Question

An unknown amount of acid can often be determined by adding an excess of base and then "back-titrating" the excess. A 0.3471 g sample of a mixture of oxalic acid, which has two ionizable protons and benzoic, which has one, is treated with 94. mL of 0.1060 M NaOH. The excess NaOH is titrated with 21.00 mL of .280 M HCl. Find the mass % of benzoic acid.

Answer 1

no of moles of HCl needed to titrate the excess NaOH = 0.280 M x 0.021 L=0.00588 moles
no of moles of NaOH titrated = 0.00588 moles
moles NaOH initially used = 0.094 L x 0.1060 M = 0.009964 moles
no of moles NaOH actually required to titrate mixture of acids = 0.009964 - 0.00588= 0.004084 moles

The reactions:
C₆H₅COOH + NaOH = C₆H₅COONa + H₂O                  ; equivalent weight of benzoic acid = 122.1 g/mol

C₂H₂O₄ + 2 NaOH = C₂O₄Na₂ + 2 H₂O          ;equivalent weight of oxalic acid = 90/2 = 45 g/mol


let x = mass benzoic acid

let y = mass oxalic acid

x + y = 0.3471                                   .....eqn 1

(x /122.1)+ (y/45) = 0.004084

or, 45 x + 122.1 y=22.44                .....eqn 2

or, 45 x + 122.1 (0.3471-x) =22.44             
or, 45x + 42.38 - 122.1x = 22.44

or, 77.1 x = 19.94

x = 0.2586 g

% of benzoic acid = 0.2586 * 100 /0.3471 = 74.51 %