Question

Consider a weak monoacidic base $BOH$ having a concentration C with dissociation $\alpha <<1.$ The pH of this weak base in terms of base dissociation constant $K_b$ and concentration $C$ is :

• A. $14+lo{g}_{10}\left(\sqrt{\frac{K_b}{C}}\right)$
• B. $14+lo{g}_{10}\sqrt{K_b.C}$
• C. $14-lo{g}_{10}\sqrt{K_b.C}$
• D. $14-lo{g}_{10}\left(\sqrt{\frac{K_b}{C}}\right)$

Answer 1

The correct option is B $14+lo{g}_{10}\sqrt{K_b.C}$

For weak base $BOH$ with concentration $‘C^{\prime }$, ionization constant $K_b$ and $\alpha <<1$.
In aqueous solution,
$BOH\left(aq\right)\rightleftharpoons B^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
At $t=0$ $C$ $0$ $0$
At $t=t\left(eq\right)$ $C-C\alpha$ $C\alpha$ $C\alpha$

$K_b=\frac{C{\alpha }^2}{1-\alpha }$

As $\alpha <<1(1-\alpha )\approx 1$

$K_b=C{\alpha }^2$
Multiplying both sides by C
$K_b\times C=C{\alpha }^2\times C{K}_b\times C=[O{H}^{-}{]}^2$
$\left[O{H}^{-}\right]=\sqrt{K_b.C}$
$pOH=-lo{g}_{10}\sqrt{K_b.C}pH+pOH=14pH=14+lo{g}_{10}\sqrt{K_b.C}$