Calculate the pH of 0.0100 mol dm^–3 methanoic acid stating any assumption you make. K_a = 1.6 × 10^–4.

(i) Sketch a graph of pH against volume of a strong base added to a weak acid showing how you would determine pK_a for the weak acid.

(ii) Explain, using an equation, why the pH increases very little in the buffer region when a small amount of alkali is added.

Calculation:

ALTERNATIVE 1:
[H⁺] = (K_a × [HA])^1/2 / (1.6 × 10^–4 × 0.0100)^1/2 / 1.3 × 10^–3 «mol dm^–3»

pH = «–log₁₀[H⁺] ≈» 2.9

ALTERNATIVE 2:
pH = 0.5(pK_a - log₁₀[HA])
pH = 2.9

Award [2] for correct final answer

Assumption:
ionisation is << 0.0100 so 0.0100 - [A^–] ≈ 0.0100
OR
[HA]_eqm = [HA]_initial
OR
all H⁺ ions in the solution come from the acid «and not from the self-ionisation of water»
OR
[H⁺] = [HCOO^–]

Do not accept partial dissociation

i

correct shape of graph
pH at half neutralization/equivalence

M1: must show buffer region at pH < 7 and equivalence at pH > 7.
Accept graph starting from where two axes meet as pH scale is not specified.

ii

ALTERNATIVE 1:

HCOOH  HCOO^– + H⁺
H⁺ ions consumed in reaction with OH^– are produced again as equilibrium moves to the right «so [H⁺] remains almost unchanged»

ALTERNATIVE 2:
HCOOH + OH^–  HCOO^– + H₂O
added OH^- are neutralized by HCOOH
OR
strong base replaced by weak base 

Accept HA or any other weak acid in equations.
Equilibrium sign must be included in equation for M1