a) A weak acid, HX, is 1.3% ionized in .20 M solution. What percent of HX is ionized in a .030 M solution? b) Does the percent ionization increase or decrease upon dilution? c) Does the H3O+ concentration of the above weak acid increase or decrease?

Answer 1

Please help, struggling like crazy.

Answer 2

HX <-> H+ + X-

Concentration = mol/vol
Since we only know mol and not concentration, we can just use mol since it is directly proportional to concentration
HX H+ X-
I 0.2 0 0
C -y +y +y
E 0.2-y y y

Since y is very small(weak acid) Equilibrium mol for HX is 0.2

Ionization % = [H+]eq/[HX]i *100%
For 0.2M, [H+] = 1.3% / 100% * 0.2= 2.6*10^-5
y=2.6*10^-5 M

Now we can calculate Ka of HX
Ka = [H+][X-] / [HX]
=(2.6*10^-5)^2/0.2
=1.3*10^-4

ICE table for 0.03M
HX H+ X-
I 0.03 0 0
C -z +z +z
E 0.03-z z z
(Once again, HX at equi = 0.03)

Using Ka = [H+][X-] / [HX]
1.3*10^-4 = z^2/0.03
z=1.97*10^-3 = [H+]

Ionization % for 0.03M =1.97*10^-3/0.03 *100%
=6.58%

I'm not too sure if this is correct, haven't done this is awhile and quite rusty
(Do check the calculations too)