Calculate the pH of a solution obtained by mixing 30 mL of 0.75 M CH3COOH with 15 mL of 1.5 M NaOH.

Answer 1

millimoles NaOH = mL x M = 15 x 1.5 = 22.5
mmoles CH3COOH = 30 x 0.75 = 22.5 where CH3COOH = HAc.
.........HAc + NaOH ==> NaAc + H2O
initial..22.5...22.5.....0......0
change..-22.5..-22.5....+22.5..22.5
equil.....0......0.......22.5....

So you can see that the CH3COOH is exactly neutralized by the NaOH and neither free CH3COOH nor free NaOH come through the reaction. We have just a solution of CH3COONa (NaAc) and that is hydrolyzed. The pH is determined by that. (Ac^-) = mmoles/mL = 22.5/45 = 0.5M
..........Ac^- + HOH ==> HAc + OH^-
initial..0.50....0.......0.....0
change.....-x..............x......x
equil......0.50-x..........x.......x

Kb for Ac^- = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)
Kw = 1E-14 for water.
Ka = 1.8E-5 for acetic acid
x = (HAc) = (OH^-)
(Ac^-) = 0.50-x
Substitute, solve for x (which = OH^-), convert to pOH, then to pH.

Answer 2

gibbe that answer b0ss