20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (pK$$_a$$ value of acetic acid is 4.75).

The unknown molarity of the acetic-acid stock solution is calculated first.

Neutralisation data: $$28.4\text{ mL} = 0.0284\text{ L}$$ of $$0.1\text{ M }NaOH$$ is required for $$20\text{ mL} = 0.020\text{ L}$$ of the acid.

Moles of $$NaOH$$ used $$= 0.0284 \times 0.1 = 0.00284\text{ mol}$$.

Each mole of $$NaOH$$ reacts with one mole of acetic acid $$\bigl(CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O\bigr)$$, so moles of acetic acid present in $$20\text{ mL}$$ of the stock solution are also $$0.00284\text{ mol}$$.

Molarity of the acetic-acid stock solution: $$\displaystyle C_{acid} = \frac{0.00284}{0.020} = 0.142\text{ M}$$.

Solution X preparation: $$20\text{ mL}$$ of this acid is mixed with $$14.2\text{ mL} = 0.0142\text{ L}$$ of $$0.1\text{ M }NaOH$$.

Moles of reagents placed in the mixture:
Acetic acid: still $$0.00284\text{ mol}$$ (because the same $$20\text{ mL}$$ is taken).
$$NaOH$$: $$0.0142 \times 0.1 = 0.00142\text{ mol}$$.

The neutralisation reaction consumes $$NaOH$$ completely (it is the limiting reagent):

Remaining moles of acetic acid $$= 0.00284 - 0.00142 = 0.00142\text{ mol}$$.
Moles of acetate ion formed $$= 0.00142\text{ mol}$$.

Thus in solution X, $$[CH_3COOH]$$ and $$[CH_3COO^-]$$ are equal, giving a buffer with $$\displaystyle \frac{[A^-]}{[HA]} = 1$$.

Using the Henderson-Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_a + \log\!\left(\frac{[A^-]}{[HA]}\right) = 4.75 + \log(1) = 4.75$$.

Total volume (34.2 mL) cancels out in the ratio, so no further adjustment is needed.

Hence, the pH of solution X is $$4.75$$.

Option B which is: $$4.75$$