Question

At $25^{\circ }C$, the dissociation constant of a base, $BOH$, is $1.0\times 10^{-12}$. The concentration of hydroxyl ions in $0.01M$ aqueous solution of the base would be

• A. $2.0\times 10^{-6}mol{L}^{-1}$
• B. $1.0\times 10^{-5}mol{L}^{-1}$
• C. $1.0\times 10^{-6}mol{L}^{-1}$
• D. $1.0\times 10^{-7}mol{L}^{-1}$

Answer 1

Answer: (D)

Base $BOH$ is dissociated as follows
$BOH\rightleftharpoons B^{+}+O{H}^{-}$
So, the dissociation constant of $BOH$ base
$K_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{[BOH]}...(i)$
At equilibrium $\left[B^{+}\right]=\left[O{H}^{-}\right]$
$\therefore K_b=\frac{{\left[O{H}^{-}\right]}^2}{[BOH]}$
Given that $K_b=1.0\times 10^{-12}$ and
$[BOH]=0.01M$
Thus $1.0\times 10^{-12}=\frac{{\left[O{H}^{-}\right]}^2}{0.01}$
${\left[O{H}^{-}\right]}^2=1\times 10^{-14}$
$\left[O{H}^{-}\right]=1.0\times 10^{-7}mol{L}^{-1}$