Question

$BOH$ is a weak base. Molar concentration of $BOH$ that provides a $\left[O{H}^{-}\right]$ of $1.5\times {10}^{-3}M$ is:

Given: $K_b$ of $\left[BOH\right]=1.5\times {10}^{-5}M$

• A. $0.15M$
• B. $0.1515M$
• C. $0.0015M$
• D. $15\times {10}^{-5}M$

Answer 1

The correct option is A $0.15M$

The dissociation of $BOH$ may be represented as,

$BOH(aq.)\rightleftharpoons B^{+}(aq.)+O{H}^{-}(aq.)$

$K_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[BOH\right]}$

Since $\left[B^{+}\right]=\left[O{H}^{-}\right]$,

$K_b=\frac{[O{H}^{-}{]}^2}{\left[BOH\right]}$

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

$\Rightarrow \left[BOH\right]=0.15M$

Hence, option $A$ is the correct answer