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Q. At $25^{\circ} C$, the ionization constant for anilinium hydroxide is $5.00 \times 10^{-10}$. Thehydrolysis constant of anilinium chloride is

TS EAMCET 2020

Solution:

Given, $K_{b}$ (ammonium hydroxide)
$=5.00 \times 10^{-10} $
$K_{w}=10^{-14} $ at $ 298\, K$
$K_{h}=\frac{K_{W}}{K_{b}}=\frac{10^{-14}}{5 \times 10^{-10}}$
$=2 \times 10^{-5}$