Terephthalic( C8H6O4) is a diprotic acid with pka's 3.51 and 4.82.

Question

a. What is the pH of a 0.15M solution of this acid?
b. What is the concentration of C8H4O4-2 in a 0.15M solution f this acid?

Answer 1

Let's call terephthalic acid H2T to make it simpler to write. You will have (H^+) from both ka1 and ka2. Take them one at a time.
............H2T ==> H^+ HT^-
I............0.15........0......0
C............-x............x......x
E,,,,,,,,0.15-x..........x......x

Ka1 = (H^+)(HT^-)/(H2T)
Plug in the E line, solve for x = (H^+) and convert to pH. I've estimated x = 0.007 M but that's just an estimate. You need to calculate it yourself and make sure you can use the assumption that x is about zero if you choose to go that route. Then for Ka2.
............HT^- ==> H^+ T^2-
I...........0.007......0.007...0
C...........-x.............x..........x
E........0.007-x....0.007+x....x

plug the E line into Ka2 expression and solve for x = H^+ from Ka2.
Total H^+ = H from ka1 + H from ka2.

For part B that is x from Ka2 calculation.

Post your work if you get stuck.