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 $



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