Question

Dissociation constant of $N{H}_{4}OH$ is $1.8\times {10}^{-5}$. The hydrolysis constant of $N{H}_{4}Cl$ would be:

• A. $1.80\times {10}^{-19}$
• B. $5.55\times {10}^{-10}$
• C. $5.55\times {10}^{-9}$
• D. $1.8\times {10}^{-5}$

Answer 1

The correct option is B $5.55\times {10}^{-10}$

${NH}_{4}OH\rightleftharpoons {NH}_4^{+}+O{H}^{-}$

$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{N{H}_{4}OH}$

$N{H}_{4}Cl+H_{2}O\to HCl+N{H}_{4}OH$

$K_h=\frac{\left[HCl\right]\left[N{H}_{4}OH\right]}{N{H}_{4}Cl}$

$K_w=K_b\times K_h$

$K_h=\frac{K_w}{K_b}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}}=5.55\times {10}^{-10}$

Hence, option C is correct.