Question

Assume you dissolve 0.281 g of the weak acid benzoic acid in enough water to make 1.00x10^2 mL of solution and then titrate the solution with 0.171 M NaOH. Ka=6.3x10^-5.

what are the concentrations of Na+ H3O+ OH- and C6H5CO2-

Answer 1

molarity of weak acid = 0.281 g/122.123 g/mol x 0.1 L

                                   = 0.023 M

Titrated with 0.171 M NaOH

Volume of NaOH needed to reach equivalence point = 0.023 M x 0.1 L/0.171 M

                                                                                     = 0.01345 L

molarity of [C6H5COONa] formed = 0.023 M x 0.1 L/(0.1 L + 0.01345 L)

                                                       = 0.0203 M

C6H5COO- + H2O <==> C6H5COOH + OH-

let x amount has hysrolyzed

Kb = Kw/Ka = [C6H5COOH][OH-]/[C6H5COO-]

1 x 10^-14/6.3 x 10^-5 = x^2/0.0203

x = [OH-] = 1.80 x 10^-6 M

pOH = -log[OH-] = 5.746

pH = 14 - pOH = 8.254

[H3O+] = Kw/[OH-] = 1 x 10^-14/1.80 x 10^-6 = 5.572 x 10^-9 M

[C6HCOO-] = 0.0203 - 1.80 x 10^-6 = 0.020282 M