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(D) The $pH$ of an acidic buffer solution is given by the Henderson-Hasselbalch equation:$pH = pK_{a} + \log \frac{[salt]}{[acid]}$For $pH = pK_{a}$,

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Both $(1)$ and $(2)$ are correct

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(D) The $pH$ of an acidic buffer solution is given by the Henderson-Hasselbalch equation:$pH = pK_{a} + \log \frac{[salt]}{[acid]}$For $pH = pK_{a}$, we must have $\log \frac{[salt]}{[acid]} = 0$, which implies $[salt] = [acid]$.$(1)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $100 \ mL$ of $0.1 \ M \ CH_{3}COONa$ $(10 \ mmol)$.Here, $[salt] = [acid] = 10 \ mmol$. Thus, $pH = pK_{a} + \log(1) = pK_{a}$.$(2)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $50 \ mL$ of $0.1 \ M \ NaOH$ $(5 \ mmol)$.The reaction is: $CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$.Initial: $10 \ mmol \ CH_{3}COOH, 5 \ mmol \ NaOH$.Final: $5 \ mmol \ CH_{3}COOH, 5 \ mmol \ CH_{3}COONa$.Since $[salt] = [acid] = 5 \ mmol$, $pH = pK_{a} + \log(1) = pK_{a}$.$(3)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $100 \ mL$ of $0.1 \ M \ NaOH$ $(10 \ mmol)$.This results in complete neutralization to form $CH_{3}COONa$. No buffer is formed.$(4)$ $CH_{3}COOH + NH_{3} \rightarrow CH_{3}COONH_{4}$.This forms a salt of a weak acid and a weak base, not a buffer solution.Therefore, both $(1)$ and $(2)$ satisfy the condition $pH = pK_{a}$.

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