI) dominates, in acidic solution orange dichromate(VI). All dichromate(VI) salts are soluble; addition of dichromate(VI) ions to solutions of ions of metals which have insoluble chromate(VI) salts leads to the precipitation of chromates. This means that the only dichromates that can exist are those of group 1 metals, ammonium, magnesium, calcium and strontium.
Barium chloride solution added to a chromate(VI) or dichromate(VI) solution precipitates bright yellow barium chromate(VI):
Ba2+(aq) + CrO42-(aq) " BaCrO4(s)
The addition of a heavy metal ion to potassium dichromate solution precipitates the chromate and therefore moves the equilibrium to the right hand side. If the chromate is sparingly soluble (e.g. strontium) the supernatant liquid will remain yellow. Very insoluble chromates, such as lead, remove all the colour from the supernatant liquid.
Dichromate(VI) ion solution in sulphuric acid is an oxidising agent; oxidation is shown by the solution turning from orange to green (Cr(III)). The following can be oxidised:
(a) iron(II) to iron(III):
Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) " 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
(b) iodide to iodine (the solution turns murky greenish-brown):
Cr2O72-(aq) + 14H+(aq) + 6I-(aq) " 2Cr3+(aq) + 7H2O(aq) + 3I2(aq)
(c) sulphite to sulphate:
Cr2O72-(aq) + 8H+(aq) + 3SO32-(aq) " 2Cr3+(aq) + 4H2O(l) + 3SO42-(aq)
(d) nitrite to nitrate:
Cr2O72-(aq) + 8H+(aq) + 3NO2-(aq) " 2Cr3+(aq) + 4H2O(l) + 3NO3-(aq)
(e) hydrogen peroxide reacts with acidified dichromate(VI) solutions to give a blue compound that rapidly turns green and evolves oxygen. The blue compound can be extracted into an organic solvent such as butan-1-ol. The blue compound is CrO5, which contains a peroxy structure. It is covalent, and is stable in organic solvents though not in water.
Cr2O72-(aq) + 8H+(aq) + 3H2O2 (aq) " 2Cr3+(aq) + 7H2O(l) + 3 O2(g)
Alcohols are oxidised by acidified potassium dichromate(VI) solution. Primary alcohols give aldehydes and then acids, secondary alcohols give ketones. Ethanol can be used to test for dichromate(VI) ions, therefore, the solution turning green and the apple smell of ethanal being evident.
Cation Analysis
Sodium Na+
Flame test: covered earlier. There are no simple reagents that will precipitate sodium compounds.
Potassium K+
Flame test: covered earlier. There are no simple reagents that will precipitate potassium compounds.
Magnesium Mg2+
Ions
Sodium hydroxide solution precipitates white magnesium hydroxide, insoluble in excess NaOH:
Mg2+(aq) + 2OH-(aq) " Mg(OH)2(s)
Ammonia solution only partially precipitates magnesium hydroxide, and not at all in the presence of ammonium ions. This is because magnesium hydroxide is fairly soluble, and the small concentration of OH- ions in ammonia becomes even smaller in the presence of ammonium ions and is not sufficient to produce the precipitate.
Sodium or potassium carbonate solution gives a white gelatinous precipitate of the basic carbonate Mg(OH)2.4MgCO3.5H2O.
Calcium Ca2+
Dilute sulphuric acid gives a white precipitate of calcium sulphate if the original solution is fairly concentrated:
Ca2+(aq) + SO42-(aq) " CaSO4(s)
Calcium sulphate solution is permanently hard water; its solubility is about 0.05 mol dm-3 at 25oC. Sodium or potassium carbonate solution precipitates white calcium carbonate:
Ca2+(aq) + CO32-(aq) " CaCO3(s)
Ammonium ethanedioate solution precipitates white calcium ethanedioate from neutral or alkaline solutions. The precipitate dissolves readily in dilute acid:
Ca2+(aq) + C2O42-(aq) " CaC2O4(s)
Calcium ethanedioate (calcium oxalate) is found in rhubarb leaves - it is what makes them poisonous.
