Q38 / 150 · MHT CET 2012 · Chemistry
The degree of ionization of $0.4 \mathrm{M}$ acetic acid will be $\left(K_{a}=1.8 \times 10^{-5}\right)$
- Correct answerA$6.71 \times 10^{-3}$
- B$1.6 \times 10^{-3}$
- C$0.4 \times 1.8 \times 10^{-5}$
- D$1.8 \times 10^{-5}$
Solution
For a weak acid (acetic acid),
$$
\begin{aligned}
K_{a} &=C \alpha^{2} \\
1.8 \times 10^{-5} &=0.4 \times \alpha^{2} \\
\alpha &=\sqrt{\frac{1.8 \times 10^{-5}}{0.4}} \\
&=6.71 \times 10^{-3}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.