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Column I(solutions) Column II(Characteristics)
A $\begin{array}{l} 0.1 M C H_{3} C O O H \\ (p K_{a} = 4.74) + 0.1 M \\ C H_{3} C O O K \end{array}$ p Acidic buffer at its maximum capacity
B $0.1 M C H_{3} C O O H + 0.1 M H C l$ q Buffer solution
C $\begin{array}{l} 0.1 M C H_{3} C O O H \\ (p K_{a} = 4.74) + 0.1 M \\ N H_{4} O H (p K_{b} = 4.74) \end{array}$ r $p H < 7 a t 25^{0} C$
D
300 mL of 0.1 M
$C H_{3} C O O N a + 100 m L o f 0.05 M H_{2} S O_{4}$
s $p H = 7 a t 25^{0} C$

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A-s B-p C-r D-q

A-q B-r C-s D-r

A-s B-p C-q D-r

A-p B-r C-s D-q

answer is A.

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Detailed Solution

$\begin{array}{l} (a \to p) p H = p K_{a} + \log (\frac{s a l t}{A c i d}) \\ = 4.74 + \log (\frac{0.1}{0.1}) = 4.74 \end{array}$
$(b \to r)$ It is mixture of $W_{A}$ and $S_{A} .$ So pH of 0.01 MHCI will predominate and $p H < 7$
$(c \to s)$ It will forms salts of $W_{A} / W_{B} .$ Since $p K_{a} = p K_{b},$ Thus $p H = 7. (p H = \frac{1}{2} (p K_{w} + p K_{a} + p K_{b}))$
$(d \to q)$ m. mol of $C H_{3} C O O N a = 300 \times 0.1 = 30$
m. mol of $H_{2} S O_{4} = 100 \times 0.05 \times 2 = 10$
$\begin{array}{l} 2 C H_{3} C O O N a + H_{2} S O_{4} \to 2 C H_{3} C O O H + N a_{2} S O_{4} \\ I n i t i a l 30 m . m o l 10 m . m o l 0 0 \\ F I n a l (30 - 10) = 20 (10 - 10) = 0 10 - \end{array}$
$\begin{array}{l} P H = p K_{a} + \log (\frac{s a l t}{A c i d}) \\ = 4.74 + \log [\frac{20 / 400}{10 / 400}] \\ = 4.74 + 0.3 = 5.03 \end{array}$

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