Question

Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3NH2(aq) solution. (Kb for CH3NH2 = 4.4 × 10-4)

Answer 1

Answer:  The percent ionization of $CH_3NH_2$ in a 0.050 M $CH_3NH_2(aq)$ solution is 8.9 %

Explanation:

$CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+$

 cM                            0             0

$c-c\alpha$                       $c\alpha$            $c\alpha$

So dissociation constant will be:

$K_b=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= concentration = 0.050 M and $\alpha$ = degree of ionisation = ?

$K_b=4.4\times 10^{-4}$

Putting in the values we get:

$4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}$

$(\alpha)=0.089$

percent ionisation =$0.089\times 100=8.9\%$