Given: A certain weak acid, HA, has a Ka value of 9.5E-7.

**Answer 1**

9.5 x 10^-7 =x^2/ 0.10-x

x = 0.00031

% = 0.00031 x 100/ 0.10=0.31

9.5 x 10^-7= x^2/0.01-x

x =0.000097

% = 0.000097 x 100/0.01=0.97

**Answer 2**

We see that the acid ionizes as follows: HA <==> H+ + A-. Then, we’ve that: ok(a) = [H+][A-]/[HA]. when you consider that H+ and A- are produced in equivalent quantities, we are able to enable: [H+] = [A-] = x. Then, because of the fact the preliminary concentration of HA is (9.40 two x 10^-2)/2.30 = 4.09 x 10^-2 M, we see that: [HA] = 4.09 x 10^-2 – x. this provides: ok(a) = x^2/(4.09 x 10^-2 – x) ==> 2.39 x 10^-6 = x^2/(4.09 x 10^-2 – x). fixing for x yields: x = 3.a million * 10^-4. to that end, [H+] = 3.a million x 10^-4 and (3.a million x 10^-4)/(4.09 x 10^-2) * one hundred = seventy six% of the H+ ions have ionized. i wish this helps!