Question

The amount of I3–(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32–(aq) (thiosulfate ion). The determination is based on the net ionic equation

2S₂O₃^2-(aq) + I₃^- (aq)---------> S₄O₆^2-(aq) + 3I^-(aq)

Given that it requires 40.1 mL of 0.340 M Na2S2O3(aq) to titrate a 15.0-mL sample of I3–(aq), calculate the molarity of I3–(aq) in the solution.

Answer 1

Answer – We are given, reaction - 2S₂O₃^2-(aq) + I₃^- (aq)---------> S₄O₆^2-(aq) + 3I^-(aq)

Volume = 40.1 mL , [Na₂S₂O₃] = 0.340 M , volume of I₃^- = 15.0 mL

First we need to c calculate the moles of Na₂S₂O₃ = 0.340 M * 0.0401 L

                                                                          = 0.0136 moles

We know,

1 moles of Na₂S₂O₃ = 1 moles of S₂O₃^2-

So, moles o S₂O₃^2- = 0.0136 moles

From the above reaction

2 moles of S₂O₃^2- = 1 moles of I₃^-

So, 0.0136 moles of S₂O₃^2- = ?

= 0.00682 moles of I₃^-

So molarity of I₃^- = 0.00682 moles / 0.015 L

                            = 0.454 M