June 14th, 2007
Hardness is a term to describe water that doesn’t allow soaping action, in other words no bubbles (lather) and no cleaning. The term ’soaping action’ refers to the formation of micelles by negative ions joined to long carbon chains that dissolve in the fat or grease leaving the negative charges outside the micelle structure. These can then bond to the water molecules. In this way the hydrophobic grease is made to dissolve and lifts off the fabric or skin.
Hardness is due to Ca2+ or Mg2+ ions dissolved in the water. These ions react with the long negative ions of soap (stearate ions C17 H35 COO-) to make an insoluble grey compound which we call ’scum’. As the reaction removes the necessary soap ions from the water it prevents lathering and the soap cannot perform its function.
Hardness can be removed by various means depending on the other ion that is dissolved with the Mg2+ or Ca2+ ions.
There are two subcategories of hardness:
1. Temporary
2. Permanent
Temporary hardness is due to dissolved calcium (or magnesium) hydrogen carbonate Ca(HCO3)2, it can be removed by boiling the water, when the hydrogen carbonate ions decompose forming insoluble calcium (or magnesium) carbonate:
Ca(HCO3)2 –> CaCO3 + CO2 + H20
Permanent hardness (due to dissolved calcium and magnesium sulphates and chlorides) must be removed by another method either:
1. ion exchange using resins etc.
2. chemical precipitation
Ca2+ + CO3(2-) –> CaCO3(ppt)
use of substances such as washing soda (sodium carbonate) do just that.
August 3rd, 2007
Manganese (VI) is usually in the form of the MnO42- ion. This ion disproportionates in acid solution but not in base. What is going on here?
The Mn(VI) state is not stable under acid conditions because it CAN disproportionate (notice that this is NOT a half-equation - it is a reaction):
3MnO4(2-) + 4H+ –> MnO2 + 2MnO4(-) + 2H2O
The Mn(VI) state cannot do this under alkaline conditions, and the alkaline disproportionation reaction (theoretical) would be:
3MnO4(2-) + 2H2O –> MnO2 + 2MnO4(-) + 4OH-
This is now in direct competition with the reverse reaction (as all reactions are) which we know DOES occur.
However, what we are really comparing is the Gibbs free energy of the two disproportionation equations. In the case of acid conditions:
3MnO4(2-) + 4H+ –> MnO2 + 2MnO4(-) + 2H2O
The Gibbs Free energy makes the forward reaction feasible. In the case of alkaline conditions however, Gibbs Free energy makes the backward reaction equally feasible.
3MnO4(2-) + 2H2O <– MnO2 + 2MnO4(-) + 4OH-
This can be predicted by reference to the electrode potential of the half equations involved. If you break down each disporportionation into two half-equations and compare the redox potentials using E = E(red) - E(ox).
OK, let’s do it!
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Acidic conditions:
3MnO4(2-) + 4H+ –> MnO2 + 2MnO4(-) + 2H2O
This disproportionation can be though of as two half equations:
1. MnO4(2-) + 4H+ + 2e –> MnO2 + 2H2O ……….. Eº = +2.26V ( the Manganate ion is reduced)
2. MnO4(2-) –> MnO4(-) + 1e ……………………….Eº = +0.56V ( the Manganate ion is oxidised
Calculating Eº = E(reduced state) - E(oxidised state) = +2.26V - +0.56V = + 1.70V
This is a large positive value (remember that any Eº value greater than 0.3v means that the reaction is spontaneous as shown) so the forward reaction proceeds.
Alkaline conditions:
3MnO4(2-) + 2H2O –> MnO2 + 2MnO4(-) + 4OH-
This disproportionation can be though of as two half equations:
MnO4(2-) + 2H2O + 2e –> MnO2 + 4OH- ……….. Eº = +0.67V (reduction)
MnO4(2-) –> MnO4(-) + 1e…………………………Eº = +0.56V (oxidation)
Calculating Eº = E(red) - E(ox) = 0.67 - 0.56 = +0.11V
Although this is a positive value, it is very small (under 0.3V) indicating that an equilibrium will be established (under standard conditions) and that the reaction will proceed in the forward direction by only an immeasurable amount, and as the equation is in equilibrium, by Le Chatelier, addition of base will drive the equilibrium in the reverse direction.
