Question

The weak acid, HA has a $K_a$​ of $1.00\times {10}^{-5}$. If $0.2mol$ of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

• A. 0.7
• B. 0.05
• C. 0.07
• D. 0.5

Answer 1

The correct option is A 0.7

$0.2$ mol of acid is dissolved in 1 litre of water means $\left[HA\right]=0.2M$
Let $\alpha$ be degree of dissocition
$HA\rightleftharpoons H^{+}+A-$
0.2 0 0
$0.2(1-\alpha )0.2\alpha 0.2\alpha$
$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}=\frac{(0.2\alpha {)}^2}{0.2(1-\alpha )}$

Since, $\alpha <<1,\therefore (1-\alpha )\approx 1$
$K_a=0.2(\alpha {)}^2$
or $\alpha =\sqrt{K_a/C}$
Putting values,
$1.00\times {10}^{-5}=0.2(\alpha {)}^2$
$1.00\times {10}^{-4}=2.0(\alpha {)}^2$
$\alpha =0.7\times {10}^{-2}$
% of acid dissociated = $0.7\times {10}^{-2}\times 100=0.70\%$