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Question

The ionisation constant of Propanoic acid is 1.32×105. Calculate the degree of ionisation of the acid in its 0.05 M solution and also its PH. What will be its degree of ionisation if the solution is 0.01 M in HCl also?

A
1.25×103
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B
1.32×105
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C
1.32×103
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D
None of these
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Solution

The correct option is C 1.32×103
CH3CH2COOH+H2OCH3CH2COO+H3O+
0.05Cα Cα Cα
from Ostwald's Dilution law,
α=Ka/C=1.35×1050.05
α=0.016248
[H3O+]=Cα=0.05×0.016248
=0.0008124=8.124×104
pH=log[H3O+]=log(8.124×104)
=40.9098=3.09023.09
When the soln contains 0.01M of HCl,
Ka=[CH3COO[H+][CH3CH2COOH]
1.32×105=Cα×0.01(0.05Cα)
Now, [H3O+]=0.01M from HCl. In the presence of 0.01M HCl dissociation of propanoic acid decreases.
Cα1.32×105×(0.05Cα)0.01 (as 0.05Cα=0.05)
Cα=6.6×105
degree of ionization, α=6.6×1050.05
1.32×103

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