Asked by chem1
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?
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?
Answers
Answered by
chem1
nevermind, I figured it out
Answered by
Chemilleterate
you figured it out then what was the solution how did you solve it then because others are having the same problems and the same errors
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