Question

Assume you dissolve 0.216 g of the weak acid benzoic acid in enough water to make 100 mL of solution and then titrate the solution with 0.144 M . ( Ka for benzoic acid = 6.3 times 10^-5.) a) What was the pH of the original benzoic acid solution? b) What are the concentrations of all of the following ions at the equivalence point: Na, H3O, OH, and C6H5CO2? c) What is the pH of the solution at the equivalence point?

Answer 1

Moles benzoic acid = 0.216 g / 122.12 g/mol = 0.00177
Concentration benzoic acid = 0.00177 mol/ 0.100 L = 0.0177M

at equivalence point moles NaOH needed = 0.00177
Volume NaOH = 0.00177 / 0.144 = 0.0123 L
total volume = 0.0123 + 0.100 =0.1123L

At equivalence point we find only C6H5COO-
[C6H5COO-] = 0.00177 / 0.1123 = 0.01576 M

C6H5COO- + H2O <----> C6H5COOH + OH-
K = Kw/Ka = 10^-14 / 6.3 x 10^-5 = 1.6 x 10^-10 =
= x^2 /0.01576 -x

1.6*10^-10*0.01576 - 1.6*10^-10*x = x^2
x = [OH-] = [CH3COOH] = 2.7 x 10^-6 M
[H+] = 10^-14 / 2.7 x 10^-6 =3.7 x 10^-9 M
pH = 8.43
Na+ = 0.100 L x 0.144 =0.0144 M