PH of solution?
I thought I could just take the amount of excess acid and divide by total volume to get [H3O+] since the acid is in surplus, then take the -log([H3O+]) to get pH but the answer came out wrong.
They are telling me to take the amount excess acid/total volume, which gives you acid conc. then dividing that by amount of base added / total volume, which gives you [H3O+].
This is in sharp contrast to a problem I just did that was extremely similar and the answers are coming out different.
Answer:
I think you're forgetting to take into account the fact that the acid is a weak one and does not dissociate completely.
Although the conc. of excess acid is 0.0035mol/0.065L =0.0538M, only part of is becomes H3O+ in solution as indicated by its small Ka.
Eqn: HOCl ~> H+ + OCl-
Orig: 0.0538 0 0
change: -x +x +x
Equlib: 0.0538-x x x
The equlibrium quotient equation is Ka=[H+][OCl-]/[HOCl]
or 3e-8 = x^2/0.0538
I omitted the x on the bottom b/c I expect it to be much much smaller than 0.0538 and it will not make much of a difference. The general rule is that you can omit the x if your original acid concentration is more than 1000-fold greater than Ka (here, it's about 1000,000-fold greater).
Solve that equation for x (H3O+): x = 4.02e-5 M
pH = 4.4
Take the negative log of the hydrogen ion concentration to get the pH. Once you figured that out, you will learn how to do your own homework
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: