Equilibrium Question 803
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:
8
10
4
5
Show Answer
Answer:
Correct Answer: C
Solution:
Hydrolysis of a salt is the reverse reaction of an acid-base neutralization reaction. $ \therefore K_{h}=\frac{K_{w}}{K_{a}}=\frac{{10^{-14}}}{{10^{-7}}}={10^{-7}} $
$ [OH^{-}]=ch=c\times \sqrt{\frac{K_{h}}{c}}=\sqrt{c\times K_{h}}=\sqrt{{10^{-8}}}={10^{-4}} $
$ \Rightarrow pOH=-\log[ OH^{-} ]=-\log [{10^{-4}}]=4 $
BETA
