Question

Degree of dissociation of 0.1 N $C{H}_{3}COOH$ is $(K_{acid}=1\times {10}^{-5})$

• A. ${10}^{-5}$
• B. ${10}^{-4}$
• C. ${10}^{-3}$
• D. ${10}^{-2}$

Answer 1

The correct option is D ${10}^{-2}$

Degree of dissociation $\alpha =?$

Normality of solution $=0.1N=\frac{1}{10}N$

Volume $=10litre$

Dissociation constant $K=1\times {10}^{-5}$

$K=\frac{{\alpha }^2}{V};\alpha =\sqrt{KV}=\sqrt{1\times {10}^{-5}\times 10};\alpha =1\times {10}^{-2}$