1. Complete table
Calculate Kc with units if any
A + B = 2C
component A B C
Initial amount (mol)
0.4
0.2
Equilibrium amount (mol)
0.25

2. pH for strong and weak acids
L.O.:
Calculate pH of strong and weak acids.

3. 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)

4. 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
Leave out

5. 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
Leave out

6. 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)

7. 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)]

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)
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)

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)]
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
Leave out

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)
Rearranging (3) gives [H+(aq)]2 = [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka
pH = [H+(aq)]

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
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.

12. 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)

13. 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)]

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)
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

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)]
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

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 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)] =

17. 6. What is the pH of the following weak acids?
a) Ka = 3.7 × 10-8 mol dm-3 [HA] = 0.01 mol dm-3
b) Ka = 5.8 × mol dm-3 [HA] = 0.01 mol dm-3
c) Ka = 5.6 × 10-4 mol dm-3 [HA] = 0.01 mol dm-3
d) Ka = 1.74 × 10-5 mol dm-3 [HA] = 0.10 mol dm-3
e) Ka = 5.62 × 10-4 mol dm-3 [HA] = 0.20 mol dm-3
6. a) 4.72 b) 5.62 c) 2.68 d) 2.88 e) 1.97
From chemfact 25

18. 0.1 mol dm-3 solution of HA has a pH of 5.10. What is its Ka value?

20. Exercise

21. Finding the pH of weak acids
What is a “weak acid”?
What is a “weak base”?
Calculate the pH of a weak acid.
4.3 pH questions

22. whether the acids and alkalis are STRONG or WEAK
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