Animated Solution for Chemistry - Ionic Equilibrium: Solubility of ${\text{Ca(OH)}}_2$ is $S{\text{ mol L}}^{-1}$. The solubility product ($K_{sp}$) under the same condition is

Imagine a bustling dance floor representing a beaker of water. When we introduce solid calcium hydroxide, ${\text{Ca(OH)}}_2$, it's like a group of tightly knit trios entering the room. Because ${\text{Ca(OH)}}_2$ is a sparingly soluble salt, only a few of these trios decide to break apart and mingle.

When they do break apart, the chemistry is precise. Each trio consists of one calcium ion (${\text{Ca}}^{2+}$) and two hydroxide ions (${\text{OH}}^{-}$). The chemical equation perfectly captures this breakup:

Let's define the solubility of the salt as $S{\text{ mol L}}^{-1}$. This simply means that $S$ moles of the solid ${\text{Ca(OH)}}_2$ successfully dissolve in one liter of water.

Now, we must look at the stoichiometry—the ratio of the ions produced. For every $S$ moles of ${\text{Ca(OH)}}_2$ that dissolve, we get exactly $S$ moles of ${\text{Ca}}^{2+}$ ions. However, because each molecule releases two hydroxide ions, the concentration of ${\text{OH}}^{-}$ in the solution will be exactly double, which is $2S$.

Solubility Product

The solubility product constant, $K_{sp}$, is the ultimate mathematical rule governing this equilibrium. It is defined as the product of the concentrations of the dissolved ions, with each concentration raised to the power of its stoichiometric coefficient from the balanced equation.

Notice the square on the hydroxide concentration? That is the most critical part of the formula, directly inherited from the coefficient '2' in our balanced equation.

Now, we substitute our equilibrium concentrations into the $K_{sp}$ expression. We plug in $S$ for calcium and $2S$ for hydroxide:

This is where mistakes happen! You must square the entire term $(2S)$, which means squaring both the number 2 and the variable $S$.

Finally, multiplying these terms together gives us the elegant final relationship:

This result, $4S^3$, is a universal mathematical signature for any $A{B}_2$ type sparingly soluble salt. Once you understand the logic behind the stoichiometry and the $K_{sp}$ expression, you can derive this relationship for any salt imaginable!