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Ellie (Parent) February 22 2017
Firstly, we must consider the thermodynamics involved
It is known that a reaction is spontaneous due to thermodynamic favourability if the Gibbs Free Energy is negative. This is further broken down to entropy and enthalpy terms.
Upon dissolution of a salt, e.g. a hydroxide, the entropy increases (a favourable term) in all cases as the solid is separated into ions, which are free to move in solution.
However, in order for the ions to separate, the lattice must be broken. This requires breakage of the electrostatic forces of attraction between oppositely charged ions. Thus, energy must put in equal tot he lattice enthalpy - this is a very large positive term and so is unfavourable.
The above statements have only considered forced between the ions in the lattice, however, upon dissolution, the free ions interact with solvent molecules. In our example, we are using water. The polar water molecules can interact favourably with both types of ions and this introduces a further enthalpy term, known as the hydration enthalpy. If this interaction is favourable enough (large negative enthalpy) to overcome the energy required to break the lattice, the salt will dissolve.
Now we can consider the group 2 hudroxides and since the anion is identical in each case, we will only examine the cations. The early hydroxides, e.g. CaOH, are comprised of smaller cations (with a larger charge density) and thus have a very large lattice enthalpy. Interactions with water molecules are not great due to their relative sizes and so the hydration enthalpy is not large enough to overcome the electrostatic forces within the lattice and so the salt does not dissolve.
The opposite holds true for larger cations, e.g. Barium, and so later hydroxides are more readily dissolved in water
Note: HSAB theory has not been discussed as it is not taught at A Level (tot he best of my knowledge)see more