Q. Degree of dissociation of $NH_{4}OH$ in water is $1.8 \times 10^{- 5}$, then hydrolysis constant of $NH_{4}Cl$ is
A) $1.8 \times 10^{- 5}$
B) $1.8 \times 10^{- 10}$
C) $5.55 \times 10^{- 5}$
D) $5.55 \times 10^{- 10}$
Explanation
(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$\ NH_{4}OH = 1.8 \times 10^{- 5}$
$K_{h} = \frac{1 \times 10^{- 14}}{1.8 \times 10^{- 5}}$
$= 0.555 \times 10^{- 9}$
$= 5.55 \times 10^{- 10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.