12 mL x 0.01M HCl = 0.12 mmoles.
25 mL x 0.08M Ac^0 = 2.00 mmoles.
25 mL x 0.01M HAc = 0.25 mmoles.
...........Ac^- + H^+ ==> HAc
initial....2.00....0.......0
added.............0.12...........
change....-0.12...-0.12....0.12
equil.....1.88......0.....0.12
Substitute the ICE values into the Henderson-Hasselbalch equation and solve for pH.
12 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.08 M C2H3O2-. What is the pH of the resulting solution?
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