| Date | November 2016 | Marks available | 3 | Reference code | 16N.2.hl.TZ0.7 |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Calculate | Question number | 7 | Adapted from | N/A |
Question
This question is about the weak acid methanoic acid, HCOOH.
Calculate the pH of 0.0100 mol dm–3 methanoic acid stating any assumption you make. Ka = 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 pKa 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.
Markscheme
Calculation:
ALTERNATIVE 1:
[H+] = (Ka × [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(pKa - 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
