Question

A student followed the procedure of this experiment to determine the percent NaOCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00mL of commercial bleaching solution to 250mL in a volumetric flask, and titrated a 20-mL aliquot of the diluted bleaching solution. The titration required 35.46mL of 0.1052M Na2S2O3 solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10g/mL.

Calculate the number of moles of S2O3^2- ion required for the titration.

Answer 1

his reaction must be carried out in a basic solution - It is necessary to add NaOH to the solution before titration.
Equation:
4 NaOCl + Na2S2O3 + 2 NaOH → 4 NaCl + 2 Na2SO4 + H2O
4mol NaOCl react with 1 mol Na2S2O3

Calculate the number of mol Na2S2O3:
Mol Na2S2O3 in 35.46mL of 0.152M solution = 35.46/1000*0.152 = 0.00539 mol Na2S2O3 used in the titration.

1) How many mol of NaOCl is in the 20mL titration sample:
From the balanced equation: 4* 0.00539 = 0.0216 mol NaOCl in 20mL

2) mol NaOCl in the 250mL of solution = 250/20* 0.0216 = 0.269 mol NaOCl

3) Therefore 50mL of the original commercial bleaching solution contains 0.269 mol NaOCl

4) Calculate mol NaOCl in 1 litre of original commercial solution = 1000/50*0.269 = 5.38 mol NaOCl

5) molar mass NaOCl = 23+35.5+16 = 74.5g/mol.
5.38mol = 5.38*74.5 = 400.81 g NaOCl in 1 litre of solution.