Equilibrium Question 745
Question: pH for the solution of salt undergoing anionic hydrolysis (say $ CH_3COONa $ ) is given by:
Options:
A) $ pH=1/2[ pK_{w}+pK_{a}+\log C ] $
B) $ pH=1/2[ pK_{w}+pK_{a}-\log C ] $
C) $ pH=1/2[ pK_{w}+pK_{b}-\log C ] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ CH_3CO{O^{-}}+H_2O\rightarrow CH_3COOH+O{H^{-}} $
$ \therefore [O{H^{-}}]=C.h=C\sqrt{\frac{K_{H}}{C}}=\sqrt{K_{H}.C}=\sqrt{\frac{K_{w}}{K_{a}}C} $ Or $ -\log OH=-\frac{1}{2}[logK_{w}+logC-logK_{a}] $ Or $ pOH=\frac{1}{2}[pK_{w}-logC-pK_{a}] $ Now $ pH+pOH=pK_{w} $
$ \therefore pH=\frac{1}{2}[pK_{w}+logC+pK_{a}] $
