Question

I have a reaction that has three elementary steps:

3I^- +S₂O₈^2- ---> I₃^- + 2 SO₄^2- (this is the RLS)
I₃ + 2 S₂O₃^2- ---> 3 I- + S₄O₆^2-
starch + I₃^- ---> starch - I₃^- complex. (color change to indicate rxn is over)

In multiple trials I change the concentrations to see its effect on the rate of the rxn. But the reagents I'm using in this rxn are: (1) starch, Na₂S₂O₃ , KI, KNO₃ , and (2) (NH₄)₂S₂O₈ , (NH₄)₂SO₄

(1) and (2) are mixed together in separate vials before we mix both to start the rxn. Different trials have different volumes of the above reagents so there's a difference in concentrations between trials and we can therefore see the effect this has on rate. We time each rxn with a stopwatch.

I need to calculate the initial [I^-] and [S₂O₈^2-] by using the equation M₁V₁ = M₂V₂. However I'm completely confused how to do this. V₁ is initial volume and V₂ is total volume. I'm also given the molarity of KI and (NH₄)₂S₂O_8. Do I need to factor in that my reagents are not actually S₂O₈^2- or I^- but actually (NH₄)₂S₂O₈ and KI?

For example, trial 1's reagents are:

Starch - .2% - .10 mL
Na₂S₂O₃ - .20 mL of .012 M
KI - .80 mL of .20 M
KNO₃ - 0 mL of .20 M

(NH₄)₂S₂O₈ - .40 mL of .20 M
(NH₄)₂SO₄ - .40 mL of .20 M

What is the intial [I^-] and [S₂O₈^2-] for this trial?

Answer 1

We know,

Molarity = moles / Litres of solution

Work out how many moles of each reagent you add to the initial solution.

Then use the total volume of the initial solution to determine the molarity.

eg trial 1

total volume = 0.10 ml + 0.12 ml + 0.80 ml + 0 mL + 0.40 mL + 0.40 mL

                     = 1.82 mL = 0.00182 L

moles of [S2O8^2-] = moles in the 0.40 ml (or 0.0004 L) of solution used             

                          = molarity x Litres

                          = 0.2 M x 0.0004 L

                          = 8 x 10^-5 moles S2O8^2-

Initial concentration of [S2O8^2-] = moles added / total volume initial solution

                                   = 8 x 10^-5 mol / 0.00182 L

                                   = 0.044 M [S2O8^2-]

Similarly,

moles of [I-] = moles in the 0.80 ml (or 0.0008 L) of solution used             

                          = molarity x Litres

                          = 0.2 M x 0.0008 L

                          = 1.6 x 10^-4 moles [I-]

Initial concentration of [I-] = moles added / total volume initial solution

                                   = 1.6 x 10^-4 mol / 0.00182 L

                                   = 0.088 M [I-]

Thus the intial concentrations of [S2O82-] = 0.044 M and [I-] = 0.088 M

Answer 2

No, you don't need to factor in that for the reagents as they are not actually S₂O₈^2- or I^- but actually (NH₄)₂S₂O₈ and KI. That is not required.

From the RLS

3I^- +S₂O₈^2- ---> I₃^- + 2 SO₄^2-

But, assuming that the concentrations of reagents for the trial=1 are what is present there in the start of the reaction, i.e., the amount you have taken, it will have the same initial concentration of I^- and S₂O₈^2- .

Thus, the initial concentrations of I^- and S₂O₈^2- are 0.2 M each.