Equilibrium Question 838
Question: If the equilibrium constant of the reaction of weak acid HA with strong base is $ {10^{-7}} $ , then pOH of the aqueous solution of $ 0.1MNaA $ is
Options:
A) 8
B) 10
C) 4
D) 5
Show Answer
Answer:
Correct Answer: C
Solution:
Hydrolysis of a salt is reverse reaction of acid-base neutralization reaction.
$ \therefore K_{h}=\frac{K_{w}}{K_{a}}=\frac{{10^{-14}}}{{10^{-7}}}={10^{-7}} $
$ [O{H^{-}}]=ch=c\times \sqrt{\frac{K_{h}}{c}}=\sqrt{c\times K_{h}}=\sqrt{{10^{-8}}}={10^{-4}} $
$ \Rightarrow pO{H^{-}}=-log[ O{H^{-}} ]=-\log [{10^{-4}}]=4 $
