posted by howie on .
Sacchirin is a monoprotic acid. If the pH of a 1.50x10^-2M solution of this acid is 5.53, what is the Ka of saccharin?
(the answer is 5.8x10^-10)
I know we're supposed to use the Henderson Hasselbach equation to solve for the [H+] from pH, but what's next?
Thanks Dr. Bob.
The HH equation won't work for this. Let's call saccharin HS, then the ionization is
HS ==> H^+ + S^-
Ka = (H^+)(S^-)/(HS)
So you know pH, convert that to (H^+). The (S^-) is the same thing. The (HS) = the unionized saccharin, is 0.015 - (H^+). Solve for Ka. You may have a quadratic equation. I didn't examine it closely enough to see.