Chromium(III) [Cr(H2O)6]3+
Sodium hydroxide solution precipitates grey-green chromium(III) hydroxide, which reacts with excess NaOH to give a deep green solution of the chromite ion:
[Cr(H2O)6]3+(aq) + 3OH-(aq) " Cr(OH)3(s) + 6H2O(l)
Cr(OH)3(s) + 3OH-(aq) " [Cr(OH)6]3-(aq)
Ammonia solution precipitates grey-green chromium(III) hydroxide, which with excess ammonia very slowly (see below) forms pinkish solutions of ammines. Ammines are transition metal complexes with ammonia, NH3. Do not confuse them with amines RNH2.The initial equation is as the first one above; then
Cr(OH)3(s) + 4NH3(aq) + 2H2O(l) " [Cr(NH3)4(H2O)2]3+(aq) + 3OH-(aq)
The exact composition of the ammine will depend on the amount of ammonia used. However, the reaction is so slow that several days are needed to see anything at all.
Oxidising agents convert chromium(III) to yellow chromate(VI) in the presence of alkali. Hydrogen peroxide is favoured:
2[Cr(H2O)6]3+(aq) + 3H2O2(aq) + 10OH-(aq) " 2CrO42-(aq) + 14H2O(l)
The test solution is made alkaline with NaOH solution, then an equal volume of ‘20 volume’ hydrogen peroxide solution is added and the mixture boiled. 20-volume hydrogen peroxide solution gives 20cm3 of oxygen per cm3 of solution that is decomposed to oxygen and water.
Manganese(II) [Mn(H2O)6]2+
Sodium hydroxide solution precipitates beige manganese(II) hydroxide. This is insoluble in excess reagent, and rapidly darkens owing to oxidation to hydrated manganese(IV) oxide:
[Mn(H2O)6]2+(aq) + 2OH-(aq) " Mn(OH)2(s) + 6H2O(l)
4Mn(OH)2(s) + 2H2O(l) + O2(aq) " 4MnO2.H2O(s)
Sodium or potassium carbonate solution gives a white precipitate of manganese(II) carbonate:
[Mn(H2O)6]2+(aq) + CO32-(aq) " MnCO3(s) + 6H2O(l)
Lead(IV) oxide in the presence of nitric acid converts manganese(II) into purple manganate(VII); an alternative oxidising agent is sodium bismuthate(V), NaBiO3:
5PbO2(s) + 2Mn2+(aq) + 6H+(aq) " 2MnO4-(aq) + 5Pb2+(aq) + 2H2O(l)
5BiO3-(aq) + 14H+(aq) + 2Mn2+(aq) " 5Bi3+(aq) + 2MnO4-(aq) + 7H2O(l)
A small amount of lead(IV) oxide or of sodium bismuthate(V) is added to the test solution, 6 or so drops of concentrated nitric acid added, and the mixture boiled. Filtration will give a purple filtrate if Mn2+ was present.
Iron(III) [Fe(H2O)6]3+
Aqueous iron(III) ions are not [Fe(H2O)6]3+; this ion is an amethyst (pale purple) colour, and is found only in solids such as aluminium iron(III) sulphate. In solution the hexaaqua ion is readily deprotonated by solvent water; the solution is acidic, and the yellow ion is [Fe(H2O)5OH]2+:
[Fe(H2O)6]3+ + H2O " [Fe(H2O)5OH]2+ + H3O+.
Sodium hydroxide solution precipitates foxy-red iron(III) hydroxide, insoluble in excess alkali:
[Fe(H2O)6]3+(aq) + 3OH-(aq) " Fe(OH)3(s) + 6H2O(l)
Ammonia solution reacts in the same way as sodium hydroxide – iron does not form complexes with ammonia.
Potassium hexacyanoferrate(II) precipitates dark blue potassium iron(III) hexacyanoferrate(II), Prussian Blue:
K+(aq) + Fe3+(aq) + [Fe(CN)6]4-(aq) " KFeIII[FeII(CN)6](s)
Prussian Blue is the pigment that is used to print 'blueprints'. It was also used for Prussian army uniforms, and for the locomotives and rolling stock of the Somerset & Dorset Joint Railway, pre-1923.
Iron(II) [Fe(H2O)6]2+
Sodium hydroxide solution precipitates dirty green iron(II) hydroxide; on standing the surface of the precipitate turns foxy-red owing to air oxidation to iron(III) hydroxide:
[Fe(H2O)6]2+(aq) + 2OH-(aq) " Fe(OH)2(s) + 6H2O(l)
Ammonia solution behaves similarly to sodium hydroxide – again there are no ammine complexes.
Potassium hexacyanoferrate(III) precipitates dark blue potassium iron(II) hexacyanoferrate(III), Turnbull’s Blue:
K+(aq) + Fe2+(aq) + [Fe(CN)6]3-(aq) " KFeII[FeIII(CN)6](s)
Although Turnbull's Blue and Prussian Blue don't have quite the same colour, they are in fact the same substance.
