Question

A $0.1$ aqueous solution of a weak acid is $2 \%$ ionised. If the ionic product of water is $1 \times 10^{-4}$, the $\left[ OH ^{-}\right]$is

• A. $ 5\times 10^{-12}M$
• B. $ 2\times 10^{-3}M$
• C. $ 1\times 10^{-14}M $
• D. None of these

Answer 1

Answer: (A)

Since, it is a weak acid, the equation to calculate $\left[H^{+}\right]$is $=C \times \alpha(\alpha=\%$ of ionisation $)$
$=0.1 \times 0.02=2 \times 10^{-3} M $
$K_{w}=\left[H^{+}\right]\left[O H^{-}\right] $
${\left[O H^{-}\right]=\frac{K_{w}}{\left[H^{+}\right]}=\frac{1 \times 10^{-14}}{2 \times \$ 10^{-3}}} $
$=5 \times 10^{-12} M$