Question

0.10 M methylamine (CH3NH2)
kb= 4.38e-4

calc the percent ionization

so i did the ICE table to find that x=6.26e-3
and then i divided by .1 and then multiplied by 100. but i got the wrong answer whats did i do wrong?

Answer 1

So far you haven't done anything wrong. But without showing your work I can't tell where you have gone wrong. The procedure you have used is correct But I get 6.62 x 10^-3 for (OH^-) and not 6.26 x 10^-3. That may be the error.

Answer 2

CH3NH2(aq) + H2O <=> CH3NH3+(aq) + OH-(aq)
Kb = [CH3NH3+][OH-] / [CH3NH2] = 4.38x10^-4
Let [OH-] = x, and [CH3NH2] = 1.00M
x^2 / (1.00-x) = 4.38x10^-4
If x is much smaller than 1.00, the above equation can be simplified to
x^2/1.00 = 4.38x10^-4
x = sqrt(4.38x10^-4)
solve for x
% ionization = (x/1.00)(100)
NOTE: if the above answer is not precise enough, the equation:
x^2 / (1.00-x) = 4.38x10^-4 must be simplified to a quadratic trinomial which is equal to 0. That must be solved. Of the two roots, only one would work.

Answer 3

A clarifying note here:
GK has used 1.00 M for the initial concentration but the problem gives 0.10 M.