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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 $K_\mathrm{sp}=\ce{[Mg^2+][OH−]}=s \times (2s)2$

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    $\begingroup$ 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. $\endgroup$ – Ivan Neretin Apr 1 '16 at 14:56
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    $\begingroup$ Here is a chart with some data for potassium alum en.wikipedia.org/wiki/Alum#Solubility $\endgroup$ – MaxW Apr 1 '16 at 15:07
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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: $$RT~\ln\left(K_\mathrm{sp}\right) = \Delta G_\mathrm{sol}^\circ = \Delta H_\mathrm{sol}^\circ - T\Delta S_\mathrm{sol}^\circ$$

Producing potassium alum is a reaction of its own:

$$\ce{K2SO4 + Al2(SO4)3 -> 2 KAl(SO4)2}$$

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.

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