The dissociation constant of a weak base is ${10}^{-4}$. The pH of $0.01\mathrm{M}$ of that weak base at ${25}^{°}C$ is

Base $\mathrm{BOH}$ is dissociated as follows:

$\mathrm{BOH}\rightleftharpoons {\mathrm{B}}^{+}+{\mathrm{OH}}^{-}$

So, the dissociation constant of base $\mathrm{BOH}$

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{B}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[\mathrm{BOH}\right]}\dots \text{ (i) }$

At equilibrium,

$\left[{\mathrm{B}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right] \therefore {\mathrm{K}}_{\mathrm{b}}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{\left[\mathrm{BOH}\right]}$

Given that

${\mathrm{K}}_{\mathrm{b}}={10}^{-4}$

and $\left[\mathrm{BOH}\right]=0.01\mathrm{M}$

Thus, ${10}^{-4}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{0.01}$

${\left[{\mathrm{OH}}^{-}\right]}^2=1\times {10}^{-6}$ 

$\left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-3} \mathrm{mol}{ \mathrm{L}}^{-1}$

$\left[H^{+}\right]\left[O{H}^{-}\right]=K_w$

Where, $K_w$ is an ionization constant of water.

$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}$

$K_w\text{ is }{10}^{-14}$

Substitute the values in the above equation.

$\left[H^{+}\right]=\frac{{10}^{-14}}{{10}^{-3}}={10}^{-11}$

pH will be:

 $pH=\log \left[H^{+}\right]=\log \left({10}^{-11}\right)=11$

Therefore, the correct option is (C).