Question

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

• A. $2.5\times {10}^{-3}$
• B. $2.5\times {10}^{-4}$
• C. $5\times {10}^{-3}$
• D. $5\times {10}^{-4}$

Answer 1

The correct option is A $2.5\times {10}^{-3}$

$\begin{array}{ccccccccc}(C_2{H}_5{)}_{2}NH& +& H_{2}O& \rightleftharpoons & (C_2{H}_5{)}_{2}N{H}_2^{+}& +& O{H}^{-}& & \\ \text{Initial conc.}1& & & & 0& & 0& & \\ \text{Eq. conc.}(1-\alpha )& & & & \alpha & & \alpha & & \end{array}$
$\left[O{H}^{-}\right]=C\alpha$
Here $C$ is concentration of base and $C=0.05M$
As $pH=12$
So $pOH=2$
$\left[O{H}^{-}\right]={10}^{-2}M$
$C\alpha ={10}^{-2}$
$0.05\times \alpha ={10}^{-2}$ (as $C=0.05$)
$\alpha =0.2$
$K_b=\frac{C{\alpha }^2}{(1-\alpha )}=\frac{0.05\times (0.2{)}^2}{(1-0.2)}$
$=\frac{0.05\times 0.04}{0.8}=2.5\times {10}^{-3}$