Animated Solution for Chemistry - Ionic Equilibrium: The solubility of ${\text{Ca(OH)}}_2$ in water is [Given: The solubility product of ${\text{Ca(OH)}}_2$ in water = $5.5\times 10^{-6}$]

Imagine you are looking at a beaker filled with water. You drop a pinch of calcium hydroxide, ${\text{Ca(OH)}}_2$, into it. You might expect it to vanish completely, like sugar or salt. But calcium hydroxide is a bit stubborn; it's what we call a sparingly soluble salt. Instead of dissolving entirely, it establishes a beautiful, dynamic equilibrium. The solid at the bottom constantly sends ions into the water, while the ions in the water constantly recombine to form the solid.

This delicate dance is governed by a strict mathematical rule, and that's exactly what we are going to decode today.

To understand this mathematically, we first need to write down the chemical equation for the dissociation process. When one molecule of solid calcium hydroxide dissolves, it breaks apart into one calcium ion and two hydroxide ions:

Let's assume the molar solubility of this salt in pure water is $S\text{ mol/L}$. This means $S$ moles of the solid dissolve in every liter of water.

According to the stoichiometry of our balanced equation, if $S$ moles of ${\text{Ca(OH)}}_2$ dissolve, they will produce $S$ moles of ${\text{Ca}}^{2+}$ ions and $2S$ moles of ${\text{OH}}^{-}$ ions.

So, at equilibrium, our concentrations are: - $[{\text{Ca}}^{2+}]=S$ - $[{\text{OH}}^{-}]=2S$

Solubility Product ($K_{sp}$)

The solubility product constant, $K_{sp}$, is the ultimate judge of how much a salt can dissolve. It is defined as the product of the equilibrium concentrations of the dissolved ions, each raised to the power of its stoichiometric coefficient.

For our reaction, the expression looks like this:

Now, let's substitute the equilibrium concentrations we found earlier into this master equation:

Be very careful here! A common trap is to forget to square the $2$ inside the bracket. Expanding this, we get:

The problem generously provides us with the value of $K_{sp}$, which is $5.5\times 10^{-6}$. Let's plug this into our simplified equation:

Our goal is to isolate $S$. First, we divide both sides by $4$:

To find the final solubility $S$, we need to take the cube root of both sides.

The cube root of $10^{-6}$ is simply $10^{-2}$. For the number part, $\sqrt[3]{1.375}$, we can estimate it. We know $1^3=1$ and $1.5^3=3.375$. The value $1.375$ is quite close to $1$. Looking at our options, $1.11$ is a perfect candidate because $1.11^3\approx 1.367$, which is incredibly close to $1.375$.

And there we have it! The solubility of calcium hydroxide is $1.11\times 10^{-2}\text{ mol/L}$, which corresponds perfectly to option (a).