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 acid which has one, is treated with 91.0 mL of 0.1010 M NaOH. The excess NaOH is titrated with 21.50 mL of 0.2060 M HCl. Find the mass % of benzoic acid

Answer 1

moles of HCl in 21.5ml of 0.2060M= 0.2060*21.5/1000=0.004429 moles

Excess of mole of NaOH= 0.004429 moles

moles of NaOH used= 0.1*91/1000=0.0091 moles

NaOH consumed = 0.0091-0.004429 =0.004671 moles

let x moles of NaOH is consumed for oxalic acid , remaining 0.004671-x moles are consumed by Benzoic acid

Reaction between oxalic acid and NaOH can be represented as

H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(i)

1 mole of oxalic acid requires 2 moles of NaOH

x moles of NaOH requires x/2 moles of oxalic acid

Molecular weight of oxalic acid (H2C2O4)= 2+22+64= 90

Mass of oxalic acid =90*x/2= 45x

The reaction between Benzoic acid and NaOH is C6H5COOH+ NaOH----> C6H5COONa+ H2O

1 mole of NaOH is consumed by 1 mole of C6H5COOH

(0.004671-x) moles of NaOH consumes (0.004671-x) moles of Benzoic acid

Molecular weight of Benzoic acid =122 , mass of Benzoic acid =122*(0.004671-x)

given the total mass of sample of Benzoic acid and oxalic acid = 0.3471 g

45x+(0.004671-x)*122= 0.3471

45x+122*0.004671- 122x= 0.3471

x*(122-45)= 0.569862-0.3471

x= 0.002893 moles of NaOH used for Oxalic acid

mass of oxalic acid =45*0.002893=0.1302 mass of Benzoic acid in the sample=0.3471-0.1302=0.2169 gm

Mass % of Benzoic acid = 100*0.2169/0.3471=62.4892%