Question

Calculate Delta [S2O82-] based on the number of moles of thiosulfate consumed. This Value is the same for all reactions.

S2O8-2 + 3I- ----> 2SO4-2 + I3

Procedure The volumes of solution required for each run are tabulated b 3 should both be clean, dry, labelled 150 mm test tu nto test tubes A and B using the calibrated automatic pipets at the bac for each run are tabulated below. Test Tube A and Test Tube dry, labelled 150 mm test tubes. The solutions can be directly delivered ated automatic pipets at the back of the lab. 0.2 M (NH), SO (mL) 0.0 امت 0.4 0.6 0.7 Set 1: Varying the Concentration of Peroxodisulfate Test Tube A Test Tube B Run Starch (mL) 1.2 x 102M 0.20 M 0.2 M 0.2 M Na,s,o, KI (ML) KNO, (mL) (NH.),s,o, (mL) (mL) 11 0.1 0.2 0.4 0.4 0.8 12 0.1 0.2 0.4 0.4 0.4 3 0.1 0.2 0.4 0.4 0.2 4 0.1 0.2 0.4 0.4 0.1 Set 2: Varying the Concentration of lodide Test Tube A Test Tube B Run Starch (mL) 1.2 x 102M0.20 M 0.2 M 0.2 M Na,s,o, KI (mL) KNO, (mL) (NH2),S,O, (mL) (mL) 5 0.1 0.2 0.8 0.0 0.4 0.1 0.2 0.4 0.4 0.4 7 7 0.1 0.2 0.2 | 0.6 0.4 8 0.11 0.21 0.1 0.7 0.2 M (NH), SO (mL) 0.4 0.4 0.4 0.4 0.4 T Set 3: Varying the Temperature Repeat Run 3 of Set 1 at 4 different temperatures: 10 °C below room temperature, room temperature, 10 °C above room temperature, and 20°C above room temperature.

Answer 1

If you can assume that the reaction occurs completely, you can use stoichiometry to calculate the number of moles of peroxodisulfate the reacted. For example, for the first reaction in the first table, the number of moles of peroxodisulfate is:

\(n=M \cdot V(L)=0.2 \frac{\mathrm{mol}}{L} \cdot 8 x 10^{-4} L=1.6 x 10^{-4} \mathrm{moles}\)

And the number of moles of iodide is:

\(n=M \cdot V(L)=0.20 \frac{\mathrm{mol}}{L} \cdot 4 x 10^{-4} L=8 \times 10^{-5} \mathrm{moles}\)

From the reaction, we can see that 3 moles of iodide react with one mole of peroxodisulfate, so the number of moles of peroxodisulfate that is \(1 / 3\) of the number of moles of iodide. In this case:

\(n_{\text {react }}=\frac{8 x 10^{-5} \text {moles }}{3}=2.67 x 10^{-5}\)

With this calculated, we can calculate the initial and final concentration of peroxodisulfate. The initial concentration is given by (the total volume of the mixture is \(1.9 \mathrm{~mL}\) ).

\(c_{i}=\frac{n_{\text {initial }}}{V}=\frac{1.6 x 10^{-4} \text {moles }}{0.0019 L}=0.0842 \mathrm{M}\)

And the final concentration is:

\(c_{f}=\frac{n_{\text {initial }}-n_{\text {react }}}{V}=\frac{\left(1.6 x 10^{-4}-2.67 x 10^{-5}\right) \text {moles }}{0.0019 L}=0.0702 M\)

So the change in concentration is:

\(\Delta c=0.0702 M-0.0842 M=-0.0140 M\)

We can calculate the value for the other reactions in the same way:

Reaction	delta c
1	-0.0140 M
2	-0.0140 M
3	-0.0140 M
4	-0.0105 M In this case, the number of moles of I- is higher than 3 times the number of moles of peroxodisulfate, so we can assume that all the peroxodisulfate will react.

Reaction	delta c
5	-0.0281 M
6	-0.0140 M
7	-0.0070 M
4	-0.0035 M