Question

A solution is made by mixing 500 ml of 0.172 M NaOh with exactly 500 ml of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below.
Ka of CH3COOH is 1.8 x10^-5

[H+]=?
{OH-]=?
[CH3COOH]=?
[Na+]=?
[CH3COO-]=?

Answer 1

mols NaOH = M x L = approx 0.0860
molw CH3COOH = approx 0.05 = HAc

........NaOH + HAc ==> NaAc + H2O
I....0.0860.....0.......0.......0
Add...........0.050..............
C.....-0.050.-0.050....0.050...0.050
E......0.036....0......0.050

So you have 0.0360 mols NaOH
0 mols HAc
0.050 mols NaAc = CH3COONa
all in 500 + 500 = 1000 mL H2O(assuming the volumes are additive).

M = mols/L each except OH^-
You find that from the hydrolysis of the Ac^- which is done as follows:
(Ac^-) = 0.05mols/L = 0.05
.........Ac^- + HOH ==> HAc + OH^-
I........0.05.............0....0
C.........-x..............x.....x
E.......0.05-x............x.....x

Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05-x) and solve for x = (OH^-). From that you can obtain H^+.
Post your work if you get stuck.