Which of the following solutions will be neutral?

(A) $50\times 0.1=5m$ mol of $C{H}_{3}COOH$ will neutralize completely $50\times 0.1=5m$ mol of NaOH  to form $C{H}_{3}COONa$ and water. But acetate ion (being strong conjugate base of the weak acid $C{H}_{3}COOH$ ) of the salt hydrolyses to give basic solution. $C{H}_{3}CO{O}^{-}+H_{2}O\stackrel{}{\rightleftharpoons }C{H}_{3}COO+O{H}^{-}\left(pH>7\right)$

(B) 100 x 0.1 =10 m mol of $C{H}_{3}COOH$ and $50\times 0.2=10m$ mol of $N{H}_3$ will neutralize each other completely to give $C{H}_{3}COON{H}_4.$ Both anion and cation of the salt hydrolyse equally $\left(K_a=1.8\times {10}^5;K_b=1.8\times {10}^{-5}\right)$ to give neutral solution. $C{H}_{3}CO{O}^{-}+N{H}_4^{+}+H_{2}O\underset{}{\overset{}{\to }}C{H}_{3}COOH+N{H}_{4}OH\left(pH=7\right)$

(C) $100\times 0.1=10m$ mol of HCl and $5.0\times 0.2=10m$ mol of KOH neutralize each other completely to give KCl. Neither $K^{+}$ ion (weak acid) nor $Cl$ ion (weak base) hydrolyse ($pH=7$ )

(D) $50\times 0.1=5m$ mol of $HCl$ and $50\times 0.1=5m$ mol of $N{H}_3$ neutralize one another completely. However, the cation $\left(N{H}_4^{+}ion\right)$ of the salt hydrolyses to give acidic solution. $N{H}_4^{+}+H_{2}O\underset{}{\overset{}{\to }}N{H}_3+H_3{O}^{+}\left(pH<7\right)$