Question

At ${25}^{o}C$, 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

The correct option is D $1.0\times {10}^{-7}mol{L}^{-1}$

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