Question

Degree of dissociation of $ N{{H}_{4}}OH $ in water is $ 1.8\times {{10}^{-5}}, $ then hydrolysis constant of $ N{{H}_{4}}Cl $ 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)

$K_{h}=\frac{K_{w}}{K_{b}}$
where $K_{w}=$ ionic product of water $=1 \times 10^{-14}$
$K_{b}=$ degree of dissociation of $NH _{4} OH$ $=1.8 \times 10^{-5}$
$K_{h}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}$
$=0.555 \times 10^{-9}$
$=5.55 \times 10^{-10}$