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The ionization constant of ammonium hydroxide is $1.77\times10^{-5}$ at 298 K. Hydrolysis constant of ammonium chloride is
A
$5.65\times10^{-12}$
B
$5.65\times10^{-10}$
C
$6.50\times10^{-12}$
D
$5.65\times10^{-13}$
Detailed Solution
$NH_4Cl$ is a salt of a weak base ($NH_4OH$) and a strong acid (HCl); the cation hydrolyses: $NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$
For such a salt: $K_h = \frac{K_w}{K_b}$
$K_h = \frac{1.0\times10^{-14}}{1.77\times10^{-5}}$
$K_h = 5.65\times10^{-10}$
For such a salt: $K_h = \frac{K_w}{K_b}$
$K_h = \frac{1.0\times10^{-14}}{1.77\times10^{-5}}$
$K_h = 5.65\times10^{-10}$