Using the given information, match the solutions given column-I with corresponding pH in column-II
:
$p K_{b} (N H_{3}) = 5; l o g 3 = 0.48$
$p K_{a} (C H_{3} C O O H) = 5; l o g 2 = 0.30$

Column-I
Column-II
$P) 10 m L 0.1 M N H_{4} O H + 15 m L 0.05 M H C l$
3
$Q) 10 m L 0.1 M N H_{4} O H + 10 m L 0.1 M C H_{3} C O O H$
5
$R) 10 m L 0.1 M C H_{3} C O O H + 5 m L 0.1 M N a O H$
7
$S) 10 m L 0.2 M C H_{3} C O O N a^{+} + 10 m L 0.2 M H C l$
8.52

P - 4, Q - 3, R - 2, S - 1

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P - 2, Q - 1, R - 3, S - 4

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P - 1, Q - 2, R - 3, S - 4

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P - 4, Q - 2, R - 1, S - 3

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Solution

The correct option is B $P - 4, Q - 3, R - 2, S - 1$
(P)
$N H_{4} O H + H^{+} \to N H_{4}^{+}$
at t=0 $1$ $0.75$ $0$
at equili. $0.25$ $0$ $0.75$
Basic buffer : $p O H = p K_{b} + l o g \frac{0.75}{0.25}$
$p O H = 5 + 0.48$
$p H = 14 - 5.48 = 8.52$

(Q) Case of hydrolysis of weak acid and weak base salt.
$p H = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b}) = 7$

(R)
$C H_{3} C O O H + O H^{-} \to C H_{3} C O O^{-} + H_{2} O$
at t=0 $1$ $0.5$ $0$
at equili. $0.5$ $0$ $0.5$

(S)
$C H_{3} C O O^{-} + H^{+} \to C H_{3} C O O H$
at t=0 $2 m m o l$ $2 m m o l$ $0$
at equili. $0$ $0$ $2 m m o l$
Case of weak acid with concentration $\frac{2}{20} = 0.1 M$

$p H = \frac{1}{2} p K_{a} - \frac{1}{2} l o g C = \frac{1}{2} (5) - \frac{1}{2} (l o g \frac{1}{10}) = 3$

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Column-I	Column-II
$P) 10 m L 0.1 M N H_{4} O H + 15 m L 0.05 M H C l$	3
$Q) 10 m L 0.1 M N H_{4} O H + 10 m L 0.1 M C H_{3} C O O H$	5
$R) 10 m L 0.1 M C H_{3} C O O H + 5 m L 0.1 M N a O H$	7
$S) 10 m L 0.2 M C H_{3} C O O N a^{+} + 10 m L 0.2 M H C l$	8.52