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 .2080 M HCl. Find the mass % of benzoic acid.

Answer 1

moles HCl = 0.2080 M x 0.021 L= 0.004368

moles NaOH titrated = 0.004368

moles NaOH used = 0.94 L x 0.1060 M = 0.009964

moles NaOH used to titrate oxalic acid + benzoic acid = 0.09964 - 0.004368= 0.005596

the balanced equations are
C2H2O4 + 2 NaOH -------------------> C2O4Na2 + 2 H2O

C6H5COOH + NaOH -------------------> C6H5COONa + H2O

let x = mass oxalic acid ( molar mass =90.0 g/mol)

let y = mass benzoic acid ( molar mass = 122.1 g/mol)

x + y = 0.3471 -------------> 1

x / 90 /2 + y/ 122.1 = 0.005596

x / 45 + y / 122.1 = 0.005596

122.1 x + 45 y = 30.747 -------------> 2


x = 0.1962g

y =0.1509

% benzoic acid  = 0.1509 x 100/ 0.3471 = 43.5 %

% benzoic acid = 43.5 %