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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 
  1. 4.75
  2. 6.75
  3. 8.75
  4. 2.75


Solution

The correct option is A 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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