2013 h2 redox titration (iodometric titration) teachers' copy

CATHOLIC JUNIOR COLLEGEH2 CHEMISTRY (Syllabus 9647)

VOLUMETRIC ANALYSIS

REDOX TITRATIONS

CALCULATIONS IN REDOX TITRATIONS

The calculations that are involved in redox titrations are similar to those in acid-base

titrations (they all use the underlying concept of mole relationships).

However, unlike acid-base titrations whose stoichiometric equations are easy to write,

redox titrations tend to have more difficult stoichiometric equations to work with.

Redox titrations usingACIDIFIED POTASSIUM MANGANATE(VII), KMnO4

This will be done in the SPA session where experiment on it will be carried out and the theory

of the titration will be discussed.

Redox titrations involvingSODIUM THIOSULFATE(VI) Na2S2O3.5H2O, AND IODINE

Titrations involving sodium thiosulfate are known as iodometric titrations and are

described below.

Step 1:

Excess potassium iodide is added to an oxidising agent (O.A.) which is the limiting

reagent. The oxidising agent then oxidises the iodide ions from the potassium iodide

solution to give iodine, which gives a brown coloured solution. The amount of iodine

liberated is dependent on the amount of O.A. originally present. (The relationship is

shown in the stoichiometric equation).

Step 2:

The iodine formed is then titrated against sodium thiosulfate(VI) of a known

concentration, which is placed in the burette, using starch solution as indicator. From

the volumes of titration (volumes of sodium thiosulfate(VI) used), the concentration of

the O.A. can be determined.

1

I2(aq) + 2 Na2S2O3(aq) Na2S4O6(aq) + 2 NaI(aq)

Or ionically as:

I2(aq) + 2 S2O32-(aq) S4O6

2-(aq) + I-(aq)

Sodium thiosulfate(VI) reacts with iodine liberated to give sodium iodide and sodium

tetrathionate. This is an important equation and it must be remembered.

Experimental considerations

In iodometric titrations, the indicator used is starch solution. Starch solution is added

only when the titration is approaching the end-point; i.e. when the brown iodine

solution turns pale yellow. When starch indicator is added, the solution turns deep-blue

due to the formation of the starch-iodine complex. The titration is continued until the

end-point is reached when the deep-blue solution is discharged.

Starch solution is not added at the beginning of the titration unlike other

titrations. Why is this so?

Since iodine is being ‘trapped’ in the starch molecules, adding starch solution at the

beginning may cause lots of iodine to be trapped as the starch-iodine complex. The

release of iodine from the starch molecules takes time and as such, it might

decrease the accuracy of the volumes of titration.

2

E.g., if K2Cr2O7 is the oxidising agent, it will oxidise I− ions to give I2 according to:

Step 1: Cr2O72− + 6 I− + 14 H+ 2 Cr3+ + 3 I2 + 7 H2O

The iodine liberated is then titrated with S2O32-

.

Step 2: I2 + 2 S2O32− S4O6

2− + 2 I−

so, Cr2O72− 3 I2 6 S2O3

2−

Therefore, Cr2O72− 6 S2O3

2−

We say, 1 mol of K2Cr2O7 oxidises iodide to give 3 mol of iodine which then react with 6 mol of S2O3

2−. Therefore 1 mol of Cr2O72− is stoichiometrically equivalent to 6

mol of S2O32−.

You must be able to derive this stoichiometric relationship before you can even begin

to do any calculation on this iodometric titration!

Worked Example 1

In an experiment, 20.0 cm3 of potassium iodate(V), KIO3, solution was added to excess acidified potassium iodide, KI. The iodine produced was then titrated with a solution of sodium thiosulfate(VI). In the titration, 22.45 cm3 of sodium thiosulfate(VI) was used. The concentration of the thiosulfate was 29.8 g of Na2S2O3.5H2O per dm3. [Mr of Na2S2O3.5H2O = 248, Mr of KIO3 = 214]Calculate the concentration of the potassium iodate(V) in g dm−3.

Given: IO3− + 5 I− + 6 H+ 3 I2 + 3 H2O

Solution:

IO3− + 5 I− + 6 H+ 3 I2 + 3 H2O

2 S2O32– + I2 S4O6

2− + 2 I−

so, IO3− 3 I2 6 S2O3

2−

therefore, IO3− 6 S2O3

2−

Conc of Na2S2O3.5H2O = mol dm−3 = 0.120 mol dm−3

Amount of S2O32− = 0.120 × = 0.00270 mol

Amount of IO3− = × 0.00269

= 0.000450 mol

Conc of KIO3 = (0.000450 × ) = 0.0225 mol dm−3

Conc of KIO3 = (0.0225 × 214) g dm−3 = 4.81 g dm −3

3

Worked example 2*

To about 2 g of potassium iodide dissolved in dilute sulfuric acid, 25.0 cm3 of 0.02 mol dm–3

KMnO4 were added. The iodine liberated required 24.70 cm3 of sodium thiosulfate,

Na2S2O3.5H2O, solution for titration.

What is the concentration of the sodium thiosulfate solution, in grams of hydrated salt per dm3,

and how much water must be added per dm3 of the solution to change its concentration to 0.1

mol dm–3? [Mr of Na2S2O3.5H2O = 248] [Ans: 25.1 g dm–3; 12 cm3]

Reduction: 2 [MnO4– + 8 H+ + 5 e– Mn2+ + 4 H2O]

Oxidation: 5 [2 I – I2 + 2 e–]-----------------------------------------------------------------------------------------------------------------------------------------

Overall: 2MnO4– + 10I – + 16H+ 2Mn2+ + 5I2 + 8H2O

-----------------------------------------------------------------------------------------------------------------------------------------

I2 + 2S2O32– S4O6

2– +2I–

MnO4- ≡ 5 S2O3

2-

Amount of MnO4– = 0.02 ×

= 5.00 × 10–4 mol

Amount of S2O32– = 5 × 5.00 × 10–4 = 2.50 × 10-3 mol

Conc of S2O32– = 2.50 × 10–3 × = 0.1012 mol dm–3

Conc. of Na2S2O3.5H2O = 0.1012 × 248 = 25.1 g dm –3

mol of S2O32– after dilution = mol of S2O3

2– before dilution

0.1 × Vafter dil = 0.1012 × 1000

Vafter dil = 1012 cm3

Vol. of H2O to be added = 1012 – 1000 = 12 cm 3

4

2 MnO4– ≡ 5 I2 ≡ 10 S2O3

2–

Points to note: This is an iodometric titration. The first reaction involves the oxidation of iodide ion to

iodine. The iodine liberated is then titrated with sodium thiosulfate in the second reaction. The iodine formed from the first reaction is used subsequently in the second reaction.

Thus the stoichiometric relationship wrt iodine is important. Do not start off by calculating the number of moles present in 2 g of KI. This is incorrect

since KI is added in excess.