pH of a buffer solution decreases by $0.02$ units when $0.12 \mathrm{g}$ of acetic acid is added to $250\mathrm{ml}$ of a buffer solution of acetic acid and potassium acetate at $27°\mathrm{C}$. The buffer capacity of the solution is?

To $250\mathrm{ml}$ of Buffer solution $0.12 \mathrm{g}$ of acetic acid is added.

To $1000\mathrm{ml}$ of Buffer solution $\frac{0.12\times 1000}{250}=0.48 \mathrm{g}$ of acetic acid is added.

Now number of moles of acetic acid added will be:

$\frac{0.48}{60}=8\times {10}^{-3}\mathrm{moles}$

We know that Buffer capacity =$\frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{acid} \mathrm{added} \mathrm{to} 1\mathrm{L} \mathrm{solution}}{\mathrm{Change} \mathrm{in} \mathrm{pH}}=0.48 \mathrm{g}$

$=\frac{8\times {10}^{-3}}{0.02}=0.4$

Therefore, the correct option is (D).