1. A guide for A level students KNOCKHARDY PUBLISHING
pH calculations
A guide for A level students
2008 SPECIFICATIONS
KNOCKHARDY PUBLISHING

2. pH calculations www.knockhardy.org.uk/sci.htm INTRODUCTION
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3. pH calculations CONTENTS What is pH? - a reminder
Calculating the pH of strong acids and bases
Calculating the pH of weak acids
Calculating the pH of mixtures - strong acid and strong alkali

4. Before you start it would be helpful to…
pH calculations
Before you start it would be helpful to…
know the differences between strong and weak acid and bases
be able to calculate pH from hydrogen ion concentration
be able to calculate hydrogen ion concentration from pH
know the formula for the ionic product of water and its value at 25°C

5. where [H+] is the concentration of hydrogen ions in mol dm-3
What is pH?
pH = - log10 [H+(aq)]
where [H+] is the concentration of hydrogen ions in mol dm-3
to convert pH into
hydrogen ion concentration [H+(aq)] = antilog (-pH)
IONIC PRODUCT OF WATER Kw = [H+(aq)] [OH¯(aq)] mol2 dm-6
= 1 x mol2 dm-6 (at 25°C)

6. Calculating pH - strong acids and alkalis
WORKED
EXAMPLE
Calculating pH - strong acids and alkalis
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentration.
Calculate the pH of 0.02M HCl
HCl completely dissociates in aqueous solution HCl H+ + Cl¯
One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 mol dm-3
pH = - log [H+] = 1.7

7. Calculating pH - strong acids and alkalis
WORKED
EXAMPLE
Calculating pH - strong acids and alkalis
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentration.
Calculate the pH of 0.02M HCl
HCl completely dissociates in aqueous solution HCl H+ + Cl¯
One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 mol dm-3
pH = - log [H+] = 1.7
Calculate the pH of 0.1M NaOH
NaOH completely dissociates in aqueous solution NaOH Na+ + OH¯
One OH¯ is produced for each NaOH dissociating [OH¯] = 0.1M = 1 x 10-1 mol dm-3
The ionic product of water (at 25°C) Kw = [H+][OH¯] = 1 x mol2 dm-6
therefore [H+] = Kw / [OH¯] = 1 x mol dm-3
pH = - log [H+] = 13

8. Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows HA(aq) H+(aq) A¯(aq) (1)

9. Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows HA(aq) H+(aq) A¯(aq) (1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]

10. Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows HA(aq) H+(aq) A¯(aq) (1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)

11. Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows HA(aq) H+(aq) A¯(aq) (1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
Rearranging (3) gives [H+(aq)]2 = [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka

12. Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows HA(aq) H+(aq) A¯(aq) (1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
Rearranging (3) gives [H+(aq)]2 = [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka
pH = [H+(aq)]

13. Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows HA(aq) H+(aq) A¯(aq) (1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
Rearranging (3) gives [H+(aq)]2 = [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka
pH = [H+(aq)]
ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that
its equilibrium concentration is approximately that of the original concentration.

14. Calculating pH - weak acids
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) X¯(aq)

15. Calculating pH - weak acids
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) X¯(aq)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]

16. Calculating pH - weak acids
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) X¯(aq)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and then rearrange equation

17. Calculating pH - weak acids
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) X¯(aq)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and the rearrange equation
ASSUMPTION
HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration

18. Calculating pH - weak acids
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows HX(aq) H+(aq) X¯(aq)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and the rearrange equation
ASSUMPTION
HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration
[H+(aq)] = x 4 x mol dm-3
= x mol dm-3
= x mol dm-3
ANSWER pH = - log [H+(aq)] =

19. CALCULATING THE pH OF MIXTURES
The method used to calculate the pH of a mixture of an acid and an alkali depends on...
whether the acids and alkalis are STRONG or WEAK
which substance is present in excess
STRONG ACID and STRONG BASE - EITHER IN EXCESS

