Iodometry Titration. Redox titration of iodine and sodium thiosulphate ( Iodometry)

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1 Iodometry Titration Redox titration of iodine and sodium thiosulphate ( Iodometry) Page: 1 Iodometry: When an oxidizing agent (e.g. KMnO 4, H 2 O 2 etc.) is added to excess iodide (I ) to produce iodine and iodine so produced is determined by titration with sodium thiosulphate ( Na 2 S 2 O 3 ) then this method is known as iodometry. In case of iodometry, reducing agent (sodium thiosulphate) is titrated directly with standard I 2 solution which is produced during a reaction (i.e. reaction of some oxidizing agents like KMnO 4, H 2 O 2 with KI). Nature of iodine: 1. Iodine is a soft oxidizing agent i.e. it get reduced to I slowly. Therefore, in presence of any (even weak) oxidizing agent it changes back into iodine from I. In other words I is a strong reducing agent, since it changes fast to I 2. I 2 + 2e 2I (Slow) It is important to note that F 2, Cl 2, Br 2 are stronger oxidizing agent than I 2. Due to its ( I 2 ) soft oxidizing nature (i.e. less tendency to get reduce or less tendency to accept electrons), mild oxidizing agents like Fe +3 (aq) can oxidize aqueous I to I 2. In other words, Fe +3 (aq) can oxidize (i.e. acts as an oxidizing agent or accepts electrons or undergoes reduction) aqueous I to I 2. Therefore, I acts as strong reducing agent. The redox reaction is shown as below. Reduction/ oxidizing agent 2Fe I 2Fe 2+ + I 2 Oxidation/reducing agent Because of this iodine/ thiosulphate titrations are used in the quantitative estimation of oxidizing agents. The oxidizing agent (like KMnO 4 /H 2 O 2 ) to be estimated is first added to excess of KI this liberates iodine. 2 I (aq) I 2 (aq) + 2e Amount of iodine liberated is then determined by titration against a solution of Na 2 S 2 O 3 of known concentration. Here oxidizing agents like KMnO 4 or H 2 O 2 Oxidises KI into I 2. I 2 (aq) + 2e 2 I (aq) 2S 2 O 3 (aq) S 4 O 2 6 (aq) +2e thiosulphate tetrathionate I 2 (aq) + 2S 2 O 3 (aq) 2 I (aq) + S 4 O 6 2 (aq)

2 Page: 2 The ionic equation of the oxidation of I to I 2 by KMnO 4 in acidic medium is as follows: MnO 4 + 8H + + 5e Mn H 2 O ] 2 2 I I 2 + 2e ] 5 10 I + 2 MnO H + 2Mn I 2 + 8H 2 O Same way, the ionic equation of the oxidation of I to I 2 by H 2 O 2 in acidic medium is as follows: 2 I + H 2 O 2 + 2H + I 2 + H 2 O 2. I 2 is not very soluble. It can evaporate from titration flask during titration. Problem with this titration is ph must be controlled. End point The end point of the titration is detected by adding the starch solution. The starch gives a blue solution as long as any iodine is present. By calculating the amount of thiosulphate used in the titration we can determine amount of iodine (I 2 ) liberated and hence the amount of oxidizing agent taken. It is important to note that KI is also an oxidizing agent since on adding KI solution to the test solution until it is in excess then following colour ppt. are obtained. (i) Yellow ppt. in case of Pb 2+ (ii) Green/ cream ppt. in case of Cu 2+ (iii) Brown ppt. in case of Fe 3+ The end point of this titration can also be determined without using starch solution. In such a case when sodium thiosulphate solution is added to I 2 solution, then brown colour of the iodine disappears at the end point.

3 Exercise: Page: 3 FA 1 is moldm- 3 potassium manganate (VII). FA 2 is an aqueous solution of sodium thiosulphate. You are required to determine the concentration of thiosulphate ions in FA 2. Procedure: Pipette out 25.0 cm 3 of FA 1 into a conical flask, add about 10 cm 3 of dilute sulphuric acid and about 10 cm 3 of aqueous potassium iodide provided. Titrate the iodine liberated with FA2. Starch indicator is not necessary. Repeat the titration as many times as you think necessary to obtain accurate results. Result: Rough Accurate Final burette reading / cm 3 Initial burette reading/ cm 3 Volume of FA 2 used Summary: cm 3 of FA 1 liberated iodine requiring cm 3 of FA2 for reduction. Questions: (a) Calculate the number of moles of mangante (VII), MnO4, ions in the 25.0 cm 3 of FA 1 pipetted out. (Turn over)

4 (b) Page: 4 Given that one mole of mangante (VII) ions liberates sufficient iodine from potassium iodide to react with five moles of thiosulphate S2O3 2 ions. Calculate the number of moles of thiosulphate ions present in the volume of FA2 that you found necessary in the titration. (c) Calculate the number of moles of thiosulphate ions in 1.00 dm 3 of FA2.

5 Questions: SOLUTION Page: 5 (a) Calculate the number of moles of mangante (VII), MnO4, ions in the 25.0 cm 3 of FA 1 pipetted out. No of moles of of mangante (VII), MnO4, ions = Molarity (M MnO4 - x V MnO4 - ) = moldm- 3 x dm 3 = mol. (b) Given that one mole of mangante (VII) ions liberates sufficient iodine from potassium iodide to react with five moles of thiosulphate S2O3 2 ions. Calculate the number of moles of thiosulphate ions present in the volume of FA2 that you found necessary in the titration. Given; One mole of iodine/ mangante (VII) ions reacts with 5 moles of of thiosulphate S2O3 2 ions. Using formula No of moles of thiosulphate S2O3 2 ions = 5 x mol. = mol (c) Calculate the number of moles of thiosulphate ions in 1.00 dm 3 of FA2. No of moles of thiosulphate ions in 1.00 dm 3 of FA 2 means it is Molarity we need to find out. Molarity = Molarity = ******