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

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.

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.

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, $pH = 4.92$.

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

• pH does change while buffering – canadianer Sep 6 '14 at 5:13