Q. If at 25$℃$ the ionization constant of acetic acid is $2 \times 10^{- 5}$, the hydrolysis constant of sodium acetate will be
A) $5 \times 10^{- 8}$
B) $5 \times 10^{- 9}$
C) $5 \times 10^{- 10}$
D) $4 \times 10^{- 10}$
Explanation
(c)
Hydrolysis constant of sodium acetate
$K_{h} = \frac{K_{w}}{K_{a}}$
$\because$ $K_{w} = 10^{- 14}\ and\ \ K_{a} = 2 \times 10^{- 5}(given)$
$\therefore$ $K_{h} = \frac{10^{- 14}}{2 \times 10^{- 5}} = \frac{10}{2} \times 10^{- 10}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5 \times 10^{- 10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.