Please show me the light..... I don\'t know what I need to do. The experiment do

Question

Please show me the light..... I don't know what I need to do.

The experiment done this week in the lab involved calculating the rate of the reaction below at various concentrations of the reactants. 2I^-(aq) + S_2O_8^2- rightarrow I_2(aq) + SO_4^2- (aq) In order to calculate the rate of reaction, it was necessary to calculate the change in the iodine concentration. In order to do this a known small amount of another sulfur-containing ion was added to the solution. This ion reacted with the iodine formed. When it was consumed, the solution turned blue. Which of the ions below is the one that is added in small amount? S_4O_62^- S_2O_8^2- S_2O_3^2- SO_4^2-

Explanation / Answer

We have the reaction

2I- + S2O8 ^2- ---> I2 + SO42-

now formed I2 is made to react with S2O3^2-

we have reaction I2 + 2S2O3^2- ---> S4O6 ^2- + 2I-

This reaction is much faster than given one. We use kow amount of S2O3^2-

Thus we can measure I2 which is formed reacted with S2O3^2-. From this we can find I- presnet.

Thus overall due to 2nd reaction we can measure I- present

Answer is S2O3^2-

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.