# Solubility of Alum

What is solubility product of $$\ce{K2SO4. Al2(SO4)3.24H2O}$$?

Can we consider $$K_\mathrm{sp}$$ of salts $$\ce{K2SO4}$$ and $$\ce{Al2(SO4)3}$$ or something else to find solubility?

I know that Sulphate ions are coming from two different salts. That's were my brain stop working. Is it possible to solve this question something like $$K_\mathrm{sp}=\ce{[Mg^2+][OH−]}=s \times (2s)2$$

• The very idea of using $K_{sp}$ for well-soluble salts does not sit well with me, but whatever. Now, there are no two different salts here; $\ce{K2SO4. Al2(SO4)3.24H2O}$ is one compound and should be treated as such. The data for $\ce{K2SO4}$ and $\ce{Al2(SO4)3}$ have nothing to do with it. Commented Apr 1, 2016 at 14:56
• Here is a chart with some data for potassium alum en.wikipedia.org/wiki/Alum#Solubility
– MaxW
Commented Apr 1, 2016 at 15:07

No you should not because $$\ce{KAl(SO4)2}$$ has a different crystal structure and thus different enthalphy and entropy of formation values than the pure independent salts.
The $$K_\mathrm{sp}$$ of a substance is given by: $$RT~\ln\left(K_\mathrm{sp}\right) = \Delta G_\mathrm{sol}^\circ = \Delta H_\mathrm{sol}^\circ - T\Delta S_\mathrm{sol}^\circ$$
$$\ce{K2SO4 + Al2(SO4)3 -> 2 KAl(SO4)2}$$
This independent reaction has its own $$\Delta G^\circ$$ of formation and thus the reactants to product solution have a different energy and thus different solubility.
Further this double salt affects your cation and anionic species of solution. For example the potassium may now dissociate to form a $$\ce{K+}$$ and a more stable $$\ce{[Al(SO4)2]-}$$ counter ion which will affect solubility as well.