Question

The pH of $0.05$ $M$ aqueous solution of diethyl amine is $12.0$. The value of $K_b$ is:

• A. $K_b=3.5\times {10}^{-3}$
• B. $K_b=2.0\times {10}^{-3}$
• C. $K_b=2.1\times {10}^{-3}$
• D. $K_b=2.5\times {10}^{-3}$

Answer 1

Answer: (D)

$K_b=2.5\times {10}^{-3}$

$\underset{11-\alpha }{{\left(C_2{H}_5\right)}_{2}NH}+H_{2}O⟶\underset{0\alpha }{{{\left(C_2{H}_5\right)}_{2}NH}_2^{+}}+\underset{0\alpha }{{OH}^{-}}$
$\therefore$ $\left[{OH}^{-}\right]=c\alpha$; where $c$ is concentration of base $=0.05M$
$\because$ $pH=12$ $\therefore$ $pOH=2$
or $\left[{OH}^{-}\right]={10}^{-2}M$
$\therefore$ ${10}^{-2}=c\alpha$
${10}^{-2}=0.05\times \alpha$
or $\alpha =0.2$
Since, $\alpha$ is not small and thus, do not use $K_b=c{\alpha }^2$
for base
$K_b=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.05\times {\left(0.2\right)}^2}{(1-0.2)}=2.5\times {10}^{-3}$