Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.00 ✕ 102 mL of solution and then titrate the solution with 0.168 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-

M Na+ ?

M H3O+ ?

M OH- ?

M C6H5CO2- ?

What is the pH of the solution?

Answer 1

Given mass of benzoic acid is 0.235 g and the molecular weight of benzoic acid is 122 g/mol.

Calculate the moles of benzoic acid as follows:

Moles of benzoic acid =mass of acid/molecular weight of acid

Moles of benzoic acid = 0.235 g / 122 gmol^-1 = 0.00192 mol

Molarity of acid =moles of acid /volume of solution

Molarity of benzoic acid = 0.00192 mol/ 102 mL*10^-3 = 0.018 M .

C₆H₅CO₂H(aq) + OH-(aq) ------> C₆H₅CO₂^-(aq) + H₂O(ℓ)

One mole of acid requires one mole of acid, so 0.00192 moles of acid requires 0.00192 moles of NaOH

So at equivalence point, moles of NaOH needed = 0.00192
Volume of NaOH = 0.00192 mol / 0.168mol L^-1= 0.0114 L
Total volume = 0.0114 + 0.102 = 0.1134 L
Na⁺ = 0.102 L x 0.168 =0.017 M
At equivalence point ,C₆H₅COO^-
[C₆H₅COO^-] = 0.00192 mol / 0.1134 L= 0.0169 M
The dissociation constant Ka is  6.3 x 10^-5 .

C₆H₅COO^- + H₂O <----> C₆H₅COOH + OH-

K = Kw/Ka = 10^-14 / 6.3 x 10^-5 = 1.6 x 10^-10
Let x be the moles of reacted

At equilbrium,

K = x² /(0.0169 -x)

1.6 x 10^-10= x² /(0.0169 -x)

x = [OH^-] = [C₆H₅COOH] = 1.64 x 10^-6 M
[H⁺] = 10^-14 /1.64 x 10^-6 =6.09 x 10^-9 M

pH=-log[6.09*10^-9]
pH = 8.21