posted by chem1 on .
Enough of a monoprotic acid is dissolved in water to produce a 0.0195 M solution. The pH of the resulting solution is 6.24. Calculate the Ka for the acid.
Also, the answer is not 1.7 * 10 ^ -11
The response is: You treated the initial concentration as 0. In this case, the initial 10^-7 M H+ from water is not negligible. If [H+]initial = 10 ^ -7 and [H+]final = 5.8 * 10^-7, by how much did the concentration change? What does that say about how much A- was produced?