Potassium manganate(VII) oxidises iron(II) to iron(III) in acidic solution. The purple manganate colour is lost and the resulting solution of iron(III) is yellow:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) " Mn2+(aq) + 5Fe3+(aq) + 4H2O(aq)
Potassium dichromate(VI) oxidises iron(II) to iron(III) in acidic solution. The orange dichromate colour is lost and the resulting solution is green owing to presence of chromium(III) which masks the iron(III) colour:
Cr2O72-(aq) + 6Fe2+(aq) + 14H+(aq) " 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
Copper(II) [Cu(H2O)6]2+
Sodium hydroxide solution gives a pale blue precipitate usually described as copper(II) hydroxide:
[Cu(H2O)6]2+(aq) + 2OH-(aq) " Cu(OH)2(s) + 6H2O(l)
In fact this precipitate is the basic salt – thus from copper(II) sulphate solution the product is basic copper sulphate:
2[Cu(H2O)6]2+(aq) + 2OH-(aq) + SO42-(aq) " Cu(OH)2.CuSO4(s) + 12H2O(l)
Basic copper sulphate or basic copper carbonate is responsible for the green patina - verdigris - found on weathered copper roofs. The sulphate is present in industrial surroundings. To get true copper(II) hydroxide, the solution of the copper(II) salt must be added to excess sodium hydroxide; the precipitate is then blue rather than turquoise. In both cases CAUTIOUS warming of the mixture causes the precipitate to lose water and turn black as CuO(s) is formed. The loss of water by heating in an aqueous medium is unusual.
Ammonia solution initially gives a blue precipitate as for sodium hydroxide. Further addition of ammonia gives deep blue soluble cuprammine complexes whose composition depends on the amount of ammonia present. The reaction is usually represented:
Cu(OH)2(s) + 4NH3(aq) + 2H2O(l) " [Cu(NH3)4(H2O)2]2+(aq) +2OH-(aq)
The cuprammine solution has the bizarre property of being able to dissolve cellulose - bits of filter paper, for e example. The resulting viscous substance can be extruded into a bath of dilute sulphuric acid to give viscose fibre, which is a form of artificial silk.
Concentrated HCl or a saturated solution of sodium chloride gives a bright green solution containing CuCl42- ions; dilution with water causes the solution to turn pale blue as the [Cu(H2O)6]2+ ion re-forms.
[Cu(H2O)6]2+(aq) + 4Cl-(aq) " CuCl42-(aq) + 6H2O(l)
Sodium carbonate solution precipitates greenish-blue basic carbonates of indefinite composition.
Potassium iodide solution precipitates iodine and copper(I) iodide; this reaction is used volumetrically for the estimation of copper, the liberated iodine being titrated with sodium thiosulphate solution:
2[Cu(H2O)6]2+(aq) + 4I-(aq) " 2CuI(s) + I2(aq) + 6H2O(l)
The mixture turns a sludgy brown colour; addition of sodium thiosulphate solution leaves a creamy-pink precipitate of copper(I) iodide.
Zinc(II) [Zn(H2O)6]2+
Sodium hydroxide solution precipitates white zinc(II) hydroxide, easily soluble in excess sodium hydroxide to give sodium zincate(II), because the hydroxide is amphoteric:
[Zn(H2O)6]2+(aq) + 2OH-(aq) " Zn(OH)2(s) + 6H2O(l)
Zn(OH)2(s) + 2OH-(aq) " Zn(OH)42-(aq)
Ammonia solution initially precipitates zinc(II) hydroxide, as with sodium hydroxide. Excess ammonia causes the precipitate to disappear owing to the formation of ammine complexes – a different reason from that in 10.1:
Zn(OH)2(s) + 4NH3(aq) " [Zn(NH3)4]2+(aq) + 2OH-(aq)
Sodium carbonate solution precipitates a white basic carbonate:
5Zn2+(aq) + 2CO32-(aq) + 6OH-(aq) " 2ZnCO3.3Zn(OH)2(s)
This latter substance is what you get when you buy zinc carbonate – which is why on heating it gives off water vapour as well as carbon dioxide.
Aluminium Al[H2O] 3+
Sodium hydroxide solution precipitates white gelatinous aluminium hydroxide. This reacts with excess NaOH to give a colourless solution of sodium aluminate:
[Al(H2O)6] 3+(aq) + 3OH-(aq) " Al(OH)3(s) + 6H2O(l)
Al(OH)3(s) + 3OH-(aq) " [Al(OH)6]3-(aq)
Ammonia solution precipitates white aluminium hydroxide, as with NaOH; however it does not react further with excess ammonia.
