Question

Calculate the pH of a $0.1$M solution of $H_{2}NC{H}_{2}C{H}_{2}N{H}_2$; ethylenediamine(en). Determine the en $H_2^{2+}$. Concentration in the solution $K_{b_1}$ and $K_{b_2}$ values of ethylenediamine are $9\times {10}^{-5}$ and $7.1\times {10}^{-8}$ respectively.

Answer 1

Ethylenediamine $\left(en\right)$ is dibasic base, the 1st ionisation is quite feasible. Consideration of 1st ionisation is sufficient in order of calculate the $pH$ as $K_2$ is $1000$ smaller than $K_1$ and we can treat en as monobasic base.

$\therefore$ $\left[{OH}^{-}\right]=\sqrt{K_{1}C}=\sqrt{9\times {10}^{-5}\times 0.1}=3\times {10}^{-3}M$

$\therefore$ $pH=14-pOH=14-(-log{3NO}^{-3})=11.47$

$\therefore$ $\left[{enH}^{+}\right]=3\times {10}^{-3}M$

$\therefore$ $\left[{enH}_2^{+}\right]=K_2\frac{\left[{enH}^{+}\right]}{\left[{OH}^{(-)}\right]}=K_2\frac{3\times {10}^{-3}}{3\times {10}^{-3}}=7.1\times {10}^{-8}M$