Question

I have to do lab report and I'm stuck with the discussion only. If someone can help to write it it will be very nice.

EXPERIMENT 3

THE DETERMINATION OF HYDROGEN PEROXIDE BY IODOMETRIC TITRATION

___________________________________________________________

LEARNING AIMS

• To improve titration technique
• To determine the concentration of a hydrogen peroxide solution

LEARNING OUTCOMES

• To evaluate the use of titrimetric analysis
• To manipulate common volumetric apparatus
• To carry out standard calculations involving concentrations and moles

DIRECTED READING

Vogel’s Textbook of Quantitative Chemical Analysis 6P^thP Ed; Mendham et al (2000) Prentice Hall, ISBN: 0582226287; Chapter 10, pp: 312 – 314, 428 – 436

INTRODUCTION

Peroxides occur in a wide variety of situations: organic peroxides, having the general formula ROOR, where R is an organic group, are often used to start polymerization reactions, such as the ones needed to make commonly occurring polymers like polystyrene and polythene.

Hydrogen peroxide, HB_2BOB_2B is readily available in supermarkets and pharmacies since it is a principal component in bleaches for hair and most recently and controversially it was one of the components used to make the bombs which exploded in London on 7P^thP July 2005.

Peroxides are also thought to contribute to aging of skin and they are also a source of free radicals thought to be partly responsible for some cancers.

In this experiment you will use a titrimetric method to determine the amount of hydrogen peroxide in a given solution. Hydrogen peroxide can act as both an oxidizing agent and a reducing agent depending upon the conditions of the reaction. In this experiment hydrogen peroxide will oxidize the iodide ion, IP^-P, obtained from potassium iodide to iodine, IB_2B. The iodine produced reacts with sodium thiosulphate and the amount of thiosulphate required can be used to calculate the amount of iodine present and hence the amount of hydrogen peroxide in the solution.

THEORY

Titrimetry, or titrimetric analysis, is an example of a so-called classical method of analysis. Titrimetry is a convenient method of getting very small amounts of chemicals in to a reaction because the method involves the use of dilute solutions of reactants.

Hydrogen peroxide reacts with potassium iodide according to the reaction given below:

                                        HB_2BOB_2B + 2HP^+P + 2IP^-P = IB_2B + 2HB_2BO                (equation 1)

Equation 1 shows that 1 mole of hydrogen peroxide gives 1 mole of IB_2B.

Note: we have omitted the potassium ion, KP^+P, from this reaction since it would occur on both sides of the equation and so it does not take part in the reaction.

The iodine produced reacts with sodium thiosulphate, NaB_2BSB_2BOB_3B as shown in equation 2 below:

                                        IB_2B + 2S₂O₃^2- à 2IP^- + S₄O_6B^2-                             (equation 2)

Equation 2 shows that 1 mole of iodine reacts exactly with 2 moles of thiosulphate.

Combining equations 1 and 2:

1 mole HB_2BOB_2B gives 1 mole iodine AND 1 mole iodine reacts with 2 moles thiosulphate.

This shows that 1 mole HB_2BOB_2B is equivalent to 2 moles thiosulphate.

This means that if we know how much (the number of moles) thiosulphate we need to react with the iodine we can calculate how much hydrogen peroxide was in the solution we started with.

EXPERIMENTAL

In this experiment you will use titrimetric analysis to determine the concentration of hydrogen peroxide in a solution. The titration is relatively easy to follow because the iodine produced by reaction of hydrogen peroxide with iodide is iodine, which is purple in colour. Addition of thiosulphate solution gives a colourless solution when all of the iodine has reacted (the end point). You can make the end point even sharper by adding a few drops of starch solution when the colour of the solution is pale yellow. The starch changes the colour of the solution to black but the solution will become colourless at the end point.

The experiment is quite short and so you have plenty of opportunity to repeat the titrations until you get consistent results.

You have already carried out an acid-base titration and so you should know that “consistent results” means that the volumes required to completely decolourise the iodine produced should be consistent to ± 0.1 cmP^3P.

You are provided with a solution of hydrogen peroxide and a solution of sodium thiosulphate as well as 1.25 mol dm^-3 sulphuric acid and solid potassium iodide.

Throughout your experiment, note any colours that are observed in your lab book.

1. Note the concentration of the sodium thiosulphate solution supplied.

0.1 m

2. Measure out 10 cmP³ of 1.25 mol dm^-3 sulphuric acid using a measuring cylinder and place it in a 250 cmP^3P conical flask.

3. Add about 1 g of potassium iodide. There is NO NEED TO WEIGH THIS ACCURATELY so just use a top pan balance. Swirl the flask in order to dissolve the potassium iodide in the sulphuric acid.

1* -1.00g /2*- 1.11g /3*- 1.5g

4. When the potassium iodide has all dissolved use pipette and measure out 25.00 cmP^3P of the hydrogen peroxide solution.

