Consider the titration of 50.0 mL of 0.200 M HOCl by 0.200 M KOH for the next five questions (Ka for HOCl = 3.5 x 10-8).

Question

Calculate the pH after 150.0 mL of KOH has been added.

Answer 1

KOH + HOCl => H2O + KCl

On these problems it helps to know where the equivalence point is located.
mL HOCl x M HOCl = mL KOH x M KOH
Substitute and solve for mL KOH. I get 50 mL NaOH needed. Therefore, you know this 150 is well past the equivalence point so the pH is determined by the excess KOH present
You started with 50 x 0.2 = 10 mmoles HOCl
You added 150 mL x 0.2 = 30 mmols KOH.

..........HOCl + KOH ==> KCl + H2O
I..........10.....0.......0.....0
add .............30...............
C.........-10...-10......+10....+10
E..........0.....20......+10....+10

Therefore, you have 20 mmols KOH in 150 + 50 = 200 mL solution.
(OH^-) = 20/200 = ?
pOH = -log(OH^-)
Convert pOH to pH.