A 20.0 mL sample of 0.687 M HBr is titrated with a 0.216 M NaOH solution. The pH after the addition of 10.0 mL of NaOH is __________.

Answer 1

initial mols HBr = M x L = about 0.014
mols NaOH added = about 0.002
but you need to do them more accurately.
........NaOH + HBr ==> NaBr + H2O
I.........0....0.014....0.......0
added....0.002.....................
C.......-0.002..-0.002..0.002..0.002
E.........0.....0.0120...0.002.0.002

So what you have at 10.0 mL is a solution of HBr which is a strong acid (100% ionized); therefore, (H^+) = (HBr) so pH = -log(HBr)
Note: (HBr) = mols/L = 0.012/(0.010+0.020) = ?

Answer 2

Thank you so much that cleared up a lot!