What is the concentration of acetate ions at the stiochiometric point in the titration of

Question

0.018 M CH3COOH(aq) with 0.036M NaOH(aq) ? acetic acid Ka is 1.8 x 10^ -5

Answer 1

CH3COOH + NaOH => NaCH3COO + H2O

If you start with x mL of 0.018 M acid it will take 0.5x mL of 0.036 M NaOH to titrate it. You will produce 0.018 / 1.5 = 0.0135 M NaCH3COO.

At the endpoint you have a solution of sodium acetate. To find the pH, consider the hydrolysis of acetate ion:

CH3COO-  +  H2O  <=> CH3COOH  +  OH-

0.0135 M C2H3O2- will produce x M each of

CH3COOH and OH-. There will be 0.0135-x M CH3COO- left.

Kb CH3COO- = Kw / Ka = 1 x 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10

Ka = [CH3COOH][OH-] / [CH3COO-]

= (x)(x) / (0.0135 - x)

Since -x will be small compared to 0.0135, delete it.

Ka = 5.6 x 10^-10 = x^2 / 0.0135

7.6 x 10^-12 = x^2; 2.7 x 10^-6 = x = [OH-]

pOH = -log [OH-] = -log (2.7 x 10^-6) = 5.6

pH = 14 - pOH = 14 - 5.6 = 8.4