Prior to going into detail we must first understand what a100-volume solution is. By definition this means that 1cm3 of H2O2 will decompose to produce 100cm3 of O2 at STP. In simple terms it is just another way of indicating the strength of H2O2. 2H2O2 (aq) 2H2O(l) + O2(g) Equation 1. Consequently, we can understand that 100-volume represents a very strong concentration of H2O2. Thus, before carrying out the titration the solution of H2O2 must be diluted. A reasonable dilution factor for this experiment is 100; hence from a 100-volume to a 1-volume solution.
As I know the strength of the original (C1) and the new solution (C2), I can use the below formula: C1V1= C2V2 Formula 1. where V1 is the original volume (dm3) V2 is final volume (dm3) To make up a 1-volume H2O2, we would need to use a graduated pipette to transfer 0. 01dm3 of the 100-volume H2O2 into a 1dm3 volumetric flask. As we know 3 of the 4 quantities in formula 1 we can work out V2: 100(0. 01) = 1(V2) V2 = 1 dm3 As the final volume is 1 dm3, the volumetric flask needs to be filled up to the top with distilled water.
The principle behind a redox titration is that if a solution contains a substance that can be oxidized, then the concentration of that substance can be analyzed by titrating it with a standard solution of a strong oxidizing agent. The reaction can be balanced by presuming that it occurs through two separate half-reaction. 2H2O2 (aq) O2(g) + 2H+(aq) + 2e- Equation 2. MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O (l) Equation 3. Potassium manganate (VII) is very useful in redox titrations, due to its capability to be self indicating. However, for it to be able to function properly it must be in acid solution.
Therefore, we must add excess dilute sulphuric acid to the solution. If we combine both equation 2 and 3, we can get the full reaction between H2O2 and KMnO4. Note, that sulphuric acid is also part of the reaction. 2KMnO4(aq) + 5H2O2(aq) + 3H2SO4(aq) 2MnSO4(aq) + K2SO4(aq) + 5O2(g) + 8H20(l) The stoichiometric ratio provided from the above equation, along with experimental data obtained from the titration will be used to determine the concentration of H2O2. Method: 1. Measure out 25cm3 of the diluted H2O2 in a graduated pipette and pour this into a conical flask. Record the volume in a data table.
2. Measure 10cm3 of 2 molar sulphuric acid into a graduated cylinder and carefully add the acid to the conical flask. 3. Set up a clamp, boss and stand in order to fix the burette on to it. 4. Place a white tile on the bottom of the stand (underneath the conical flask) in order to make the colour change easier to recognise. 5. Fill up the burette with 0. 1 molar KMnO4 just above 0cm3, whilst having the tap closed. Record the morality in a data table. 6. Open the stopcock on the burette to allow any air bubbles to escape from the tip. Close the stopcock when the liquid level in the burette is 0cm3.
Record the initial volume, remembering to always read from bottom of meniscus. 7. Start by opening the tap of the burette slightly and gently shaking the conical flask as the KMnO4 is poured in. 8. The KMnO4 is a purple colour, however as it reacts with the H2O2 it becomes colourless. The endpoint of the titration is when all the H2O2 has reacted and a further drop of KMnO4 causes the solution to remain pink/purple. This volume of KMnO4 added should be recorded. 9. Repeat the titration until the results only differ by 0. 5cm3. The results obtained will be recorded in a table; below is a specimen: Titre 1 Titre 2
Titre 3 Molarity of KMnO4 0. 1 0. 1 0. 1 Initial volume of KMnO4 solution (cm3) 0 22. 4 44. 9 Final volume of KMnO4 solution (cm3) 22. 4 44. 9 67. 3 Volume of KMnO4 solution (cm3) used 22. 4 22. 5 22. 4 Mean volume of KMnO4 solution (cm3) 22. 43 24. 43 22. 43 By using the data collected from the titration and the stoichiometric ratio, it is then possible to work out the concentration of H2O2. Gas Collection In order to determine the concentration of H2O2 I will be carrying out two separate experiments; the titration explained above and a gas collection. H2O2 always decomposes into oxygen and water, as in equation 1.
2H2O2 (aq) 2H2O(l) + O2(g) Equation 1. However, the rate of decomposition of H2O2 is very slow without the use of a catalyst. As a result, I will be using manganese dioxide as a catalyst; this provides an alternative pathway with a lower activation energy. Thus, the rate of decomposition and production of O2 is faster. The below arrangement will be used for the decomposition: In this experiment we are provided with a 100-volume H2O2. As explained earlier a 100-volume solution means that 1cm3 of H2O2 decomposes to 100cm3 of O2 gas. Hence, 1cm3 of H2O2 would fill up the graduated syringe to its maximum.
As a result, I have chosen to use the 1-volume H2O2 which was also used in the titration. As illustrated in figure 1. 0, the manganese dioxide will be kept apart from the H2O2 until the start of reaction. The MnO2 will be delivered using an inner test tube, which is turn is connected to a thread. Once the bung is attached to the conical flask the thread can be released and the reaction can proceed. The purpose of this is to increase the accuracy. Method: 1. Measure 20cm3 of 1-volume H2O2 using a measuring cylinder and add to a conical flask. 2. Weight 15g of MnO2 powder using an electronic balance and add to a small test tube.
3. Attach a thread to the test tube; make sure it is secured tightly. 4. Place the test tube inside the conical flask using the attached thread; remember to keep part of the thread outside the conical flask. 5. Secure the bung, both to the conical flask and the gas syringe as illustrated in figure 1. 0. Release the thread and measure the volume of gas collected. Repeat the experiment 3 times to get reliable results. Trail 1 Trail 2 Trail 3 Volume of O2 gas collected (cm3) 20. 2 19. 8 20. 1 Mean volume of O2 gas collected (cm3) 20. 03 20. 03 20. 03 6. Record the volume of gas collected in both reactions in a data table:
Below is a specimen calculation: 20. 03cm3= 0. 02003dm3 number of moles of O2 = 0. 02003/22. 4 = 0. 000894 3sf. Equation 1. 0 states that H2O2 and O2 are in stoichiometric ratio of, 2:1 respectively. number of moles of H2O2 = 0. 0008942 = 0. 00179 3sf. The volume of H2O2 used was 0. 02dm3, therefore the concentration of H2O2 is: number of moles/ volume = concentration 0. 00179/0. 02= 0. 0894 mol dm-3 As the solution used was diluted by a factor of 100; form 100 to 1 volume, the actual concentration of 100-volume H2O2 is: 0. 0894 mol dm-3 100= 8. 94 mol dm-3.