20. Strong acids and strong alkalis (either in excess)
pH of mixtures
Strong acids and strong alkalis (either in excess)
1. Calculate the initial number of moles of H+ and OH¯ ions in the solutions
2. As H+ and OH¯ ions react in a 1:1 ratio; calculate unreacted moles species in excess
3. Calculate the volume of solution by adding the two original volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess the ion in mol dm-3
6. Convert concentration to pH
If the excess is H+ pH = - log[H+]
If the excess is OH¯ pOH = - log[OH¯] then
pH + pOH = 14
or use Kw = [H+] [OH¯] = 1 x at 25°C therefore
[H+] = Kw / [OH¯] then
pH = - log[H+]

21. Strong acids and alkalis (either in excess)
pH of mixtures
Strong acids and alkalis (either in excess)
WORKED
EXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl

22. Strong acids and alkalis (either in excess)
pH of mixtures
Strong acids and alkalis (either in excess)
WORKED
EXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
25cm3 of
0.1M NaOH
20cm3 of
0.1M HCl
2.5 x 10-3 moles
2.0 x 10-3 moles
moles of OH ¯
= 0.1 x 25/1000
= 2.5 x 10-3
moles of H+
= 20 x 20/1000
= 2.0 x 10-3

23. Strong acids and alkalis (either in excess)
pH of mixtures
Strong acids and alkalis (either in excess)
WORKED
EXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
25cm3 of
0.1M NaOH
20cm3 of
0.1M HCl
2.5 x 10-3 moles
2.0 x 10-3 moles
The reaction taking place is… HCl NaOH NaCl H2O
or in its ionic form H OH¯ H2O (1:1 molar ratio)

24. Strong acids and alkalis (either in excess)
pH of mixtures
Strong acids and alkalis (either in excess)
WORKED
EXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
5.0 x 10-4
moles of OH¯
UNREACTED
25cm3 of
0.1M NaOH
20cm3 of
0.1M HCl
2.5 x 10-3 moles
2.0 x 10-3 moles
The reaction taking place is… HCl NaOH NaCl H2O
or in its ionic form H OH¯ H2O (1:1 molar ratio)
2.0 x 10-3 moles of H+ will react with the same number of moles of OH¯
this leaves x x = x 10-4 moles of OH¯ in excess

25. Strong acids and alkalis (either in excess)
pH of mixtures
Strong acids and alkalis (either in excess)
WORKED
EXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
the volume of the solution is = 45cm3

26. Strong acids and alkalis (either in excess)
pH of mixtures
Strong acids and alkalis (either in excess)
WORKED
EXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
the volume of the solution is = 45cm3
there are 1000 cm3 in 1 dm3
volume = 45/ = dm3

27. Strong acids and alkalis (either in excess)
pH of mixtures
Strong acids and alkalis (either in excess)
WORKED
EXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess ion in mol dm-3
[OH¯] = x 10-4 / = x mol dm-3

28. Strong acids and alkalis (either in excess)
pH of mixtures
Strong acids and alkalis (either in excess)
WORKED
EXAMPLE
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess ion in mol dm-3
6. As the excess is OH¯ use pOH = - log[OH¯] then pH + pOH = 14
or Kw = [H+][OH¯] so [H+] = Kw / [OH¯]
[OH¯] = x 10-4 / = x mol dm-3
[H+] = Kw / [OH¯] = x mol dm-3
pH = - log[H+] =
Kw = 1 x mol2 dm-6 (at 25°C)

29. What should you be able to do?
REVISION CHECK
What should you be able to do?
Calculate pH from hydrogen ion concentration
Calculate hydrogen ion concentration from pH
Write equations to show the ionisation in strong and weak acids
Calculate the pH of strong acids and bases knowing their molar concentration
Calculate the pH of weak acids knowing their Ka and molar concentration
Calculate the pH of mixtures of strong acids and strong bases
CAN YOU DO ALL OF THESE? YES NO

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pH calculations
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