6. Iodine Thiosulfate Titrations

Question 1

what kind of titration is an Iodine thiosulfate titration

Answer 1

a redox titration

Question 2

what can iodine thiosulfate titrations find the concentration of

Answer 2

oxidising agents

oxidising agents accept electrons and get reduced

Question 3

the more concentrated the oxidising agent=

Answer 3

the more ions will be oxidised

Question 4

3 stages involved in using iodine thiosulfate titration to find the concentration of an oxidising agent potassium iodate (V)

Answer 4

1. use sample of oxidising agent to oxidise as much iodine as possible
2. find out how many moles of iodine have been produced
3. calculate the concentration of the oxidising agent

Question 5

e.g. using oxidising agent potassium iodate (V)

stage 1. use sample of oxidising agent to oxidise as much iodine as possible

Answer 5

1. measure out a certain volume of potassium iodate (V) solution (KIO3) (the oxidising agent - 25.0 cm3
2. add to an excess of acidifies potassium iodide solution (KI). the iodate (V) ions in potassium iodate(V) solution oxidise some of the iodide ions to iodine

IO3(aq) + 5I-(aq) + 6H+(aq) -> 3I2(aq) + 3H2O(l)

Question 6

e. g. using oxidising agent potassium iodate (V)

stage2. find out how many moles of iodine have been produced

Answer 6

1. titrate resulting solution with sodium thiosulfate (Na2S2O3).
   Iodine in solution reacts with thiosulfate ions
   I2(aq) + 2S2O32-(aq) -> 2I-(aq) + S4O62- (aq)

Question 7

titration of iodine with Sodium Thiosulfate

experiment steps

Answer 7

1. take flask containing solution produced in stage 1
2. from burette add sodium thiosulfte solution to flask drop by drop
3. when iodene colour fades to pale yellow addd 2cm3 of starch solution (to detect presence of iodene). solution in conical flask will go dark blue showing iodine still present
4. add sodium thiosulfate one drop at a time until blue colour disappears
5. means all iodine has been reacted
6. now calculate number of moles of iodine in solution

Question 8

calculating number of moles of iodine produced in stage 1

e.g. the iodine in solution produced in stage 1 reacted fully with 11.0 cm3 of 1.20 mol dm-3 thiosulfate solution.

work out number of moles of iodene present in starting solution.

Answer 8

I2(aq) + 2S2O32-(aq) -> 2I-(aq) + S4O62- (aq)

number of moles of thiosulfate= conc x vol (cm3) / 1000
0.120 x 11.0 / 1000 = 1.32 x 10-3 moles

1 mole of iodine reacts with 2 moles of thiosulfate

number of moles of iodine in solution = 1.32 x 10-3 / 2 = 6.60 x 10-4 moles

Question 9

e.g. using oxidising agent potassium iodate (V)

stage 3 . calculate the concentration of the oxidising agent

Answer 9

1. original equation IO3(aq) + 5I-(aq) + 6H+(aq) -> 3I2(aq) + 3H2O(l)
2. 25.0 cm3 of potassium iodate (V) solution produced 6.60 x 10-4 moles of iodine. equation shows one mole of iodate (V) ions will produce 3 moles of iodine
3. 6.60 x10-4 / 3 = 2.2 x 10-4 moles of iodate (V) ions in original solution
4. moles = c x v(cm3) / 1000 = 2.20 x 10-4 = c x 25.0 / 1000
   concentration of potassium iodate (V) solution = 0.00880 moldm-3