# Question 06929

Apr 27, 2016

Any substance that can deliver calcium cations or sulfate anions to the solution.

#### Explanation:

Calcium sulfate, ${\text{CaSO}}_{4}$, is considered insoluble in aqueous solution, which means that when you dissolve it in water, an equilibrium is established between the dissolved ions and the undissolved solid.

${\text{CaSO"_ (4(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + "SO}}_{4 \left(a q\right)}^{2 -}$

Calcium sulfate's solubility product constant, ${K}_{s p}$, which essentially tells you how far to the left that equilibrium lies, is equal to $9.1 \cdot {10}^{- 6}$.

Since ${K}_{s p}$ is significantly smaller than $1$, you can expect that equilibrium to lie very far to the left, i.e. you have a lot more undissolved solid than dissolved ions.

Now, in order to decrease the solubility of calcium sulfate, you must essentially find away to push the equilibrium even further to the left.

As you know, equilibrium reactions are governed by Le Chatelier's Principle, which states placing a stress on a system at equilibrium will cause that equilibrium to shift in such a way as to reduce that stress.

In your case, you can place a stress on the position of the equilibrium by increasing the concentration of either calcium cations, ${\text{Ca}}^{2 +}$, or sulfate anions, ${\text{SO}}_{4}^{2 -}$, or both.

This will cause the equilibrium to shift to the left because the reverse reaction consumes the extra ions by reforming some of the solid.

So, take a look at the Solubility Rules for ionic compounds

Your goal here is to find substances that can dissolve in aqueous solution to produce calcium cations or sulfate anions.

Let's start with calcium cations. Notice that all the compounds that contain the nitrate anion, ${\text{NO}}_{3}^{-}$, are soluble.

This means that if you were to add calcium nitrate, "Ca"("NO"_3)_2#, to a solution of calcium sulfate, you would reduce the calcium sulfate's solubility because the calcium nitrate would deliver calcium cations to the solution and affect the position of the equilibrium

${\text{Ca"("NO"_ 3)_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"NO}}_{3 \left(a q\right)}^{-}$

Now do the same for the sulfate anion. Notice that you are given a list of cations that form insoluble compounds when paired with the sulfate anion. Pick one that is not on that list.

A good choice would be sodium sulfate, ${\text{Na"_2"SO}}_{4}$, a soluble ionic compound that will dissolve in aqueous solution to give you sulfate anions

${\text{Na"_ 2"SO"_ (4(aq)) -> 2"Na"_ ((aq))^(+) + "SO}}_{4 \left(a q\right)}^{2 -}$

Dissolving sodium sulfate in a saturated solution of calcium sulfate will decrease the latter's solubility because of the increase in the concentration of sulfate anions.