2
$\begingroup$

While removing temporary hardness through boiling, magnesium bicarbonate will give $\ce{Mg(OH)2}.$ $\ce{Mg(OH)2}$ predominates over $\ce{MgCO3}$ as the solubility product $K_\mathrm{sp}$ of $\ce{Mg(OH)2}$ is higher compared to that of $\ce{MgCO3}$.

When a compound is dissolved where it is possible to precipitate two different salts, will the one with the lower $K_\mathrm{sp}$ form a precipitate?

$\endgroup$
3
  • 1
    $\begingroup$ You can precipitate selectively, when the Ksp of one compound is exceeded and the other one is not; howver, when both Ksp are exceeded, both will precipitate and it will be a matter of kinetics which precipitates first. $\endgroup$
    – Mäßige
    Commented Apr 12 at 14:19
  • 2
    $\begingroup$ the related Q/A is probably this Is Mg(OH)2 more insoluble than MgCO3? $\endgroup$
    – Poutnik
    Commented Apr 12 at 14:53
  • $\begingroup$ Yes, thank you! $\endgroup$
    – Avaneesh B
    Commented Apr 15 at 11:12

1 Answer 1

2
$\begingroup$

Not necessarily. If the salts have different stoichiometries, the solubility products are really comparing apples with olives.

According to this table, calcium hydroxide $\ce{Ca(OH)2}$ has a solubility product constant of $5.5×10^{-6}$ and calcium sulfate $\ce{CaSO4}$ has a solubility product constant of $9.1×10^{-6}$. Now suppose that a soluble calcium salt such as calcium chloride is dissolved into a medium containing $0.01$ molar hydroxide ions and $0.001$ molar sulfate ions. The calcium ion concentration required to precipitate each of the salts given above is calculated:

$\ce{Ca(OH)2}: \dfrac{5.5×10^{-6}}{0.01^2}=0.055$ molar

$\ce{CaSO4}: \dfrac{9.1×10^{-6}}{0.001}=0.0091$ molar

Despite the calcium sulfate having a higher solubility product and the hydroxide ions being more abundant in solution, the calcium sulfate will precipitate first and only after the sulfate concentration is lowered substantially further than even the conditions given above will enough calcium ions get into solution to precipitate the hydroxide. One natural consequence is that calcium-bearing sediments are more likely to be seen as gypsum (hydrated calcium sulfate) or as calcite (calcium carbonate, which gives similar results to the sulfate) than larnite (calcium hydroxide) even with significantly alkaline waters.

The main reason the relative solubility product constants and concentrations of the ions proved unreliable here is because the products are defined on a different basis; the calcium hydroxide product, with two hydroxide ions per calcium ion in the formula, requires taking the hydroxide ions twice versus the calcium sulfate taking the sulfate ion only once. The solubility product constant comparison is apples to apples only if the salts have matching stoichiometries.

$\endgroup$

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.