Question

The dissociation constant of acetic acid is $1.6 \times 10^{-5}$. The degree of dissociation $(\alpha)$ of $0.01 \,M$ acetic acid in the presence of $0.1 \,M \,HCl$ is equal to

• A. 0.4
• B. 0.026
• C. 1.6
• D. 0.016

Answer 1

Answer: (D)

$Ka =\frac{\left[ CH _{3} COO ^{-}\right]\left[ H ^{3} O ^{+}\right]}{\left[ CH _{3} COOH \right]}=\frac{ X (0.1+ X )}{(0.01- X )}$
Assuming that $(0.1+X)=0.1$ and $(0.01-X)=0.01$
$1.6 \times 10^{-5}=\frac{ X \times 0.1}{0.01}$
$X =1.6 \times 10^{-4}$
So degree of dissociation $\alpha=\frac{X}{C}$
$=\frac{1.6 \times 10^{-4}}{0.01}=0.016$