Lead(II) Pb2+
Dilute hydrochloric acid or other soluble chlorides precipitate white lead(II) chloride from moderately concentrated solutions of lead(II) salts.
Pb2+(aq) + 2Cl-(aq) " PbCl2(s)
The precipitate dissolves in an excess of concentrated hydrochloric acid to give yellow solutions containing various species including PbCl3- and PbCl42-.
Sodium hydroxide solution gives a white precipitate of lead(II) hydroxide which is soluble in excess NaOH to give colourless sodium plumbate(II)
Pb2+(aq) + 2OH-(aq) " Pb(OH)2(s)
Pb(OH)2(s) + 2OH-(aq) " [Pb(OH)4]2-(aq)
Ammonia solution precipitates white lead(II) hydroxide; this does not dissolve in excess ammonia, because no complexes are formed and ammonia is not a strong enough base to bring out the amphoteric properties of lead(II) hydroxide.
Dilute sulphuric acid or other soluble sulphates give a white precipitate of lead(II) sulphate:
Pb2+(aq) + SO42-(aq) " PbSO4(s)
Potassium chromate(VI) solution gives a bright yellow precipitate of lead(II) chromate(VI):
Pb2+(aq) + CrO42-(aq) " PbCrO4(s)
Potassium iodide solution gives a bright yellow precipitate of lead(II) iodide. This will dissolve in boiling water to give a colourless solution – the colour comes from interactions in the crystal lattice rather than from coloured ions:
Pb2+(aq) + 2I-(aq) " PbI2(s)
Ammonium NH4+
Heat: all ammonium salts decompose on heating; ammonium nitrate may explode. Ammonium chloride and sulphate give products that recombine on cooling, so that the salts apparently sublime.
NH4Cl(s) " NH3(g) + HCl(g)
NH4NO3(s) " N2O(g) + 2H2O(g)
2(NH4)2SO4(s) D 2NH3(g) + SO3(g) + H2O(g)
Ammonium dichromate(VI) decomposes spectacularly on ignition in a reaction that is oxidation of the cation by the anion; the initially orange solid gives a fluffy green product of much larger volume:
(NH4)2Cr2O7(s) " N2(g) + Cr2O3(s) + 4H2O(g)
Alkalis (sodium hydroxide, calcium hydroxide) liberate ammonia from ammonium salts on warming with the solution, or even from a mixture of the solids. This is because OH- is a stronger base than ammonia, so removes a hydrogen ion from the ammonium ion:
NH4+(aq) + OH-(aq) " NH3(g) + H2O(l)
The test solution is warmed with sodium hydroxide solution and the vapours tested with moist red litmus paper. It is important to test the vapours immediately heating begins, since the ammonia is lost very quickly and by the time the solution boils it may well have all gone.
Cobalt(II) [Co(H2O)6]2+
The (+2) state is the most stable for simple cobalt salts; they are coloured pink or blue. The pink colour of the hydrated ion [Co(H2O)6]2+ changes to blue on heating, on dehydration, or in the presence of concentrated acids.
Sodium hydroxide solution precipitates a blue basic salt which on warming with excess sodium hydroxide forms solid pink cobalt(II) hydroxide. With a solution of cobalt(II) nitrate the equations are:
[Co(H2O)6]2+ (aq) + NO3 – (aq) + OH – (aq) "
Co(OH)NO3 (s) + 6H2O(l)
blue
Co(OH)NO3 (s) + OH – (aq) "
Co(OH)2 (s) + NO3 – (aq)
pink
Basic salts, usually hydroxides or carbonates, contain not only the expected hydroxide or carbonate ion but also the anion from the original salt. The precipitate from cobalt(II) nitrate solution is therefore different from that which would be given from cobalt(II) chloride. The latter would be Co(OH)Cl.
