Second Year Chem. pH problem?

Question

I just need a little hint for this, I would like to be able to figure it out myself.

Calculate the pH of a soln of acetic acid that is 3% ionized, Ka = 1.8 x 10^-5.

Since I don't have M of the solution, I can't solve by
% ionization = [H+}Equil / [HA]i

please help me understand what approach to take. Thank you in advance.

Answer 1

The Ka is the (H+)(C2H3O2-) / HC2H3O2

You know that the HC2H3O2 is 0.03 and you have the Ka.

THe H+ and C2H3O2- are equal, so once you have the Ka times 0.03, take the square root and you have (H+). From there you should know how to calculate the pH using the negative log.

Answer 2

This might help:

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

and % ionization is [H+] / ([A-] + [HA]) * 100%

Because most of the H+ is made from the dissociated HA, you may also make the approximation

[H+] ≈ [A-]

Making this approximation, you can eliminate [A-] from the other two equations, you get

P = [H+] / ([H+] + [HA])

or, more conveniently,

1/P = 1 + [HA]/[H+]

and Ka = [H+]^2 / [HA]

This second equation can be reformulated into

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

from which we can substitute the left side to get

1/P = 1 + [H+] / Ka

From which [H+], hence pH, can be easily found.

I calculated [H+] to be 5.8e-4, which is much greater than the 1e-7 from the water, justifying our simplifying assumption. If the acid were weak enough to endanger our assumption that [H+] ≈ [A-], then P would probably be less than 0.01 and we could instead have assumed that [A-] ≈ [KA].