Question

50 ml of 0.02M acetic acid is titrated with 0.1M NaOH.Calculate the pH of the solution when 10ml of NaOH is added

Answer 1

50 mL x 0.02M = 1 mmol HAc.
10 mL x 0.1M = 1 mmol NaOH.

HAc + NaOH ==> H2O + NaAc
1 mmol 1mmol = e.p.

So you have a solution of Ac^- in aqueous solution. The (Ac^-) is 1 mmol/60 mL = about 0.017M but you nee to do it more accurately.
........Ac^- + HOH ==> HAc + OH^-
I........0.017...........0.....0
C........-x.............x......x
E......0.017-x..........x......x

Kb for Ac^- = (Kw/Ka for HAc) = (HAc)(OH^-)/(Ac^-) and solve for x = OH^-, then convert to pH. I expect you will need to use the quadratic.

Answer 2

0.0012