20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH...

For complete neutralization of $20$ mL of acetic acid, the millimoles of NaOH required is:

$28.4 \times 0.1 = 2.84$ mmol

Thus, $20$ mL of the acetic acid solution contains $2.84$ mmol of $CH_3COOH$.

For the preparation of solution (X), the millimoles of NaOH added is:

$14.2 \times 0.1 = 1.42$ mmol

The reaction between acetic acid and sodium hydroxide is:

$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Millimoles of $CH_3COOH$ remaining after the reaction = $2.84 - 1.42 = 1.42$ mmol

Millimoles of $CH_3COONa$ formed = $1.42$ mmol

Since the solution contains a weak acid and its conjugate base, it forms an acidic buffer. Using the Henderson-Hasselbalch equation:

$pH = pK_a + \log \left( \dfrac{[\text{Salt}]}{[\text{Acid}]} \right)$

Substituting the values:

$pH = 4.75 + \log \left( \dfrac{1.42}{1.42} \right)$

$pH = 4.75 + \log(1)$

$pH = 4.75$

Answer: $4.75$