A 0.35 M solution of a weak acid is 4.2% dissociated.
(a) Calculate the [H3O+], pH, [OH -], and pOH of the solution.
(b) Calculate Ka of the acid.
2006-10-04 12:49:27 · 2 answers · asked by mariah3785 1 in Science & Mathematics ➔ Chemistry
This isn't a place to outsource your homework, refer your textbook and solve it yourself, its very easy
2006-10-04 13:06:41 · answer #1 · answered by ashwin_hariharan 3 · 0⤊ 0⤋
Well, for one thing, you have to assume that this is a monoprotic acid, such that:
HA <---> H+ + A-
If 4.2% is dissociated, that means 95.8% isn't, so the final concentration of the unknown acid is 0.958 * 0.35 = 0.335 M. [H+] and [A-] will be equal to the difference, or 0.015 M. pH will be -log [H+], or 1.82. pOH will be 12.18, and [OH-] is 6,60 * 10^-13.
Ka will be [H+][A-]/[HA] or 0.015^2/0.335 or 6.72 * 10^-4.
2006-10-04 20:23:19 · answer #2 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