Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 7.00 ✕ 102 mL of solution and then titrate the solution with 0.153 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

Number of moles of benzoic acid = 0.235 g / 122.12 g/mol=0.00192 moles

At equivalence point moles NaOH required = 0.00192 mole
volume of NaOH required = 0.00192
volume of NaOH = 0.00192 / 0.148 M = 0.0129 L
total volume = 0.0129 L + 0.100 L = 0.1129 L

moles benzoate formed = 0.00192
concentration of benzoate = 0.00192 / 0.1129 L= 0.017 M

C6H5COOH + OH - -------> C6H2COO- + H2O

As we know Kb = Kw/Ka = 1.0 x 10^-14 / 6.3 x 10^-5 = 1.6 x 10^-10 M

1.6 x 10^-10 = x^2 / 0.017- x

x = [OH-] = [C6H5COOH]= 1.64 x 10^-6 M (concentration of OH-)

Concentration of [H+]= 1.0 x 10^-14 / 1.64 x 10^-6 = 6. 09 x 10^-9
pH = - log [H+] = -log [6.09 X 10^-9 ] = 8.21

and concentration of [Na+]= 0.00192 / 0.1129 L=0.017 M