Please Help Me With Chemistry?

Question

A 0.35 M solution of a weak acid is 2.0% dissociated.

(a) Calculate the [H3O+], pH, [OH -], and pOH of the solution.
[H3O+]
_____
pH
_____
[OH -]
_____
pOH
_____
(b) Calculate Ka of the acid.
_____

Answer 1

Let the acid be called HA. Let the solution be called S.

HA <===> H+ + A-

0.35molHA/L x 2% = 0.007mol/L = [H+] = [A-]

[H3O+] = 0.007M = 7 x 10^-2M First answer

pH = -Log[H+] = 2.155 Second answer

Kw = [H+][OH-] = 10^-14

(7x10^-2)[OH-] = 1 x 10^-14

[OH-] = 1.43 x 10^-13 Third answer

pOH = 0.128 Fourth answer

Ka = [H+][A-]/[HA]

Ka = (7x10^-2)(7x10^-2)/(0.35) = 1.4 x 10^-2 Fifth answer

The denominator is 0.35, because 0.35 - 0.007 is approximately 0.35.