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Question

The degree of dissociation of acetic acid in a 0.1 M solution is 1.32×102, find out its pKa value.
Take log10(1.76)=0.25

A
8.75
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B
4.75
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C
6.75
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D
2.75
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Solution

The correct option is B 4.75
Given: degree of dissociation of CH3COOH=1.32×102=0.0132

So, according to given reaction:

CH3COOH CH3COO + H+
initially 0.1 0 0
at equilibrium 0.1(10.0132) 0.1(0.0132) 0.1(0.0132)

Ka=[CH3COO][H+][CH3COOH]=(0.1×0.0132)(0.1×0.0132)(0.1×(10.0132))=1.76×105

pKa=log10(Ka)
putting values,
pKa=log10(1.76×105)=(0.255)
pKa=4.75

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