pH OF BUFFER SOLUTION

Ethanoic acid, CH3COOH, has a Ka of 1.7 × 10–5 mol dm–3.
It can form a buffer solution when mixed with NaOH.
A student mixes equal amounts of solutions of CH3COOH and NaOH.Which mixture will give a buffer with pH = 4.77?

A 0.1 [CH3COOH] / mol dm–3
0.2 [NaOH] / mol dm–3
This would not form a buffer as the base is in excess – the pH would be >7.

B 0.1 [CH3COOH] / mol dm–3
0.1 [NaOH] / mol dm–3
This would not form a buffer as the weak acid would be neutralized. The pH would be 7.

C 0.2 [CH3COOH] / mol dm–3
0.1 [NaOH] / mol dm–3
Correct answer: –log(1.7 × 10–5 × (0.2 – 0.1)/0.1)

D 0.3 [CH3COOH] / mol dm–3
0.1[NaOH] / mol dm–3
Incorrect, this mixture would give a pH of 4.46.


I have supplied the mark scheme answers. I understand why the A and B are incorrect but my question is why did they subtract the concentration of NaOH from CH3COOH when working out the pH in option C? Isn't it supposed to be (concentration of weak acid/ concentration of salt) x Ka?