Question

Consider a weak base, $BOH$. Molar concentration of $BOH$ that provides $[OH{]}^{-}$ of $1.5\times {10}^{-3}\text{M}$ will be :
Given: $K_b\left(BOH\right)=1.5\times {10}^{-5}$

• A. $1.5\times {10}^{-2}M$
• B. $0.015M$
• C. $1.5\times {10}^{-4}M$
• D. $0.15M$

Answer 1

The correct option is D $0.15M$

The dissociation of BOH may be represented as,
$BOH(aq.)\rightleftharpoons B^{+}(aq.)+O{H}^{-}(aq.){\text{K}}_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[BOH\right]}$
Since same amount of $\left[B^{+}\right]$ and $\left[O{H}^{-}\right]$, is produced
. $\therefore \left[B^{+}\right]=\left[O{H}^{-}\right]$ at equilibrium

${\text{K}}_b=\frac{[O{H}^{-}{]}^2}{\left[BOH\right]}1.5\times {10}^{-5}=\frac{(1.5\times {10}^{-3}{)}^2}{\left[BOH\right]}$

$\left[BOH\right]=\frac{(1.5\times {10}^{-3}{)}^2}{1.5\times {10}^{-5}}=0.15\text{M}$