Asked by Jenny
A 0.490 M solution of an unknown monoprotic weak acid, HA, is 2.80% ionized. What is the value of Ka for HA?
Answers
Answered by
DrBob222
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
You know HA is 0.490. If it is ionized 2.80%, then H^+ and A^- must be 0.490 x 0.0280 = ??. Substitute into Ka expression and solve.
Ka = (H^+)(A^-)/(HA)
You know HA is 0.490. If it is ionized 2.80%, then H^+ and A^- must be 0.490 x 0.0280 = ??. Substitute into Ka expression and solve.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.