Question

The weak acid, $HA$ has a $K_a$ of $1.00\times {10}^{-5}$. If $0.1$ mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

• A. $1$%
• B. $99.9$%
• C. $0.1$%
• D. $99$%

Answer 1

The correct option is A $1$%

0.1 mole of acid is dissolved in 1 litre of water means $\left[HA\right]=0.1M$

Let '$\alpha$' be degree of dissocition

$HA\rightleftharpoons H^{+}+A^{-}$

$0.1$

$0.1(1-\alpha )$ $0.1\alpha$ $0.1\alpha$
$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}=\frac{{0.1}^2{\alpha }^2}{0.1(1-\alpha )}$

Let $\alpha <<1$ so $1-\alpha =1$

$K_a=0.1{\alpha }^2={10}^{-5}$

$\alpha ={10}^{-2}$

% of acid dissociated=$1$%