When you calculate the solubility product constant for a salt and one of its hydrates, you sometimes get drastically different numbers! I don't know how this can be because they both dissociate into the same ions. For example, sodium sulfate heptahydrate $\ce{Na2SO4.(H2O)7}$ is more soluble than anhydrous sodium sulfate $\ce{Na2SO4}$. If you work through the math, the discrepancy in solubility is not completely explained by the larger mass of the heptahydrate (see calculations below).

I can explain the differences in solubility by assuming that the water molecules within the crystal structure of sodium sulfate heptahydrate lower the enthalpy of crystallization, which makes solvation energetically more favorable than the anhydrous salt.

What confuses me is the logical series of events that would happen when you add the solid hydrate into solution. The hydrate will dissolve into $\ce{Na+}$ and $\ce{SO4^2-}$ions. The concentration of those ions will increase until the solution is saturated (with respect to the anhydrous salt). Anhydrous $\ce{Na2SO4}$ will begin to precipitate out, keeping the ion concentrations below $K_\mathrm{sp}(\ce{Na2SO4})$ The hydrate will dissolve completely, because the solution will never reach saturation (with respect to the heptahydrate). How, then, was the solubility of the hydrate ever experimentally determined?

Using $\ce{Na2SO4}$ as an example:

Solubility of anhydrous $\ce{Na2SO4}$ = $\frac{4.76\ \mathrm g}{100\ \mathrm{mL}}$ (water, $0\ \mathrm{^\circ C}$)
Solubility of $\ce{Na2SO4.(H2O)7}$ = $\frac{19.5\ \mathrm g}{100\ \mathrm{mL}}$ (water, $0\ \mathrm{^\circ C}$)

Calculating saturated concentration:

Solubility of anhydrous $\ce{Na2SO4}$ = $$\frac{4.76\ \mathrm g\ \ce{Na2SO4}}{100\ \mathrm{mL}}\times\frac{1\ \mathrm{mol}\ \ce{Na2SO4}}{142.04\ \mathrm g\ \ce{Na2SO4}}\times\frac{1000\ \mathrm{mL}}{1\ \mathrm L}=0.335\ \mathrm M\ \ce{Na2SO4}$$

Solubility of $\ce{Na2SO4.(H2O)7}$ = $$\frac{19.5\ \mathrm g\ \ce{Na2SO4.(H2O)7}}{100\ \mathrm{mL}}\times\frac{1\ \mathrm{mol}\ \ce{Na2SO4.(H2O)7}}{268.1491\ \mathrm g\ \ce{Na2SO4.(H2O)7}}\times\frac{1000\ \mathrm{mL}}{1\ \mathrm L}=0.727\ \mathrm M\ \ce{Na2SO4.(H2O)7}$$

Calculating $K_\mathrm{sp}$:

Anhydrous $\ce{Na2SO4}$: $$\ce{Na2SO4 <=>2Na+ + SO4^2-}$$ $$K_\mathrm{sp}(\ce{Na2SO4})=\left[\ce{Na+}\right]^2\left[\ce{SO4^2-}\right]=\left(0.335\times 2\right)^2\times\left(0.335\right)=0.150$$

$\ce{Na2SO4.(H2O)7}$: $$\ce{Na2SO4.(H2O)7 <=>2Na+ + SO4^2- +7H2O}$$ $$K_\mathrm{sp}\left(\ce{Na2SO4.(H2O)7}\right)=\left[\ce{Na+}\right]^2\left[\ce{SO4^2-}\right]=\left(0.666\times 2\right)^2\times\left(0.666\right)=1.18$$

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    $\begingroup$ You're whole reasoning doesn't have much sense, like these were really different things. For purpose of solubility you can treat Na2SO4⋅(H2O)7 like Na2SO4 with some water added - you're changing amount of water and don't account for it in last equation. $\endgroup$
    – Mithoron
    Aug 3, 2015 at 23:29
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    $\begingroup$ I think you don't have a good handle on what ionic hydrates are. They are solid salts with waters incorporated into the crystal structure. They can have wildly different properties, solubility is just one of them. en.m.wikipedia.org/wiki/Water_of_crystallization $\endgroup$
    – Dan Burden
    Aug 3, 2015 at 23:38
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    $\begingroup$ You're overthinking it totally - error in last equation, you can't put there 0.727 as water amount is increasing - concentration of Na and SO42- is much lower. $\endgroup$
    – Mithoron
    Aug 4, 2015 at 14:03
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    $\begingroup$ Fixed in edit. The concentrations do not change much. Ksp is still much larger for the hydrate than the anhydrous salt. $\endgroup$
    – Dan Burden
    Aug 4, 2015 at 14:43
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    $\begingroup$ I think that any precipitates that form from an aqueous saturated solution of hydratable salts will always be hydrates. $\endgroup$
    – Curt F.
    Aug 5, 2015 at 4:23

1 Answer 1


Let there be three species: a hydrated solid $H$, an anhydrous solid $A$, and a complete solvated (liquid) aqueous solution, $S$.

Your question is about differences in the two equilibria $\ce{H <=> S}$ and $\ce{A <=> S}$. Can $H$ really have a different solubility than $A$?

I think the answer is yes, but only if the equilibrium $\ce{H<=>A}$ cannot and is not established. In the presence of an "infinite" kinetic barrier that blocks conversion of the hydrate to the anhydrous form and vice versa, it makes sense to treat $\ce{H<=>S}$ and $\ce{H<=>A}$ and independent equilibria which might have different equilibrium constants.

In a comment to your question, I made it seem like "any" precipitation from a (super)saturated solution of $S$ would be the hydrate. On further reflection, I think that is true, unless (i) the equilibrium $\ce{A<=>H}$ cannot be established due to kinetic barriers, and (ii) there is solid phase $A$ present in the solution. In that case, the supersaturation could just result in "precipitation" of solid onto the existing crystal surface of already-present $A$, a process that would increase the mass of $A$ (solid) in the system but which would not require formation of any crystallites or nucleation intermediates.

How, then, was the solubility of the hydrate ever experimentally determined?

I think the answer is that the hydrate solubility can differ from the anhydrous solid solubility only in cases when there is no mechanism by which equilibrium can be established between the hydrate and the anhydrous solid.

  • $\begingroup$ Just to be sure - did you mix up any A H or S in there? I can't follow some of your statements (i.e. The solubility of S, which is an aqueous solution). $\endgroup$
    – Dan Burden
    Aug 16, 2015 at 6:13
  • $\begingroup$ I also feel like your answer may not address the real concern of the question. Once the ions are in solution, they are participating in two competing equilibria. Common sense says that the lowest Ksp solid will always precipitate out before the higher Ksp is reached, right? $\endgroup$
    – Dan Burden
    Aug 16, 2015 at 21:07
  • $\begingroup$ I don't think that is necessarily true. It depends on the height of the activation barrier for precipitation. The free energy barrier $\Delta G^{\dagger}$ for the route that leads to the more stable product can be higher than the route that leads to the less stable product. In other words, precipitation can be theoretically controlled. $\endgroup$
    – Curt F.
    Aug 17, 2015 at 3:25

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