If the pink precipitate is boiled for some time (or is warmed with hydrogen peroxide solution) it is converted to brownish-black cobalt(III) hydroxide:
4Co(OH)2 (s) + 2H2O (l) + O2 (aq) " 4Co(OH)3 (s)
Ammonia solution gives a blue basic precipitate as with sodium hydroxide; excess ammonia converts this to a brown solution of a cobaltammine which turns red on exposure to air giving an ammine of cobalt(III). The overall reaction is:
Co(OH)NO3 (s) + 28NH3(aq) + 6H2O(l) + O2(aq) " 4[Co(NH3)6](OH)3(aq) + 4NH4NO3(aq)
Nickel(II) [Ni(H2O)6]2+
Sodium hydroxide solution gives a green gelatinous precipitate of nickel(II) hydroxide, insoluble in excess of the reagent:
[Ni(H2O)6]2+ (aq) + 2OH – (aq) " Ni(OH)2 (s) + 6H2O(l)
Ammonia solution gives a green precipitate of a basic salt which dissolves readily in excess ammonia solution to form a blue-violet solution of complex nickel ammines. With nickel(II) chloride the reactions are:
[Ni(H2O)6]2+(aq) + Cl – (aq) + NH3(aq) " Ni(OH)Cl(s) + NH4+(aq) + 5H2O(l)
Ni(OH)Cl(s) + 5NH3(aq) + H2O(l) " [Ni(NH3)4](OH)2(aq) + NH4+(aq) + Cl – (aq)
Ni(OH)Cl(s) + 7NH3(aq) + H2O(l) " [Ni(NH3)6](OH)2(aq) + NH4+(aq) + Cl – (aq)
Flame test: covered earlier. There are no simple reagents that will precipitate sodium compounds.
Potassium K+
Flame test: covered earlier. There are no simple reagents that will precipitate potassium compounds.
Magnesium Mg2+
Ions
Sodium hydroxide solution precipitates white magnesium hydroxide, insoluble in excess NaOH:
Mg2+(aq) + 2OH-(aq) " Mg(OH)2(s)
Ammonia solution only partially precipitates magnesium hydroxide, and not at all in the presence of ammonium ions. This is because magnesium hydroxide is fairly soluble, and the small concentration of OH- ions in ammonia becomes even smaller in the presence of ammonium ions and is not sufficient to produce the precipitate.
Sodium or potassium carbonate solution gives a white gelatinous precipitate of the basic carbonate Mg(OH)2.4MgCO3.5H2O.
Calcium Ca2+
Dilute sulphuric acid gives a white precipitate of calcium sulphate if the original solution is fairly concentrated:
Ca2+(aq) + SO42-(aq) " CaSO4(s)
Calcium sulphate solution is permanently hard water; its solubility is about 0.05 mol dm-3 at 25oC. Sodium or potassium carbonate solution precipitates white calcium carbonate:
Ca2+(aq) + CO32-(aq) " CaCO3(s)
Ammonium ethanedioate solution precipitates white calcium ethanedioate from neutral or alkaline solutions. The precipitate dissolves readily in dilute acid:
Ca2+(aq) + C2O42-(aq) " CaC2O4(s)
Calcium ethanedioate (calcium oxalate) is found in rhubarb leaves - it is what makes them poisonous.
Chromium(III) [Cr(H2O)6]3+
Sodium hydroxide solution precipitates grey-green chromium(III) hydroxide, which reacts with excess NaOH to give a deep green solution of the chromite ion:
[Cr(H2O)6]3+(aq) + 3OH-(aq) " Cr(OH)3(s) + 6H2O(l)
Cr(OH)3(s) + 3OH-(aq) " [Cr(OH)6]3-(aq)
Ammonia solution precipitates grey-green chromium(III) hydroxide, which with excess ammonia very slowly (see below) forms pinkish solutions of ammines. Ammines are transition metal complexes with ammonia, NH3. Do not confuse them with amines RNH2.The initial equation is as the first one above; then
Cr(OH)3(s) + 4NH3(aq) + 2H2O(l) " [Cr(NH3)4(H2O)2]3+(aq) + 3OH-(aq)
The exact composition of the ammine will depend on the amount of ammonia used. However, the reaction is so slow that several days are needed to see anything at all.
Oxidising agents convert chromium(III) to yellow chromate(VI) in the presence of alkali. Hydrogen peroxide is favoured:
2[Cr(H2O)6]3+(aq) + 3H2O2(aq) + 10OH-(aq) " 2CrO42-(aq) + 14H2O(l)
The test solution is made alkaline with NaOH solution, then an equal volume of ‘20 volume’ hydrogen peroxide solution is added and the mixture boiled. 20-volume hydrogen peroxide solution gives 20cm3 of oxygen per cm3 of solution that is decomposed to oxygen and water.
