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Question

Calculate the hydrolysis constant of analinium acetate if $K_{a} = 1 \times 10^{- 5}$ and $K_{b} = 1 \times 10^{10}$ .

1.24

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10

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0.1

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0.01

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Solution

The correct option is C $0.1$
$H = 8.26$
$O H = 14 - p_{H}$
$= 14 - 8.26$
$= 5.74$
$k_{p} = 14 - 9.26$
$= 4.74$
$p O H = p k_{p} + log \frac{[N H_{4}^{+}]}{[N H_{4} O H]}$
$5.74 = 4.74 + log \frac{[{N H}_{4}^{+}]}{0.01}$
$1 = log [{N H}_{4}^{+}] - log (0.01)$
$1 = log [{N H}_{4}^{+}] + 2$
$log [{N H}_{4}^{+}] = - 1$
Taing antilog $[{N H}_{4}^{+}] = 0.1$

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