Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

Degree of dissociation of 0.1 N CH3COOH0.1 \,N \,CH_{3}COOH is (Kacid=1×10−5)(K_{\text{acid}}=1\times10^{-5})

A

10−510^{-5}

B

10−410^{-4}

C

10−310^{-3}

D

10−210^{-2}

Answer

10−210^{-2}

Explanation

Solution

For a weak acid, degree of dissociation, α=KaC \alpha=\sqrt{\frac{K_{a}}{C}} =1×10−510−1=\sqrt{\frac{1\times10^{-5}}{10^{-1}}} =10−2=10^{-2}