Manganese(II) [Mn(H2O)6]2+
Sodium hydroxide solution precipitates beige manganese(II) hydroxide. This is insoluble in excess reagent, and rapidly darkens owing to oxidation to hydrated manganese(IV) oxide:
[Mn(H2O)6]2+(aq) + 2OH-(aq) " Mn(OH)2(s) + 6H2O(l)
4Mn(OH)2(s) + 2H2O(l) + O2(aq) " 4MnO2.H2O(s)
Sodium or potassium carbonate solution gives a white precipitate of manganese(II) carbonate:
[Mn(H2O)6]2+(aq) + CO32-(aq) " MnCO3(s) + 6H2O(l)
Lead(IV) oxide in the presence of nitric acid converts manganese(II) into purple manganate(VII); an alternative oxidising agent is sodium bismuthate(V), NaBiO3:
5PbO2(s) + 2Mn2+(aq) + 6H+(aq) " 2MnO4-(aq) + 5Pb2+(aq) + 2H2O(l)
5BiO3-(aq) + 14H+(aq) + 2Mn2+(aq) " 5Bi3+(aq) + 2MnO4-(aq) + 7H2O(l)
A small amount of lead(IV) oxide or of sodium bismuthate(V) is added to the test solution, 6 or so drops of concentrated nitric acid added, and the mixture boiled. Filtration will give a purple filtrate if Mn2+ was present.
Iron(III) [Fe(H2O)6]3+
Aqueous iron(III) ions are not [Fe(H2O)6]3+; this ion is an amethyst (pale purple) colour, and is found only in solids such as aluminium iron(III) sulphate. In solution the hexaaqua ion is readily deprotonated by solvent water; the solution is acidic, and the yellow ion is [Fe(H2O)5OH]2+:
[Fe(H2O)6]3+ + H2O " [Fe(H2O)5OH]2+ + H3O+.
Sodium hydroxide solution precipitates foxy-red iron(III) hydroxide, insoluble in excess alkali:
[Fe(H2O)6]3+(aq) + 3OH-(aq) " Fe(OH)3(s) + 6H2O(l)
Ammonia solution reacts in the same way as sodium hydroxide – iron does not form complexes with ammonia.
Potassium hexacyanoferrate(II) precipitates dark blue potassium iron(III) hexacyanoferrate(II), Prussian Blue:
K+(aq) + Fe3+(aq) + [Fe(CN)6]4-(aq) " KFeIII[FeII(CN)6](s)
Prussian Blue is the pigment that is used to print 'blueprints'. It was also used for Prussian army uniforms, and for the locomotives and rolling stock of the Somerset & Dorset Joint Railway, pre-1923.
Iron(II) [Fe(H2O)6]2+
Sodium hydroxide solution precipitates dirty green iron(II) hydroxide; on standing the surface of the precipitate turns foxy-red owing to air oxidation to iron(III) hydroxide:
[Fe(H2O)6]2+(aq) + 2OH-(aq) " Fe(OH)2(s) + 6H2O(l)
Ammonia solution behaves similarly to sodium hydroxide – again there are no ammine complexes.
Potassium hexacyanoferrate(III) precipitates dark blue potassium iron(II) hexacyanoferrate(III), Turnbull’s Blue:
K+(aq) + Fe2+(aq) + [Fe(CN)6]3-(aq) " KFeII[FeIII(CN)6](s)
Although Turnbull's Blue and Prussian Blue don't have quite the same colour, they are in fact the same substance.
Potassium manganate(VII) oxidises iron(II) to iron(III) in acidic solution. The purple manganate colour is lost and the resulting solution of iron(III) is yellow:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) " Mn2+(aq) + 5Fe3+(aq) + 4H2O(aq)
Potassium dichromate(VI) oxidises iron(II) to iron(III) in acidic solution. The orange dichromate colour is lost and the resulting solution is green owing to presence of chromium(III) which masks the iron(III) colour:
Cr2O72-(aq) + 6Fe2+(aq) + 14H+(aq) " 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
Copper(II) [Cu(H2O)6]2+
Sodium hydroxide solution gives a pale blue precipitate usually described as copper(II) hydroxide:
[Cu(H2O)6]2+(aq) + 2OH-(aq) " Cu(OH)2(s) + 6H2O(l)
In fact this precipitate is the basic salt – thus from copper(II) sulphate solution the product is basic copper sulphate:
2[Cu(H2O)6]2+(aq) + 2OH-(aq) + SO42-(aq) " Cu(OH)2.CuSO4(s) + 12H2O(l)
Basic copper sulphate or basic copper carbonate is responsible for the green patina - verdigris - found on weathered copper roofs. The sulphate is present in industrial surroundings. To get true copper(II) hydroxide, the solution of the copper(II) salt must be added to excess sodium hydroxide; the precipitate is then blue rather than turquoise. In both cases CAUTIOUS warming of the mixture causes the precipitate to lose water and turn black as CuO(s) is formed. The loss of water by heating in an aqueous medium is unusual.
