Question

Which of the following expressions for $\text{\%}$ ionization of a monoacidic base $\left(BOH\right)$ in aqueous solution is not correct at appreciable concentration?

• A. $100\times \sqrt{\frac{K_b}{c}}$
• B. $\frac{1}{1+10(p{K}_b-pOH)}$
• C. $\frac{K_w\left[H^{+}\right]}{K_b+K_w}$
• D. $\frac{K_b}{K_b+\left[O{H}^{-}\right]}$

Answer 1

Answer: (C)

$\frac{K_w\left[H^{+}\right]}{K_b+K_w}$

The expression for the dissociation of the base is $BOH\left(aq\right)\rightleftharpoons B^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$.
Let $h$ be the degree of dissociation.
At equilibrium, the concentrations of $BOH,B^{+}andO{H}^{-}$ are $c(1-h),ch$ and $ch$ respectively.
$K_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[BOH\right]}=\frac{c{h}^2}{1-h}$
Since $h$ is small, $(1-h)\approx 1$.
Hence, $h=\sqrt{\frac{k_b}{C}}$.
or, $h=100\times \sqrt{\frac{k_b}{C}}\%$.
Since total concentration of base is the sum of the concentration of $BOH$ and $B^{+}$.
Hence, $h=\frac{\left[B^{+}\right]}{\left[B^{+}\right]+\left[BOH\right]}=\frac{1}{1+\frac{\left[BOH\right]}{\left[B^{+}\right]}}=\frac{1}{1+\frac{\left[O{H}^{-}\right]}{K_b}}.$
or, $\frac{K_b}{K_b+\left[O{H}^{-}\right]}=\frac{K_b\left[H^{+}\right]}{K_b\left[H^{+}\right]+K_w}$
Also, $pOH=-log\left[O{H}^{-}\right]$.
$\left[O{H}^{-}\right]={10}^{-pOH}$
$K_b={10}^{-p{K}_b}$
$h=\frac{1}{1+\frac{{10}^{-pOH}}{{10}^{-p{K}_b}}}=\frac{1}{1+{10}^{p{K}_b-pOH}}$