Question

A $0.1M$ aqueous solution of a weak acid is $2\%$ ionized. If the ionic product of water is $1\times {10}^{-14}$, the $\left[O{H}^{-}\right]$ is:

• A. $5\times {10}^{-12}$M
• B. $2\times {10}^{-3}$M
• C. $1\times 0^{-14}$M
• D. none of these

Answer 1

Answer: (A)

$5\times {10}^{-12}$M

Degree of dissociation of weak acid, $\alpha =\frac{2}{100}=0.02$

Concentration of acid is $\left[H^{+}\right]=C\alpha =0.1\times 0.02=0.002$

Ionic product of water $=1\times {10}^{-14}$

$\left[H^{+}\right]\left[{OH}^{-}\right]=1\times {10}^{-14}$

$\left[{OH}^{-}\right]=\frac{1\times {10}^{-14}}{0.002}=5\times {10}^{-12}$