Solubility (hydration and lattice enthalpies)

Lithium iodide is generally much more soluble in organic solvents than lithium chloride. Explain this observation using values of lattice energies and your knowledge of the trend in ionic radii down group 7.

So I understand that LiI has a greater ionic radii so it's more polarisable than chloride ions so LiI has a greater degree of covalent character but I don't understand why that makes it more soluble?

Surely more polarisation increases the energy required to split the lattice and therefore makes solubility more endothermic and therefore less likely to be feasible?

Mark scheme:

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Thank you. :smile:

Reply 1

username913907

Original post by LeanneAmyx

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Thank you. :smile:

It's to do with the relative values of the enthalpies of hydration compared to the lattice enthalpy. A trend you will find is that ionic compounds with large differences in the radii of their ions are more soluble than than those of similar radii.
Try to work out why this might be. You need to consider the equations for the lattice enthalpy and solvation enthalpy.
These are Lattice H = k/(r1*r2)
solvation H = k/r1 + k/r2
where the k's are just constants, its the scaling with the radii which is important.

Reply 2

tigerboobs

Don't forget that there are two different kind of lattice enthalpies:
Theoretical lattice enthalpies (calculated using Coulombs law) and experimental lattice enthalpies (calculated using born haber cycles). Now, I believe that the lattice enthalpies they have given are theoretical lattice enthalpies.

Theoretical lattice enthalpies do not account for covalent character. This is because the calculation using Coulombs law treats the ions as point charges based on the assumption that the ions are perfect spheres. Covalent character occurs when there is distortion of the electron clouds i.e the ions are no longer perfect spheres.
Anyway, I'll get to the point. Theoretical lattice enthalpies will therefore show a decreasing trend down the group of their constituent ions. This is because if the ions are larger then they will be further away from one another and according to Coulombs law, this will decrease the force between the ions. If there is less of an attractive force between the ions, then the lattice enthalpy, (of formation or dissociation), will decrease.
Based on the trends of theoretical lattice enthalpies, the enthalpy of solution (dissolving one mole of solid ionic compound) will also decrease as ions get larger. This can be shown mathematically, as dHsolution = dH hydration(Cl+Li) - dHlatticeformation where dH represents change in enthalpy. Remember that lattice enthalpies of formation are exothermic (-) so in effect dHsolution = dH hydration(Cl+Li) + dHlatticeformation. Therefore if dHlatticeformation is getting smaller, so will dHsolution.

However, these are only theoretical values. As you correctly commented, the experimental values should show the opposite trend. This is because experimental values, unlike theoretical values, include the additional attraction between the ions caused by the covalent character. Larger ions are more likely to polarize and distort, therefore larger ions will introduce more covalent character into a lattice. This increased attraction will increase the lattice enthalpy of formation/dissociation.

I hope that this has helped clarify things.

Solubility (hydration and lattice enthalpies)

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