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 %

Step-by-step 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=((c\alpha)^(2))/(c-c\alpha)`

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

`K_b=4.4* 10^(-4)`

Putting in the values we get:

`4.4* 10^(-4)=((0.050* \alpha)^2)/((0.050-0.050* \alpha))`

`(\alpha)=0.089`

percent ionisation =
`0.089* 100=8.9\%`