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JiD-O April 3rd, 2005 10:53

help on PbCl2 soluability?
 
gday. im in year 12 chem and i need help to explain moleculelly and alsod raw why lead chloride does not dissolve in water compared with sodium chloride. i kow all about sodium chloride and i can draw/explain that but teh pbcl2 has got me stuck. i have done heps o freasearch btu can t seem to find it :?: . i know itll be simple for some one out there and id really appriciate ur help.
cheers
Jid-o :wink: [/b]

charco April 5th, 2005 18:02

solubility is a complex process and there is no hard and fast rule to decide why one substance dissolves and another doesn't...

The process is a balance between

1. the lattice enthalpy of the salt (high lattice enthalpy - large amounts of energy needed to break it - not favourable)
2. The charge density on the ions (high charge density = high hydration enthalpy = favourable process)

[as 1 and 2 are fundamentally caused by the same thing = charge density of the ions it is difficult to predict which is the most important.]

3. The increase in entropy when the substance dissolves (although this is also very variable due to hydration of the ions)

so between factors 1,2,and 3 there are difficulties in prediction.


In your question you are comparing sodium Na+ and Pb2+

The lead ion has a greater charge than the sodium ion Na+ but also a greater radius therefore without actuually measuring the ion it would be difficult to give its charge density. And even with this value the overall delta G (Gibbs free energy = prediction of feasibility) of the process of dissolution is very difficult to predict.

The dissolution process is endothermic in both cases as the solubilty increases with temperature. This suggests that the lattice enthalpy (breaking of which is an endothermic process) is the decisive factor.

In conclusion, I would say that the lower solubility of PbCl2 is down to the greater lattice enthalpy suggesting that the Pb2+ ion has a greater charge density than the Na+ ion.

JiD-O April 7th, 2005 01:24

cheers. thanx 4 ur reply.


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