The correct option is C Kh=KwKa
Theory:
Salt of weak acid WA and a strong base SB
Consider, weak acid (CH3COOH) and strong base (NaOH) together form a Salt of (CH2COONa)
On complete dissociation CH3COONa(aq)→CH3COO−(aq)+Na+(aq)
On hydrolysis CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq))
Note:- Only the anion will undergo hydrolysis and the solution will be alkaline in nature due to OH− formation.
So here, the anion is considered responsible for the alkaline nature of the solution.
⇒pH>7
Consider a salt (MX) added to water and produce cations (M+)and anions (X−) in water. After dissociation (OH−) concentration increases.
For example :- CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)
C 0 0
C−Ch Ch Ch
In this reaction the (Kh) hydrolysis constant is:
Kh=[CH3COOH][OH−][CH3COO−]=(Ch)2C(1−h)
CH3COOH(aq)⇌CH3COO−(aq)+OH−(aq)
Ka=[CH3COO−][H+][CH3COOH]
H2O(l)⇌H+(aq)+OH−(aq)
In this reaction the Kw constant is:
Kw=[H+][OH−]
KwKa=[H+][OH−][CH3COO−][H+][CH3COOH]
KwKa=[CH3COOH][OH−][CH3COO−]
Kh=KwKa
Form earlier equation of Kh
Kh=(Ch)2C(1−h) Generally h<<1. So, neglect(1-h)
If h<<1
Ch2=Kh=KwKa
Then h=√KhC=√KwKaC
[OH−]=C×√KwKaC=√KwCKa
[H+]=Kw[OH−]=√KwKaC
pH=12[pKw+pKa+log C].
Only valid if h<0.1 or 10%
pKa=−logKa
pH=7+12[pKa+log C] at 25∘C