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Q.
At $25^{\circ} C$, the dissociation constant of a base, $BOH$, is $1.0 \times 10^{-12}$. The concentration of hydroxyl ions in $0.01\, M$ aqueous solution of the base would be
Base $BOH$ is dissociated as follows
$BOH \rightleftharpoons B ^{+}+ OH ^{-}$
So, the dissociation constant of $B O H$ base
$K_{b}=\frac{\left[B^{+}\right]\left[O H^{-}\right]}{[B O H]}\,\,\,\,...(i)$
At equilibrium $\left[B^{+}\right]=\left[ OH ^{-}\right]$
$\therefore K_{b}=\frac{\left[ OH ^{-}\right]^{2}}{[ BOH ]}$
Given that $ K_{b} =1.0 \times 10^{-12}$ and
$[B O H] =0.01\,M $
Thus $1.0 \times 10^{-12} =\frac{\left[ OH ^{-}\right]^{2}}{0.01} $
$\left[ OH ^{-}\right]^{2} =1 \times 10^{-14} $
$\left[ OH ^{-}\right] =1.0 \times 10^{-7} \,mol\, L ^{-1} $