AP chem stochiometry

Question

When 25.0 mL of a solution of 0.200 M NAIO3 was added to acidified iodide ions, the iodine produced reacted with 20.3 mL of Na2S2O3. Calculate the concentration of the Na2S2O3 solution given the complete, balanced redox equation below.

IO3^-1 + 5 I + 6 H^-1 --> 3 I2 + 3 H2O

I2 + 2 S2O3^-2 --> 2 I^-1 + S4O6^-2

Answer 1

So, to approach this problem, you want to go step by step and follow the procedure of the experiment.

To begin, we find the mols of IO₃^- being used:

25.0 ml x 1 L / 1000 mls x 0.200 mol / L = 0.005 mols IO₃^-

We then use the oxidation reaction to find mols of I₂ produce:

IO₃^- + 5I^- + 6H⁺ ==> 3I₂ + 3H₂O (correct equation; the one you provided is incorrect)

0.005 mols IO3- x 3 mol I2 / 1 mol IO3- = 0.015 mols I₂ produced

Now we use the reduction reaction to find the mols of S₂O₃^2-:

I₂ + 2S₂O₃^2- ==> 2I^- + S₄O₆^2-

0.015 mol I₂ x 2 mols S₂O₃^2- / mol I₂ = 0.030 mols S₂O₃^2-

Finally, to obtain the concentration of S₂O₃^2- we simply divide the mols by the volume in liters...

0.030 mols S₂O₃^2- / 0.0203 L = 1.478 mols / L = 1.48 M (3 sig. figs.)