Question

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

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

Answer 1

The correct option is C $\frac{K_w\left[H^{+}\right]}{K_b+K_w}$

$BOH\leftrightharpoons B^{+}+O{H}^{-}$
$K_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[BOH\right]}=\frac{c{\alpha }^2}{(1-\alpha )}$
$\therefore \alpha =\sqrt{\frac{K_b}{c}}$ (i)
$\therefore percentagedissociation=100\times \sqrt{\frac{K_b}{c}}$
For option b lets start from option itself
$(p{K}_b-pOH)=-log{K}_b-(-log[O{H}^{-}\left]\right)$
Now we know that $\left[O{H}^{-}\right]=c\alpha =\sqrt{K_b\times c}$ now
$(p{K}_b-pOH)=-log{K}_b+1/2log{K}_b+1/2logc=1/2log\frac{c}{K_b}$ from eq(i) we will get
$(p{K}_b-pOH)=log\frac{1}{\alpha }$
$\therefore {10}^{(p{K}_b-pOH)}={10}^{log\frac{1}{\alpha }}=\frac{1}{\alpha }$ so we will get
$\frac{1\times 100}{1+{10}^{(p{K}_b-pOH)}}=\frac{\alpha }{1+\alpha }\times 100=100\times \alpha$
For option c
$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}$ so
$\frac{K_w\left[H^{+}\right]}{K_b+K_w}=\frac{K_w^2}{(K_b+K_w)\times (K_{b}c{)}^{1/2}}=\frac{K_w^2}{(K_b+K_w)c\times \alpha }$
For option d
$\frac{K_b\times 100}{K_b+\left[O{H}^{-}\right]}=\frac{c{\alpha }^2\times 100}{(c{\alpha }^2+c\alpha )}=\frac{\alpha \times 100}{1+\alpha }=percentagedissociation$