What volume of 0.50 mol/L Na2S2O3 is required to titrate the I2? Give your answer in mL to 1 decimal place.

Question

A 30 mL sample of 0.40 mol/L bleach solution, NaOCl, is reacted with an excess of KI solution according to the following reaction equation:

2H⁺ + OCl^- + 2I^- → Cl^- + H₂O + I₂

The I₂ produced is subsequently reacted according to the equation

2 S₂O₃^2- + I₂ → S₄O₆^2- + 2I^-

What volume of 0.50 mol/L Na₂S₂O₃ is required to titrate the I₂? Give your answer in mL to 1 decimal place. 

Answer 1

mols NaOCl = 30 ml x 1 L/1000 ml x 0.40 mol/L = 0.012 mols NaOCl

mols I₂ produced in first reaction = 0.012 mol NaOCl x 1 mol I₂ / mol NaOCl = 0.012 mols I₂

mols S₂O₃^2- required in 2nd reaction = 0.012 mols I₂ x 2 mols S₂O₃^2- / mol I₂ = 0.024 mols S₂O₃^2-

Volume of 0.50 mol/L Na₂S₂O₃ required:

(x L)(0.50 mol/L) = 0.024 mols

x = 0.048 L x 1000 ml/L = 48.0 mls