Question

Degree of dissociation of ${NH}_{4}OH$ in water is $1.8\times {10}^{-5}$, then hydrolysis constant of ${NH}_{4}OH$ is:

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

Answer 1

Answer: (D)

$5.55\times {10}^{-10}$

Ionic product of water is $K_w={10}^{-14}$

Degree of dissocation of $N{H}_{4}OH$ is $K_b=1.8\times {10}^{-5}$

Hydrolysis constant of $N{H}_{4}OH$ is given as $K_h=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}=5.55\times {10}^{-10}$