Practical 3
Titration of Vanadium in Various Oxidation States
Method 1:
Preparing The Solution For A Change in Oxidation State:
Preparing The Potassium Manganate (VII) In The Conical Flask:
Titration of KMnO4(aq) Against The Vanadate Solution:
Method 2:
Preparing The Solution For A Change in Oxidation State:
Titration of KMnO4(aq) Against The Vanadate Solution:
Health & Safety:
Other Safety Measures:
- Hair tied back at all times.
- Any spillages reported and cleaned up appropriately.
- Protective gloves.
- No loose clothing (wear a lab. coat).
- Be aware of other peoples’ equipment and that they may be filling a burette at floor-level (ie. don’t trip over them!).
- Let other people know what substances you are working with, especially in the communal fume cupboard.
Measures to Increase Accuracy:
Calculating The Oxidation States of The Vanadium Ions:
In Method 1:
- 25.00cm3 of 0.02molar ammonium vanadate contains 0.0005moles of vanadium ions. ...
This is a preview of the whole essay
Other Safety Measures:
- Hair tied back at all times.
- Any spillages reported and cleaned up appropriately.
- Protective gloves.
- No loose clothing (wear a lab. coat).
- Be aware of other peoples’ equipment and that they may be filling a burette at floor-level (ie. don’t trip over them!).
- Let other people know what substances you are working with, especially in the communal fume cupboard.
Measures to Increase Accuracy:
Calculating The Oxidation States of The Vanadium Ions:
In Method 1:
- 25.00cm3 of 0.02molar ammonium vanadate contains 0.0005moles of vanadium ions.
- 30.70cm3 of 0.01molar KMnO4 is required to return the vanadium ions to their 5+ oxidation state.
- Mr of KMnO4 = 158g therefore;
in 1dm3 of water there are (158/100)g of KMnO4
and in 30.70cm of water there are [(158/100000) x 30.70]g of KMnO4
--- This is 0.048506g which in turn = (0.048506/158)moles
= 0.000307moles KMnO4
- The equation;
MnO4-(aq) + 8H+(aq) + 5e-(aq) Mn2+(aq) + 4H2O(l)
shows that for one mole of MnO4- ions to be reduced, 5 moles of electrons are needed – in this case, from the vanadium ions.
- Therefore the number of moles of electrons lost by the vanadium ions equals five times the number of moles of KMnO4;
moles e- = 5 x moles KMnO4
= 5 x 0.000307
= 0.001535moles
- 0.0005moles of vanadate ions gave up 0.001535moles of electrons, therefore the number of electrons given up by each ion;
= 0.001535
0.0005
= 3.07
= 3 electrons per ion.
- If the final oxidation state of the vanadium ions is 5+ (colour shows this) then the oxidation state before the titration must have been;
(5+) + (3e-) = 5 - 3
= 2+
NB: This hypothesis is backed up by work covered in the Yr.13 module ‘Unifying Concepts’ which states that vanadium 5+ ions can be oxidised to V2+ ions which are purple in colour (thereby turning the solution pale purple):
V5+ V2+ + 3e-
In Method 2:
- 25.00cm3 of 0.02molar ammonium vanadate contains 0.0005moles of vanadium ions.
- 10.70cm3 of 0.01molar KMnO4 is required to return the vanadium ions to their 5+ oxidation state.
- Mr of KMnO4 = 158g therefore;
in 1dm3 of water there are (158/100)g of KMnO4
and in 10.70cm of water there are [(158/100000) x 10.70]g of KMnO4
--- This is 0.016906g which in turn = (0.016906/158)moles
= 0.000107moles KMnO4
- The number of moles of electrons lost by the vanadium ions equals five times the number of moles of KMnO4;
moles e- = 5 x moles KMnO4
= 5 x 0.000107
= 0.000535moles
- 0.0005moles of vanadate ions gave up 0.00535moles of electrons, therefore the number of electrons given up by each ion;
= 0.000535
0.0005
= 1.07
= 1 electron per ion.
- If the final oxidation state of the vanadium ions is 5+ (colour shows this) then the oxidation state before the titration must have been;
(5+) + (1e-) = 5 - 1
= 4+
NB: This hypothesis is backed up by work covered in the Yr.13 module ‘Unifying Concepts’ which states that vanadium 5+ ions can be oxidised to V4+ ions which are blue in colour (thereby turning the solution pale blue):
V5+ V4+ + e-
So, in ‘Method 1’ the vanadium ions were reduced from 5+ to 2+ (showing a colour change of colourless to purple) and in ‘Method 2’ from 5+ to 4+ (showing a colour change of colourless to blue).