Question

Calculate the percent dissociation of HA in a 0.10 M solution? In a 0.010 M solution?

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

Calculate the percent dissociation of HA in a 0.10 M solution?

Calculate the percent dissociation of HA in a 0.010 M solution?

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

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER

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!

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER