6 b): To determine the % w/v of sodium hypochlorite in bleach using previously standarised solution of sodium thiosulfate

Theory

• Many household bleaches contain sodium hypochlorite (NaClO) as an oxidising agent.

• A diluted version of the bleach can be used to make a solution of iodine (I2)

• The iodine can then be titrated against a previously standardised solution of sodium thiosulfate (Na2S2O3) to determine the concentration of sodium hypochlorite in the bleach

Equation

2S2O32– + I2 → S4O62– + 2I–

(Burette) + (conical flask)

Molar ratio: 2:1

Procedure

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

2) Filling the burette with the sodium thiosulfate solution

3) Diluting the bleach

4) Making a solution of iodine using the diluted bleach

5) Carrying out the titration

Reaction in the conical flask to form iodine (Will be given if required)

To obtain an aqueous solution of iodine, a set volume of the diluted bleach containing sodium hypochlorite (NaClO) is pipetted to a conical flask and is reacted with excess potassium iodide (KI) and excess sulfuric acid (H2SO4)

ClO– + 2I– + 2H+ → Cl– + I2 + H2O

Colour change at this point:

Colourless → Red/Brown

Notice: The sodium hypochlorite and iodine are in a molar ratio of 1 : 1

– If the number of moles of iodine (I2) is found from the titration, the number of moles of sodium hypochlorite in the diluted bleach can be known

i.e. the number of moles of sodium hypochlorite = the number of moles of iodine

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

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

3) Titre value will be larger than it should have been

How was it possible to have a standard solution of sodium thiosulfate to use in this titration despite the fact it is not a primary standard?

The sodium thiosulfate was previously standardised by titrating it against a standard solution of iodine (Titration 6 a))

Why is the bleach diluted before use in this titration?

• The bleach was diluted as the original bleach is too concentrated meaning the end point of the titration (and colour change) would take too long to occur.

• A large volume of sodium thiosulfate solution would be required

How is an aqueous solution of iodine obtained from the bleach?

The diluted bleach containing sodium hypochlorite is reacted with excess potassium iodide and excess sulfuric acid

ClO – + 2I – + 2H+ Cl – + I2 + H2O

(Limiting) + (Excess)

What colour change occurs in the conical flask as a result of this reaction?

Colourless → Red/Brown

Why is excess sulfuric acid added to the conical flask?

Only in an acidic environment are ClO– ions in sodium hypochlorite fully reduced to Cl– 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 sodium hypochlorite forming chlorine gas (Cl2)

Why is excess potassium iodide added to the conical flask?

1. Potassium iodide in excess ensures that the sodium hypochlorite is the limiting reagent - this ensures ALL of the sodium hypochlorite reacts – produces the maximum amount of iodine

2. The iodine that is produced will react with the extra 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

Iodine is brought into aqueous solution by reacting the acidified sodium hypochlorite 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

Explain why the use of distilled water instead of deionised water throughout this experiment would be more likely to ensure a more accurate result

• Deionised water has all ions removed but could still contain chlorine – chlorine is an oxidising agent and could take part in the reaction affecting the results

• Distilled water is the most pure form of water