Ammonia solution initially gives a blue precipitate as for sodium hydroxide. Further addition of ammonia gives deep blue soluble cuprammine complexes whose composition depends on the amount of ammonia present. The reaction is usually represented:
Cu(OH)2(s) + 4NH3(aq) + 2H2O(l) " [Cu(NH3)4(H2O)2]2+(aq) +2OH-(aq)
The cuprammine solution has the bizarre property of being able to dissolve cellulose - bits of filter paper, for e example. The resulting viscous substance can be extruded into a bath of dilute sulphuric acid to give viscose fibre, which is a form of artificial silk.
Concentrated HCl or a saturated solution of sodium chloride gives a bright green solution containing CuCl42- ions; dilution with water causes the solution to turn pale blue as the [Cu(H2O)6]2+ ion re-forms.
[Cu(H2O)6]2+(aq) + 4Cl-(aq) " CuCl42-(aq) + 6H2O(l)
Sodium carbonate solution precipitates greenish-blue basic carbonates of indefinite composition.
Potassium iodide solution precipitates iodine and copper(I) iodide; this reaction is used volumetrically for the estimation of copper, the liberated iodine being titrated with sodium thiosulphate solution:
2[Cu(H2O)6]2+(aq) + 4I-(aq) " 2CuI(s) + I2(aq) + 6H2O(l)
The mixture turns a sludgy brown colour; addition of sodium thiosulphate solution leaves a creamy-pink precipitate of copper(I) iodide.
Zinc(II) [Zn(H2O)6]2+
Sodium hydroxide solution precipitates white zinc(II) hydroxide, easily soluble in excess sodium hydroxide to give sodium zincate(II), because the hydroxide is amphoteric:
[Zn(H2O)6]2+(aq) + 2OH-(aq) " Zn(OH)2(s) + 6H2O(l)
Zn(OH)2(s) + 2OH-(aq) " Zn(OH)42-(aq)
Ammonia solution initially precipitates zinc(II) hydroxide, as with sodium hydroxide. Excess ammonia causes the precipitate to disappear owing to the formation of ammine complexes – a different reason from that in 10.1:
Zn(OH)2(s) + 4NH3(aq) " [Zn(NH3)4]2+(aq) + 2OH-(aq)
Sodium carbonate solution precipitates a white basic carbonate:
5Zn2+(aq) + 2CO32-(aq) + 6OH-(aq) " 2ZnCO3.3Zn(OH)2(s)
This latter substance is what you get when you buy zinc carbonate – which is why on heating it gives off water vapour as well as carbon dioxide.
Aluminium Al[H2O] 3+
Sodium hydroxide solution precipitates white gelatinous aluminium hydroxide. This reacts with excess NaOH to give a colourless solution of sodium aluminate:
[Al(H2O)6] 3+(aq) + 3OH-(aq) " Al(OH)3(s) + 6H2O(l)
Al(OH)3(s) + 3OH-(aq) " [Al(OH)6]3-(aq)
Ammonia solution precipitates white aluminium hydroxide, as with NaOH; however it does not react further with excess ammonia.
Lead(II) Pb2+
Dilute hydrochloric acid or other soluble chlorides precipitate white lead(II) chloride from moderately concentrated solutions of lead(II) salts.
Pb2+(aq) + 2Cl-(aq) " PbCl2(s)
The precipitate dissolves in an excess of concentrated hydrochloric acid to give yellow solutions containing various species including PbCl3- and PbCl42-.
Sodium hydroxide solution gives a white precipitate of lead(II) hydroxide which is soluble in excess NaOH to give colourless sodium plumbate(II)
Pb2+(aq) + 2OH-(aq) " Pb(OH)2(s)
Pb(OH)2(s) + 2OH-(aq) " [Pb(OH)4]2-(aq)
Ammonia solution precipitates white lead(II) hydroxide; this does not dissolve in excess ammonia, because no complexes are formed and ammonia is not a strong enough base to bring out the amphoteric properties of lead(II) hydroxide.
