What happens when 0.02 mole NaOH is added to a buffer solution?

Question

I am asking this question for my son.

(This question is related but we still have our confusions.)

We have one liter buffer solution of 4.74 pH which contains 0.1 mole $\ce{CH_3COOH}$ and 0.1 mole $\ce{CH_3COONa}$.

What will be the resulting pH when 0.02 mole $\ce{NaOH}$ is added to this solution? (Given $\ce{Ka} = 1.8 \times 10^{-5}$.)

As the definition of a buffer solution [A, B] goes, pH value should not change when small quantity of acid or base is added to it.

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So, one way of answering the question will be to simply comment that since a small quantity of base has been added, the pH value will not change.

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But, we find another solution (presented and accepted in some moderately acceptable study-sheet) in the following line.

We will have a reaction like,

$\ce{CH_3COOH} + \ce{NaOH} = \ce{CH_3COONa} + \ce{H_2O}$

So, after the reaction, we have,

$\ce{CH_3COONa}$: $(0.1 + 0.02)\ \text{mole} = 0.12\ \text{mole}$

$\ce{CH_3COOH}$: $(0.1 - 0.02)\ \text{mole} = 0.08\ \text{mole}$

Now, $pH = pKa + \log{\text{[Salt]}\over\text{[Acid]}}$

Hence, $pH = pKa + \log{{{n_\text{Salt}}\over V} \over {{n_\text{Acid}}\over V}}$

Which after plugging in the values gives, $pH = 4.92$.

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We are confused about which of the above solutions is correct.

Answer 1

The second solution using the Henderson-Hasselbach equation is correct. While buffer solutions change pH very little near the pka of their constituent weak acid, it would not be correct to say that they do not change at all. Also, an addition of 20 mmol of NaOH is not insignificant to a 100 mM buffer—a 20% change in concentration. The Henderson-Hasselback equation in the form you've shown is an acceptable estimation of the buffer pH in a moderately concentrated buffer with an acid/base ratio close to 1. Deviations from it in that form can occur for very dilute or concentrated solutions or if it's well away from the pka.