Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 9.00 ✕ 10^2 mL of solution and then titrate the solution with 0.163 M NaOH. C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O(ℓ) What are the concentrations of the following ions at the equivalence point? Na+, H3O+, OH- C6H5CO2- What is the pH of the solution?

Answer 1

Moles benzoic acid = 0.235 g / 122.12 g/mol = 0.00192 mol
Concentration benzoic acid = 0.00192 mol/ 0.900 L = 0.00213 M

at equivalence point moles NaOH needed = 0.00192 mol
Volume NaOH = 0.00192 / 0.163 = 0.0118 L
total volume = 0.0118 + 0.900 =0.9118 L

At equivalence point we find only C6H5COO-
[C6H5COO-] = 0.00192 / 0.9118 = 0.0021 M

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

1.6 x 10^-10 = x^2 /0.0021 -x

solve for x ,
x = [OH-] = [CH3COOH] = 5.79 x 10^-7 M
[H+] = 10^-14 / 5.79 x 10^-7 = 1.72 x 10^-8 M
pH = -log[H+] = 7.76