Titration Question

Question

A 30 mL sample of 0.45 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 0.80 mol/L Na₂S₂O₃ is required to titrate the I₂? Give your answer in mL to 1 decimal place. 

Answer 1

For reaction 1, NaOCl is the limiting reagent since KI is in excess:

2H⁺ + OCl^- + 2I^- → Cl^- + H₂O + I₂

30 mL NaOCl = 0.03 L NaOCl (0.45 mol NaOCl/ 1 L NaOCl) = 0.0135 mol NaOCl

Since ^–OCl: I₂ is a 1:1 ratio, 0.0135 mol NaOCl = 0.0135 mol I₂.

For reaction 2, we can calculate the amount of S₂O₃ that will react with I₂ based on stoichiometry ratios:

2 S₂O₃^2- + I₂ → S₄O₆^2- + 2I^-

0.0135 mol I₂ (2 mol S₂O₃^2- / 1 mol I₂ ) = 0.027 mol S₂O₃^2-

Must calculate the volume of S₂O₃ based on the moles required and the concentration given:

0.027 mol NaS₂O₃ (1 L Na₂S₂O₃/ 0.8 mol Na₂S₂O₃) = 0.03375 L = 33.75 mL Na₂S₂O₃