Question

The $\left[O{H}^{-}\right]$ of $0.005M$ $H_{2}S{O}_4$ is:

• A. $2\times {10}^{-2}M$
• B. $5\times {10}^{-3}M$
• C. ${10}^{-2}M$
• D. ${10}^{-12}M$

Answer 1

The correct option is D ${10}^{-12}M$

Solution:- (D) ${10}^{-12}M$

$H_{2}S{O}_4⟶2H^{+}+{S{O}_4}^{-2}$

Concentration of $H^{+}$ in $0.005M{H}_{2}S{O}_4=2\times 0.005=0.01M$

Therefore,

$\left[H^{+}\right]=0.01M$

As we know that,

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

$\Rightarrow {10}^{-2}\times \left[{OH}^{-}\right]={10}^{-14}$

$\Rightarrow \left[{OH}^{-}\right]=\frac{{10}^{-14}}{{10}^{-2}}={10}^{-12}M$