Question

I need to: DETERMINE THE CONCENTRATION OF HYDROGEN PEROXIDE IN THE SOLUTION

how ??

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.

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

0.1046 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.

FLASK 1: 1.06 g FLASK 2: 1.09 g FLASK 3: 1.09 g

4. When the potassium iodide has all dissolved use pipette and measure out 25.00 cmP^3Pof 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.

0.00

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)	0.00	0.00	0.00
Final volume (cm3)	26.6	26.3	26.4
\volume used (cmP3P)	26.6	26.3	26.4

	5	6	7	8
Initial volume (cm3)
Final volume (cm3)
\volume used (cmP3P)

15. 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:

No. of moles of thiosulphate is CxV/1000

From equations 1 and 2:

No. of moles of hydrogen peroxide = 0.5 x no. moles thiosulphate = 0.5 x CxV/1000

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):

25xZ/1000 = 0.5 x CxV/1000

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

QUESTIONS

DETERMINE THE CONCENTRATION OF HYDROGEN PEROXIDE IN THE SOLUTION

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 = 0.1046 M

25.00 cm3 of the hydrogen peroxide solution.

#1 : Volume consumed : 26.6 mL

moles of thiosulphate consumed : 0.0266 L * 0.1046 mol/L = 0.002782 moles

thus moles of hydrogen peroxide in 25 mL solution = 0.001391 moles

thus, Concentration of H₂O₂ solution : 0.001391 mole/ 0.025 L = 0.05565 M

#2 : Volume consumed : 26.3 mL

moles of thiosulphate consumed : 0.0263 L * 0.1046 mol/L = 0.002751 moles

thus moles of hydrogen peroxide in 25 mL solution = 0.001375 moles

thus, Concentration of H₂O₂ solution : 0.001391 mole/ 0.025 L = 0.05502 M

#3 : Volume consumed : 26.4 mL

moles of thiosulphate consumed : 0.00264 L * 0.1046 mol/L = 0.002761 moles

thus moles of hydrogen peroxide in 25 mL solution = 0.001381 moles

thus, Concentration of H₂O₂ solution : 0.001391 mole/ 0.025 L = 0.05523 M