Dilute sulphuric acid or other soluble sulphates give a white precipitate of lead(II) sulphate:
Pb2+(aq) + SO42-(aq) " PbSO4(s)
Potassium chromate(VI) solution gives a bright yellow precipitate of lead(II) chromate(VI):
Pb2+(aq) + CrO42-(aq) " PbCrO4(s)
Potassium iodide solution gives a bright yellow precipitate of lead(II) iodide. This will dissolve in boiling water to give a colourless solution – the colour comes from interactions in the crystal lattice rather than from coloured ions:
Pb2+(aq) + 2I-(aq) " PbI2(s)
Ammonium NH4+
Heat: all ammonium salts decompose on heating; ammonium nitrate may explode. Ammonium chloride and sulphate give products that recombine on cooling, so that the salts apparently sublime.
NH4Cl(s) " NH3(g) + HCl(g)
NH4NO3(s) " N2O(g) + 2H2O(g)
2(NH4)2SO4(s) D 2NH3(g) + SO3(g) + H2O(g)
Ammonium dichromate(VI) decomposes spectacularly on ignition in a reaction that is oxidation of the cation by the anion; the initially orange solid gives a fluffy green product of much larger volume:
(NH4)2Cr2O7(s) " N2(g) + Cr2O3(s) + 4H2O(g)
Alkalis (sodium hydroxide, calcium hydroxide) liberate ammonia from ammonium salts on warming with the solution, or even from a mixture of the solids. This is because OH- is a stronger base than ammonia, so removes a hydrogen ion from the ammonium ion:
NH4+(aq) + OH-(aq) " NH3(g) + H2O(l)
The test solution is warmed with sodium hydroxide solution and the vapours tested with moist red litmus paper. It is important to test the vapours immediately heating begins, since the ammonia is lost very quickly and by the time the solution boils it may well have all gone.
Cobalt(II) [Co(H2O)6]2+
The (+2) state is the most stable for simple cobalt salts; they are coloured pink or blue. The pink colour of the hydrated ion [Co(H2O)6]2+ changes to blue on heating, on dehydration, or in the presence of concentrated acids.
Sodium hydroxide solution precipitates a blue basic salt which on warming with excess sodium hydroxide forms solid pink cobalt(II) hydroxide. With a solution of cobalt(II) nitrate the equations are:
[Co(H2O)6]2+ (aq) + NO3 – (aq) + OH – (aq) "
Co(OH)NO3 (s) + 6H2O(l)
blue
Co(OH)NO3 (s) + OH – (aq) "
Co(OH)2 (s) + NO3 – (aq)
pink
Basic salts, usually hydroxides or carbonates, contain not only the expected hydroxide or carbonate ion but also the anion from the original salt. The precipitate from cobalt(II) nitrate solution is therefore different from that which would be given from cobalt(II) chloride. The latter would be Co(OH)Cl.
If the pink precipitate is boiled for some time (or is warmed with hydrogen peroxide solution) it is converted to brownish-black cobalt(III) hydroxide:
4Co(OH)2 (s) + 2H2O (l) + O2 (aq) " 4Co(OH)3 (s)
Ammonia solution gives a blue basic precipitate as with sodium hydroxide; excess ammonia converts this to a brown solution of a cobaltammine which turns red on exposure to air giving an ammine of cobalt(III). The overall reaction is:
Co(OH)NO3 (s) + 28NH3(aq) + 6H2O(l) + O2(aq) " 4[Co(NH3)6](OH)3(aq) + 4NH4NO3(aq)
Nickel(II) [Ni(H2O)6]2+
Sodium hydroxide solution gives a green gelatinous precipitate of nickel(II) hydroxide, insoluble in excess of the reagent:
[Ni(H2O)6]2+ (aq) + 2OH – (aq) " Ni(OH)2 (s) + 6H2O(l)
Ammonia solution gives a green precipitate of a basic salt which dissolves readily in excess ammonia solution to form a blue-violet solution of complex nickel ammines. With nickel(II) chloride the reactions are:
[Ni(H2O)6]2+(aq) + Cl – (aq) + NH3(aq) " Ni(OH)Cl(s) + NH4+(aq) + 5H2O(l)
Ni(OH)Cl(s) + 5NH3(aq) + H2O(l) " [Ni(NH3)4](OH)2(aq) + NH4+(aq) + Cl – (aq)
Ni(OH)Cl(s) + 7NH3(aq) + H2O(l) " [Ni(NH3)6](OH)2(aq) + NH4+(aq) + Cl – (aq)
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