Question

What is the pH of the solution created by combining 2.40 mL of the 0.10 M NaOH(aq) with with 8.00 mL of the 0.10 M HC2H3O2(aq)?

So, here's my working so far:

.00024 mol NaOH
.0008 mol HC2H3O2
.0008 - .00024 = .00056 mol HC2H3O2

.00056mol / 0.0104L = .05385 M HC2H3O2

The Ka of acetic acid in water at 25C = 1.76 x 10^-5.

So I find [H+] = 9.73 x 10^-4

and pH = 3.01, but this is not the right answer. Can anyone explain where I went wrong?

Answer 1

Ka is ok. (HC2H3O2) is ok.
Where is the (C2H3O2^-)? I don't see that anywhere.

NaOH + HC2H3O2 ==> NaC2H3O2 + H2O

Ka = (H^+)(C2H3O2^-)/(HC2H3O2) and solve for (H^+) = Ka*(HC2H3O2)/(C2H3O2^-)
(H^+) = ....
I get something like 4 x 10^-5 and pH about 4.4 or so. (You need to do it more accurately).