4. Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.37-M HCl to 320. mL of each of the following solutions.

Question

Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.

c) 0.115 M HC2H3O2
pH before mixing =
pH after mixing=
pH change =

My work:

Before Mixing:
HC2H3O2 + H2O --> (C2H3O2)- + (H3O)+
36.8 mmol 0 0
- x + x + x

1.8e-5 = x^2 / (36.8 -x)
x = .02572813 = [H3O+]

-log(.02572813) = 1.59 (this is not correct)

Answer 1

Nope. You used millimols in this and you should use molarity. Ka is 1.8E-5, yes, but that is Ka = (H3O^+)(Ac^-)/(HAc) and you must substitute M and not mmols. If acetic acid is HAc, then
.......HAc+ + H2O ==> Ac^- + H3O^+
I......0.15..........0......0
C......-x.............x......x
E.....0.115-x.........x.......x

This gives you (H3O^+) = ? in mols/L which is concentration and you can use that to calculate pH. .

Answer 2

Thank you so much! That makes sense.

Answer 3

I'm sure you picked it up but that's 0.115 M from the problem and not 0.15. My typo.