Question

The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 ml of 1.0 M HA solution is diluted to 32 ml, the degree of dissociation of acid and $H_3{O}^{+}$ ion concentration in the resulting solution will be respectively:

• A. $0.02$ and $3.125\times {10}^{-4}$
• B. $1.25\times {10}^{-3}$ and $0.02$
• C. $0.632$ and $3.95\times {10}^{-4}$
• D. $0.02$ and $8.0\times {10}^{-12}$

Answer 1

The correct option is C $0.632$ and $3.95\times {10}^{-4}$

Given,
$\alpha =0.5\%=0.005$ and $C=0.1M$

The expression for the degree of dissociation is $\alpha =\sqrt{\frac{K_a}{C}}$.
$0.005=\sqrt{\frac{K_a}{0.1}}$

$K_a=2.5\times {10}^{-6}$.

When the solution is diluted, the molarity of the solution is given by the following expression.

$M_1{V}_1=M_2{V}_2$

or $2\times 1=32\times M_2$

Hence, $M_2=\frac{1}{16}$.

For this diluted solution, $\alpha =\sqrt{\frac{K_a}{C}}=0.632\%$.
The hydrogen ion concentration is $\left[H^{+}\right]=c\alpha =\frac{1}{16}\times 0.00632=3.955\times {10}^{-4}$.