Equilibrium Question 288
Question: In the equilibrium $ {A^{-}}+H_2O $ ⇌ $ HA+O{H^{-}} $
$ (K_{a}=1.0\times {10^{-5}}) $ . The degree of hydrolysis of 0.001 M solution of the salt is [AMU 1999]
Options:
A) $ {10^{-3}} $
B) $ {10^{-4}} $
C) $ {10^{-5}} $
D) $ {10^{-6}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ K_{a}=1.0\times {10^{-5}} $
$ K_{h}=hydrolysisconstant $
$ K_{h}=\frac{K_{w}}{K_{a}}=\frac{{10^{-14}}}{{10^{-5}}}={10^{-9}} $ degree of hydrolysis (h ) = $ \sqrt{\frac{K_{h}}{C}} $
$ =\sqrt{\frac{{10^{-9}}}{0.001}} $ = $ \sqrt{{10^{-6}}} $ = $ {10^{-3}} $ ; $ h={10^{-3}} $