Question

A 0.1936 gram sample of primary standard K2Cr2O7 was dissolved in water, the solution was acidified, and after the addition of KI, the titration of liberated I2 required 33.61 ml of Na2S2O3 solution. Calculate the molarity of the thiosulfate solution.

Answer 1

Number of moles of K₂Cr₂O₇ is , n = mass/molar mass

                                                   = 0.1936 g / 294(g/mol)

                                                   = 6.58x10^-4 mol

K₂Cr₂O₇ + 6 KI + 7 H₂SO₄ Cr₂(SO₄)₃ + 4 K₂SO₄ + 3 I₂ + 7 H₂O

According to the balanced equation ,

1 mole of K₂Cr₂O₇ produces 3 moles of iodine

6.58x10^-4 mol of K₂Cr₂O₇ produces 3x6.58x10^-4 mol of iodine

                                                    = 1.97x10^-3 mol

2 Na₂S₂O₃ + I₂ Na₂S₄O₆ + 2 NaI

From this equation

1 mole of iodine reacts with 2 moles of Na₂S₂O₃

1.97x10^-3 mol of iodine reacts with 2x1.97x10^-3 mol of Na₂S₂O₃

                                                 = 3.95x10^-3 mol

So Molarity of Na₂S₂O₃ , M = number of moles / volume in L

                                         = 3.95x10^-3 mol / (33.61/1000) L

                                         = 0.117 M

Therefore the Molarity of Na₂S₂O₃ is 0.117 M