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  4. Degree of dissociation of 0.1 N CH3COOH is (K textacid=1×10-5)

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Q. Degree of dissociation of $0.1 \,N \,CH_{3}COOH$ is $(K_{\text{acid}}=1\times10^{-5})$

UP CPMTUP CPMT 2006

Solution:

For a weak acid, degree of dissociation,
$ \alpha=\sqrt{\frac{K_{a}}{C}}$
$=\sqrt{\frac{1\times10^{-5}}{10^{-1}}}$
$=10^{-2}$