Question

Part 1 : A solution of sodium thiosulfate, Na₂S₂O₃, is 0.4899 M. 29.65 mL of this solution reacts with 22.30 mL of I₂ solution. What is the molarity of the I₂ solution?

2(Na₂S₂O₃) + I₂↔Na₂S₄O₆ + 2(NaI)

Part 2 : 32.99 mL of the I₂ solution from above is required to titrate a sample containing As₂O₃. Calculate the mass of As₂O₃ (197.8 g/mol) in the sample.

As₂O₃ + 5(H₂O) + 2I₂ → 2(H₃AsO₄) + 4HI

Answer 1

1)

2(Na2S2O3) + I2↔Na2S4O6 + 2(NaI)

according to balanced reaction

2 moles Na2S2O3 reacts with 1 mole I2

moles of Na2SO3 = M x V in L = 0.4899 x 29.65/1000 = 0.014

0.014 / 2 = 0.007 moles I2 required

Molarity of I25 = n /V = 0.007 / 0.0223 = 0.314 M

2)

As2O3 + 5(H2O) + 2I2 → 2(H3AsO4) + 4HI

moles of I2 = 0.314 x 32.99/1000 = 0.00103

2 moles I2 reacts with 197.8 g As2O3

0.00103 moles I2 reacts with 0.00103 x 197.8 / 2 = 0.102 g

mass of As2O3 require = 0.102 g