Calculate the pH in the titration of 25 mL of 0.10 M acetic acid by sodium hydroxide after the addition to the acid solution of 35 mL of 0.10 M NaOH.

Answer 1

I will let HAc stand for acetic acid.
millimoles HAc initially = mL x M = 25 mL x 0.1 M = 2.5
millimoles NaOH added = 35 x 0.1 = 3.5
..............HAc + NaOH ==> NaAc + H2O
initial.......2.5.........0..............0.............0
add.......................3.5.................................
change....-2.5......-2.5.........+2.5..........+2.5
equilibrium..0..........1.0...........2.5..........2.5
You see that after the addition of the NaOH, all of the HAc is used and you have an excess of 1 mmol NaOH remaining. The pH is determined by the amount of NaOH. That is in a total volume of 25 mL + 35 mL = 60 mL; therefore, (NaOH) = mmols/mL = 1.0/60 = 0.0167 M.
Thus the (NaOH) = 0.0167 M = (OH-) . Convert to (H^+) from
(H^+)(OH^-) = Kw = 1E-14. You know Kw and OH^-, solve for H^+, then convert to pH from pH = -log (H^+)
Post your work if you get stuck.