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Question

For propanoic acid Ka=1.34×105 M at 25C. Then what will be the degree of dissociation for a 0.01 M solution of the acid ?

A
3.66×104
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B
3.66×102
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C
5×104
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D
5×102
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Solution

The correct option is B 3.66×102
If α is the degree of dissociation of propanoic acid of concentration c, then the concentrations of various species in solution are
CH3CH2c(1α)COOH+H2OCH3CH2cαCOO+H3O+cα
With these concentrations, the equilibrium constant becomes
Ka=[CH3CH2COO][H3O+][CH3CH2COOH]=(cα)(cα)c(1α)=cα21αcα2
or α=Kac=(1.34×105M)(0.01 M)=1.34×103
Thus, α=3.66×102

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