2
$\begingroup$

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

$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Apr 1, 2016 at 14:56
  • 2
    $\begingroup$ Here is a chart with some data for potassium alum en.wikipedia.org/wiki/Alum#Solubility $\endgroup$
    – MaxW
    Commented Apr 1, 2016 at 15:07

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.