Question

Calculate the pH of a 0.5 M solution of NaOCl. The pKa of HOCl is 7.5. (A: 10.1)
I know the answer but I don't know how to solve this problem.

Answer 1

I made an ICE chart
HOCl + H2O <==> OCl + OH-
I 0.5......-.................0........0
E 0.5 - x.................x..........x

Ka = x^2/0.5-x
Ka = 10^-pka = 10^-7.5 = 3.16e-8. then use that value to sub in the Ka formula to find x.

Answer 2

The ICE chart is OK but you don't have HOCl. You have NaOCl.
.............OCl^- + HOH --> HOCl + OH^-
I..............0.5......................0..............0
E..............0.5-x..................x..............x
Kb for OCl^- = Kw/Ka for HOCl = (HOCl)(OH^-)/(OCl^-)
Plug in the E line solve for x = (OH^-), then convert to pOH then pH.
Post your work if you get stuck.
The equation you wrote with HOCl is for the ionization of HOCl. That would be HOCl + H2O ==> H3O^+ + OCl^- but that isn't what you want for the hydrolysis of anion part of the NaOCl salt.