Calculate the (H^+) contributed by the acetic acid remembering that the HCl will suppress the ionization of the acetic acid. Then add the H^+ contributed by the HCl.. Acetic acid is HAc.
(HAc) = 0.0100 x 25/36 = approx 0.007 M
(HCl) = 0.0100 x 11/36 = 0.003 M
....................HAc ==> H^+ + Ac^-
I.................0.007.....0.003.......0
C.................-x.............x...........x
E............0.007- x....0.003+x....x
Then Ka = (H^+)(Ac^-)/(HAc) = (0.003+x)(x)/(0.007-x)
Solve for x, then total H^+ = 0.003+x
Post your work if you get stuck.
11 mL of 0.0100 M HCl are added to 25 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?
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