Question

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

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

Answer 1

Answer: (D)

$Ka=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^3{O}^{+}\right]}{\left[C{H}_{3}COOH\right]}=\frac{X(0.1+X)}{(0.01-X)}$
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 $\alpha =\frac{X}{C}$
$=\frac{1.6\times 10^{-4}}{0.01}=0.016$