5. Gradually add the hydrogen peroxide solution to the acidified potassium iodide solution with constant swirling.

6. Use a cork, rubber bung or parafilm to close the top of the conical flask and allow the mixture to stand for 15 minutes. The solution should become dark orange/brown because of the presence of iodine in solution.

7. Pour sodium thiosulphate into a burette. Run some of this solution through the burette to remove any bubbles below the tap.

8. Record the reading on the burette to the nearest 0.05 cm³ (NOTE: you will have to estimate the second decimal place). THERE IS NO NEED TO GET THE VOLUME TO EXACTLY 0.00 ON THE BURETTE.

1*- 3.20/ 2*- 3.00/ 3*- 3.10

9. Titrate the iodine solution in the conical flask with the sodium thiosulphate in the burette. You can add the thiosulphate fairly quickly in the early stages but you will need to add the thiosulphate more slowly as the colour of the iodine get more and more yellow.

10. When the iodine solution becomes straw coloured (pale yellow) add a few drops of starch solution. The iodine solution should now be much darker – almost black.

11. Continue adding the thiosulphate solution until the iodine solution becomes colourless.

12. Record the reading on the burette again to the nearest 0.05 cmP^3.P

13. The volume of thiosulphate required to react with the iodine solution is the difference between the final reading and the initial reading.
14. Repeat steps 2 – 13 until you get THREE consistent results (i.e. where the volume used agrees to ± 0.1 cmP^3P.

	1	2	3	4
Initial volume (cm3)	3.20	3.00	3.10
Final volume (cm3)	27.10	26.80	27.00
\volume used (cmP3P)	23.90	23.80	23.90

Calculate the concentration of hydrogen peroxide in the solution provided.

EVALUATION OF RESULTS

If the concentration of the sodium thiosulphate provided is C and the volume of this solution required to react with all of the iodine is V:

From equations 1 and 2:

No. of moles of hydrogen peroxide = 0.5 x no. moles thiosulphate = 0.5 x

But we started with 25.00 cmP³ of hydrogen peroxide solution of concentration, Z (the value we are trying to determine)

No. moles of hydrogen peroxide = half of the no. of moles of thiosulphate used.

Thus we can balance the following equation if we know the concentration of sodium thiosulphate used (C) and the volume of sodium thiosulphate used in the titre (V):

So as you know the concentration and volume of thiosulphate solution, you can now determine the concentration of hydrogen peroxide.

QUESTIONS

1.         Calculate a value for the concentration of the hydrogen peroxide solution for each of your titrations.

2.         Calculate the mean, standard deviation and relative standard deviation for your data.

3.         Comment on your results.

Answer 1

• H₂O₂(aq) + 2H⁺(aq) + 2I^-(aq) ===> I₂ + 2H₂O

  This equation shows that 1 mole of hydrogen peroxide gives 1 mole of   I_2.

  The iodine produced reacts with sodium thiosulphate, Na₂S₂O₃ as shown in equation below:

                                          I₂ + 2S₂O₃^2- (aq) ==> 2I^- (aq) + S₄O₆^2- (aq)

  This Equation shows that 1 mole of iodine reacts exactly with 2 moles of thiosulphate.

  Combining equations :

  1 mole H₂O₂ gives 1 mole iodine and 1 mole iodine reacts with 2 moles thiosulphate.

  1 mole H₂O₂ = 2 moles thiosulphate.

  Concentration of the sodium thiosulphate solution = 0.1 M

  25.00 cm3 of the hydrogen peroxide solution.

  #1 : Volume consumed : 23.90 mL

  moles of thiosulphate consumed : 0.0239 L * 0.1 mol/L = 0.00239 moles

  thus moles of hydrogen peroxide in 25 mL solution = 0.001195 moles

  thus, Concentration of H₂O₂ solution : 0.001195 mole/ 0.025 L = 0.0478 M

  #2 : Volume consumed : 23.80 mL

  moles of thiosulphate consumed : 0.0238 L * 0.1 mol/L = 0.00238 moles

  thus moles of hydrogen peroxide in 25 mL solution = 0.00119 moles

  thus, Concentration of H₂O₂ solution : 0.00119 mole/ 0.025 L = 0.0476 M

  #3 : Volume consumed : 23.90 mL

  moles of thiosulphate consumed : 0.00264 L * 0.1 mol/L = 0.00239 moles

  thus moles of hydrogen peroxide in 25 mL solution = 0.001195 moles

  thus, Concentration of H₂O₂ solution : 0.001195 mole/ 0.025 L = 0.0478 M

• mean, standard deviation and relative standard deviation :

Count, N:	3
Sum, Σx:	0.1432
Mean, x̄:	0.0477
Variance, s2:	1.33E-8

Standard Deviation, s: 0.000115

The relative standard deviation formula is:
100 * s / |x̄|
Where:
s = the sample standard deviation
x̄ = sample mean

% RSD = 100 * (0.000115 / 0.0477) = 0.24 %