Question

The $pH$ of $0.05M$ aqueous solution Di-ethyl amine (weak base) is 12. Calculate its $K_b$.

Answer 1

The reaction : $HN{\left(C_2{H}_5\right)}_2\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_2{N}^{+}{\left(C_2{H}_5\right)}_2\left(aq\right)+{OH}^{(-)}$

So, $K_b=\frac{\left[H_2\stackrel{(+)}{N}{\left(C_2{H}_5\right)}_2\right]\left[{OH}^{(-)}\right]}{\left[{HN\left(C_2{H}_5\right)}_2\right]}$

We are given that $pH=12$

thus $pOH=14-12=2$ and thus $\left[{OH}^{-}\right]={10}^{-2}M$

So, $\left[H_2\stackrel{(+)}{N}{\left(C_2{H}_5\right)}_2\right]={10}^{-2}M$

Undissociated concentration of $HN{\left(C_2{H}_5\right)}_2,\left[HN{\left(C_2{H}_5\right)}_2\right]=0.05M-0.01M=0.04M$

$\therefore$ $K_b=\frac{0.01\times 0.01}{0.04}=2.5\times {10}^{-3}$.