Question

In a titration, 10.4 mL of 0.0500 M Na2S2O3 was required to turn 15.0 mL of an aqueous solution containing I2, I3-, and starch from dark blue to colorless.

1. How many moles of S2O32- were added?

2.How many moles of I2 and I3- were present in the solution altogether?

3.What is the concentration (in M) of the total reactive iodine species? (I2 and I3- together)

4.From a separate titration with I2 in hexanes, you learned that the concentration of I2 in the aqueous phase is 1.4 x 10-5 M. What is the concentration of I3- (in M)?

Answer 1

1. Moles of S₂O₃^2- = molarity * volume = 0.0500*10.4*10^-3 = 0.00052.

2.

Balanced reaction

I₂ + 2 S₂O₃^2- S₄O₆ ^2- + 2 I^,-

I₃^- + 2S₂O₃^2- S₄O₆^2- + 3 I^-

hence, 4 mole S₂O₃^2- reacts with 2 mole I₂ and I₃^- together.

So, 0.00052 moles S₂O₃^2-reacts with (0.00052/2) = 0.00026 mole I₂ and I₃^- together.

3.

Concentration = moles of I₂ and I₃^- together / volume ( L)

= [(0.00026)/15*10^-3 ]

= 0.01733 M

4.

Total concentration = 0.01733 M

Concentration of I₂ = 1.4*10^-5 M

Then , concentration of I₃^- = ( 0.01733 - 1.4*10^-5)

= 0.017316 M