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).
Tuesday, April 21, 2009
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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