Consider the titration of a 35.0-mL sample of 0.175 M HBr with 0.200 M KOH. Determine the pH after adding 5.0 mL of base beyond the equivalence point.

Answer 1

Determine where the equivalence point is. Note the volume.
mmoles OH^- added in excess is 5.0 mL x M = ??
total volume = mL to arrive at the equivalence point + 5.0 from the base
mmoles/total mL = M of OH^-
pOH = -log(OH^-)
and pH + pOH = pKw = 14. Solve for pH.