Question

The dissociation constant of acetic acid at a given temperature is $1.69 \times 10^{-5}$. The degree of dissociation of $0.01 \,M$ acetic acid in the presence of $0.01 \, M \,HCl$ is equal to

• A. $0.41 \times 10^{-2}$
• B. $0.18 \times 10^{-7}$
• C. $0.169 \times 10^{-2}$
• D. $0.013 \times 10^{-4}$

Answer 1

Answer: (C)

Given, $K=1.69 \times 10^{-5}$
$\because C \alpha<<0.01$
$\therefore K=\frac{C \alpha(C \alpha+0.01)}{C(1-\alpha)}=0.01 \alpha$
or $ \alpha=\frac{1.69 \times 10^{-5}}{0.01}=0.169 \times 10^{-2}$