Consider the titration of 100.0mL of 0.10 M malonic acid with 0.10 M NaOH. What is the pH after the addition of 175 mL NaOH?

Answer 1

Let's call malonic acid H2M.
millimols H2M = mL x M = 100 x 0.1 = 10
millimols NaOH = 175 x 0,1 = 175 so you have two equivalence points. The first H is titrated by the first 100 mL of NaOH like this.
.........H2M + OH^- ==> HM^- + H2O
I........10..........0...............0............
add................10................................
C.....-10.........-10.............+10
E.........0............0..............+10
At the end of the first 100 mL you have left 10 mmols HM^- and the next 75 mL titrates the second H like this.
............HM^- + OH^- ==> M^2- + H2O
I...........10..........0..............0..................
add..................7.5....................................
C.........-7.5.....-7.5...........+7.5................
E............2.5......0...............7.5

So this gives you a buffered solution with HM^- as the acid and M^2- as the base. Substitute this into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa2 + log [(M^2-)/(HM^-)]
Post your work if you get stuck.