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Question - Determine the concentration of a saturated aqueous solution of chlorine by back titration.
1. Using measuring cylinders, measure out 25cm3 (enough for reliable results but not wasteful) of chlorine and 40cm3 (excess amount) of reducing agent Iron (II) and place into separate beakers.
Add the contents of the two beakers into one conical flask. Swirl the conical flask thoroughly for a minute.
The half-equations when chlorine and Iron (II) react are:
Fe2+--> Fe3++ 1e-
Cl2+ 2e- --> 2Cl-
doubling the first (Chlorine requires 2 electrons) and adding the 2 half equations together:
2Fe2++ Cl2 --> 2Fe3++ + 2Cl-
Therefore for this reaction to take place completely you need to have a 2:1 molar ratio.
Concentration of chlorine = mass/RMM = 7g/71 =0.1moldm-3
Concentration of Iron needs to be twice the amount of chlorine, 0.1moldm-3 x2 = 0.2moldm-3.
2. Fill a burette with KMnO4 (which contains 3.50g dm-3 of solid) and titrate into the conical flask containing Iron (II) and chlorine solution. When the solution turns pink, stop the titration and record the volume of KMnO4 used.
Since you used an excess of FeSO4, you will have some left over Fe(II) ions in solution. This will be titrated against KMnO4.
3.Repeat this procedure two more times.A mean volume value of KMnO4 is far more reliable than a one-off result.
Risk Assessment
Wear safety gloves and goggles to prevent getting KMnO4 (an irritant) on your skin. In the event of broken glass, tell a supervisor who will clear it up.
Analysis of results
Step 1
Amount of chlorine used (in moles): 0.1 moldm-3 x 0.025 = 0.0025 moles
Amount of Iron (II) used (in moles): 0.2 moldm-3 x 0.040 = 0.008 moles
Step 2
The reaction of the titration is:
5Fe2++ + MnO4- + 8H+ --> Mn2+ + 5Fe3+ + 4H2O
Therefore 5 moles of Iron (II) reacts with 1 mole of Manganate. To calculate the amount of excess Fe2+ ions, first work out the moles of indicator used.
Concentration of indicator is Mass/RMM = 3.5g/158 = 0.0222 moldm-3
Number of moles in indicator = 0.0222 x mean volume of indicator used in titration
Multiply this value by 5 to give the amount of Fe+2 moles.
Subtracting the total original moles of Fe+2 used (0.008 moles) by the moles of excess Fe2+ gives moles of iron that reacted with chlorine.
Divide this value by 2 (due to the 2:1 molar ratio) to give the moles of chlorine used
Finally, to answer the aim, the concentration of chlorine in an aqueous solution = moles of chlorine/ 0.025 dm3 (volume of chlorine)
Is it correct? Is it ok that I assumed the mass of KmnO4 to be 3.5g?
It has to be in by next week, so please help!
Question - Determine the concentration of a saturated aqueous solution of chlorine by back titration.
1. Using measuring cylinders, measure out 25cm3 (enough for reliable results but not wasteful) of chlorine and 40cm3 (excess amount) of reducing agent Iron (II) and place into separate beakers.
Add the contents of the two beakers into one conical flask. Swirl the conical flask thoroughly for a minute.
The half-equations when chlorine and Iron (II) react are:
Fe2+--> Fe3++ 1e-
Cl2+ 2e- --> 2Cl-
doubling the first (Chlorine requires 2 electrons) and adding the 2 half equations together:
2Fe2++ Cl2 --> 2Fe3++ + 2Cl-
Therefore for this reaction to take place completely you need to have a 2:1 molar ratio.
Concentration of chlorine = mass/RMM = 7g/71 =0.1moldm-3
Concentration of Iron needs to be twice the amount of chlorine, 0.1moldm-3 x2 = 0.2moldm-3.
2. Fill a burette with KMnO4 (which contains 3.50g dm-3 of solid) and titrate into the conical flask containing Iron (II) and chlorine solution. When the solution turns pink, stop the titration and record the volume of KMnO4 used.
Since you used an excess of FeSO4, you will have some left over Fe(II) ions in solution. This will be titrated against KMnO4.
3.Repeat this procedure two more times.A mean volume value of KMnO4 is far more reliable than a one-off result.
Risk Assessment
Wear safety gloves and goggles to prevent getting KMnO4 (an irritant) on your skin. In the event of broken glass, tell a supervisor who will clear it up.
Analysis of results
Step 1
Amount of chlorine used (in moles): 0.1 moldm-3 x 0.025 = 0.0025 moles
Amount of Iron (II) used (in moles): 0.2 moldm-3 x 0.040 = 0.008 moles
Step 2
The reaction of the titration is:
5Fe2++ + MnO4- + 8H+ --> Mn2+ + 5Fe3+ + 4H2O
Therefore 5 moles of Iron (II) reacts with 1 mole of Manganate. To calculate the amount of excess Fe2+ ions, first work out the moles of indicator used.
Concentration of indicator is Mass/RMM = 3.5g/158 = 0.0222 moldm-3
Number of moles in indicator = 0.0222 x mean volume of indicator used in titration
Multiply this value by 5 to give the amount of Fe+2 moles.
Subtracting the total original moles of Fe+2 used (0.008 moles) by the moles of excess Fe2+ gives moles of iron that reacted with chlorine.
Divide this value by 2 (due to the 2:1 molar ratio) to give the moles of chlorine used
Finally, to answer the aim, the concentration of chlorine in an aqueous solution = moles of chlorine/ 0.025 dm3 (volume of chlorine)
Is it correct? Is it ok that I assumed the mass of KmnO4 to be 3.5g?
It has to be in by next week, so please help!
