2. Assume that you mixed 20.00 mL of 0.040 M KI with 20.00 mL of 0.060 M (NH4)2S

Question

2. Assume that you mixed 20.00 mL of 0.040 M KI with 20.00 mL of 0.060 M (NH4)2S20s 1000 mL of 0.00070M Na2s203 and a few drops of starch. The point of mixing sets time 0. mixing but before any reaction has occurred. (1 mark) Hint: your calculated [KI] should equal 0.016 M. TOTAL VOLUME 50m L M.Le Ku z O.o2OL. O. 016 M 0.05 L o .024 M O. OSL MOLES N as (b) The basis of the "Method of Initial Rates used in this experiment, is that the concentrations of reagents are essentially unchanged during the measurement time period. Calculate the of the initial (NH4)2S208 that has reacted when the blue colour appears. (1 mark)

Explanation / Answer

The reaction will be

2 KI + (NH4)2S2O8 I2 + K2SO4 + (NH4)2SO4

The reaction produces I2 however it will not combine with starch to give blue colour as the I2 produced will be get converted by reaction with Na2S2O3.

I2 + 2 Na2S2O3 2NaI+ Na2S4O6 (sodium tetrathionate)

So as soon as Na2S2O3 gets exhausted the I2 produced will combine with starch to give blue colour.

So the amount of (NH4)2S2O8 will be equal to Na2S2O3 before the reaction starts

[Na2S2O8] =0.00014 M

Initial concentration = 0.024

So % = 0.00014 X 100/ 0.024 = 58.33 %

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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