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Question

Which of the following is correct for a salt of weak acid and strong base?
Here C is the concentration of the salt.

A
pH=7+12(pKa+logC)
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B
pH=7+12(pKb+logC)
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C
Kh=KwKa
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D
Kh×Kw=Ka
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Solution

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
CCh Ch Ch
In this reaction the (Kh) hydrolysis constant is:
Kh=[CH3COOH][OH][CH3COO]=(Ch)2C(1h)
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(1h) 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 25C

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