Which of the following solutions will be neutral?

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50 ml of 0.1 $M C H_{3} C O O H$ + 50 ml of 0.1 $M N a O H$

100 ml of 0.1 $M C H_{3} C O O H$ + 50 ml of 0.2 $M N H_{3}$

100 ml of 0.1 M HCl + 50 ml of 0.2 M KOH

50 ml of 0.1 M HCl + 50 ml of 0.1 $M N H_{3}$

answer is B, C.

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Detailed Solution

(A) $50 \times 0.1 = 5 m$ mol of $C H_{3} C O O H$ will neutralize completely $50 \times 0.1 = 5 m$ mol of NaOH to form $C H_{3} C O O N a$ and water. But acetate ion (being strong conjugate base of the weak acid $C H_{3} C O O H$ ) of the salt hydrolyses to give basic solution. $C H_{3} C O O^{-} + H_{2} O \overset{}{⇌} C H_{3} C O O + O H^{-} (p H > 7)$

(B) 100 x 0.1 =10 m mol of $C H_{3} C O O H$ and $50 \times 0.2 = 10 m$ mol of $N H_{3}$ will neutralize each other completely to give $C H_{3} C O O N H_{4} .$ Both anion and cation of the salt hydrolyse equally $(K_{a} = 1.8 \times 10^{5}; K_{b} = 1.8 \times 10^{- 5})$ to give neutral solution. $C H_{3} C O O^{-} + N H_{4}^{+} + H_{2} O \to_{}^{} C H_{3} C O O H + N H_{4} O H (p H = 7)$

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

(D) $50 \times 0.1 = 5 m$ mol of $H C l$ and $50 \times 0.1 = 5 m$ mol of $N H_{3}$ neutralize one another completely. However, the cation $(N H_{4}^{+} i o n)$ of the salt hydrolyses to give acidic solution. $N H_{4}^{+} + H_{2} O \to_{}^{} N H_{3} + H_{3} O^{+} (p H < 7)$