6 a): To use a standard solution of iodine to standardise a solution of sodium thiosulfate

Theory

By titrating a sodium thiosulfate (Na2S2O3) solution of unknown concentration against

standard solution of iodine (I2), the concentration of the sodium thiosulfate can be determined.

Equation

2S2O32– + I2 → S4O62– + 2I–

Procedure

1) Making the sodium thiosulfate up into a solution (if required)

2) Filling the burette with the sodium thiosulfate solution

3) Making a standard solution of iodine using potassium manganate (VII)

4) Carrying out the titration

Reaction in the conical flask to form iodine

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

Iodine is NOT a primary standard – it cannot be made directly into a standard solution.

To obtain a standard aqueous solution of iodine, a set volume of previously standardised solution of potassium manganate (VII) (KMnO4) is pipetted to a conical flask and is reacted with excess potassium iodide (KI) and excess sulfuric acid (H2SO4)

Colour change at this point:

Purple → red/brown

(Mn7+) → (I2)

Notice: The potassium manganate and iodine are in a molar ratio of 2 : 5 – If the concentration of potassium manganate is known, the concentration of iodine can be known i.e. the iodine is standardised in this way

Suitable indictor for this titration

Starch solution

Note: The starch is only added when close to the end point of the titration

Justification for this indicator

Iodine turns blue-black in the presence of starch

Colour change observed

During titration – Red/brown → Yellow → Pale Yellow

NOW ADD STARCH - A blue-back colour forms

At end point – Blue/Black → Colourless

Explaining the colour change

During titration: As the sodium thiosulfate is added to and reacts with the iodine, iodine’s red- brown colour becomes less intense and changes to yellow and eventually pale yellow – The iodine is being used up the reaction

At end point: Starch is added and a blue-black colour forms due to the small amount of iodine left. The sodium thiosulfate is now added slowly in drops and as soon iodine has been completely used up, the blue-black colour decolourises i.e. the reaction has reached the end point

Why is the starch only added when a pale-yellow colour forms in the conical flask i.e. when near the end point

1) Waiting until a pale-yellow colour forms before adding the starch indicator tells us that the end point is very near i.e. there is very little iodine left in the conical flask.

The sodium thiosulfate can then be added very slowly in drops, resulting in an accurate end point

2) Iodine adsorbs onto starch preventing it reacting with sodium thiosulfate- If added earlier an inaccurate end point would be obtained

Explain why a standard solution of sodium thiosulfate CANNOT be directly made up i.e. why must sodium thiosulfate be standardised by titration?

• Sodium thiosulfate is not a primary standard – it is a hydrated compound and loses some of their water of crystallisation (effloresce) in dry air – cannot be obtained in a pure state

• Therefore, a precise mass of these crystals cannot be weighed out and a standard solution cannot be directly made up

(Same as washing soda crystals)

Explain why a standard solution of iodine CANNOT be directly made up?

Iodine is not a primary standard

1. Iodine sublimes (changes directly from a solid to a vapour) at room temperature

2. Iodine is very poorly soluble in water

How is a standard aqueous solution of iodine obtained?

• A standard solution of potassium manganate (VII) is reacted with excess potassium iodide and excess sulfuric acid

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

• For every 2 moles KMnO4 that reacts, 5 moles of iodine are formed

• Using a standard solution of KMnO4 means the number of moles of I2 formed can be calculated and hence its exact concentration

What colour change occurs in the conical flask as a result of this reaction? (Not the titration colour change)

Purple → red/brown

Why is excess sulfuric acid added to the conical flask?

Only in an acidic environment are Mn7+ ions in potassium manganate fully reduced to Mn2+ ions – this means the I– ions in potassium iodide will be fully oxidised to form iodine (I2) – produces the maximum amount of iodine

(100% yield of I2)

Explain why i) nitric acid ii) hydrochloric acid cannot be used to provide an acidic environment

i) Nitric acid is itself an oxidising agent and will interfere in the reaction

ii) Hydrochloric acid would be oxidised by the potassium permanganate forming chlorine gas (Cl2)

Why is excess potassium iodide added to the conical flask?

1. Potassium iodide in excess ensures that the potassium manganate (VII) is the limiting reagent - this ensures ALL of the potassium manganate reacts – produces the maximum amount of iodine

2. The iodine that is produced will react with the potassium iodide in excess forming the triiodide ion I3 – i.e. a soluble version of iodine – keeps iodine in aqueous solution as an I3– ion

Explain how iodine, a non-polar substance of very low water solubility, is brought into aqueous solution (is made dissolve in water)

Iodine is brought into aqueous solution by reacting the acidified potassium manganate (VII) with EXCESS potassium iodide. The iodine that is produced will react with the potassium iodide in excess forming the triiodide ion I3 – i.e. a soluble version of iodine – keeps iodine in aqueous solution as an I3 – ion

Why should starch solution always be kerkly prepared

Starch is biodegradable and will be decomposed