Question

How much water must be added to 300 mL of 0.2 M solution of $C{H}_{3}COOH$ for the degree of dissociation of the acid to be doubled?
[$K_a$ for acetic acid is $1.8\times {10}^{-5}$]

• A. 900 mL
• B. 1000 mL
• C. 500 mL
• D. 300 mL

Answer 1

The correct option is A 900 mL

$C{H}_{3}COOH\leftrightharpoons C{H}_{3}CO{O}^{-}+H^{+}0.2$
$0.2-C\alpha$ $C\alpha$ $C\alpha$

We have $K_a=C{\alpha }^2$
Let initial concentration is $C_1$, thus above $e{q}^n$can be written as $K_a=C_1{\alpha }^2....\left(1\right)$
Now $K_a$ needs to remain constant while $\alpha$ needs to be doubld.
$\therefore Ka=C_2(2\alpha {)}^2$
$Ka=C_{2}4{\alpha }^2....\left(2\right)$

Dividing $e{q}^{n}1$ by $e{q}^{n}2$
$\frac{Ka}{Ka}=\frac{C_1{\alpha }^2}{4C_2{\alpha }^2}$

$\therefore C_2=\frac{C_1}{4}=\frac{1}{20}M$
Using $M_1{V}_1=M_2{V}_2$
$V=1200mL$
$\therefore$ Extra volume to be added $=(1200-300)mL$
$=900mL$