The hydrolysis constant of aniline hydrochloride in $M / 32$ solution of salt at 298 K from the following cell data at 298 K is:
$P t | H_{2} (1 a t m) | H^{+} (1 M) | | M / 32 C_{6} H_{6} N H_{3} C l | H_{2} (1 a t m) | P t; E_{c e l l} = - 0.188 V$
$h = 2.12 \times 10^{- 2}, K_{h} = 1.43 \times 10^{- 4} M$

4.2 \times 10^{- 2}

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2.1 \times 10^{- 2}

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4.1 \times 10^{- 2}

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2.6 \times 10^{- 3}

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Solution

The correct option is B $2.1 \times 10^{- 2}$
$E_{c e l l} = - 0.0591 log \frac{[H^{+}]_{A n o d e}}{[H^{+}]_{C a t h o d e}}$
$- 0.188 = - 0.0591 \frac{1}{[H^{+}]_{C a t h o d e}}$
$p H = 3.18 = 7 - \frac{1}{2} P^{K b} - \frac{1}{2} log C$
$3.18 = 7 - \frac{1}{2} P^{b} - \frac{1}{2} log (1 / 32)$
$k_{b} = 7.15 \times 10^{- 10}$
$k_{b} = \frac{k_{w}}{k_{w}} = \frac{10^{- 14}}{7.15 \times 10^{- 10}} = 1.39 \times 10^{- 5}$
$h = \sqrt{\frac{k_{h}}{C}} = \sqrt{\frac{1.39 \times 10^{- 5}}{1 / 32}} = 2.1 \times 10^{- 2}$

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P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (s) E_{c e l l} = 0.092 V w h e n \frac{[M^{2 +} (a q)]}{[M^{4 +} (a q)]} = 10^{x} G i v e n : E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

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