Animated Solution for Chemistry - Ionic Equilibrium: The solubility in water of a sparingly soluble salt $A{B}_2$ is $1.0\times 10^{-5}{\text{ mol L}}^{-1}$. Its solubility product will be

Visualizing the Dissociation

Imagine you are looking at a beaker of pure water. When we drop a sparingly soluble salt, $A{B}_2$, into it, a fascinating microscopic dance begins. A very tiny fraction of the solid manages to break free from its crystal lattice and dissolve into the water.

For every single molecule of $A{B}_2$ that dissolves, it shatters into three distinct pieces: one $A^{2+}$ ion and two $B^{-}$ ions. We can represent this dynamic equilibrium with a balanced chemical equation:

Now, let's define a crucial term: Molar Solubility ($S$). This represents the maximum number of moles of the solid $A{B}_2$ that can dissolve in one liter of water.

If $S$ moles of $A{B}_2$ dissolve, the stoichiometry of our balanced equation tells us exactly how many ions are produced. We will get $S$ moles per liter of $A^{2+}$ ions. However, because of the coefficient '2' in front of the $B^{-}$ ion, we will get $2S$ moles per liter of $B^{-}$ ions. This factor of two is the most common place students make a silly mistake, so watch out for it!

Solubility Product

To quantify this equilibrium, we use the Solubility Product Constant ($K_{sp}$). The law of mass action dictates that $K_{sp}$ is the product of the equilibrium concentrations of the dissolved ions, with each concentration raised to the power of its stoichiometric coefficient.

For our salt, the expression looks like this:

Notice that the solid $A{B}_2$ does not appear in this equation. The concentration of a pure solid is constant and is already baked into the $K_{sp}$ value itself.

Now, let's substitute our equilibrium concentrations (in terms of $S$) into this master equation:

Here is where we need to be careful with our algebra. We must square the entire $(2S)$ term, not just the $S$.

This elegant little formula, $K_{sp}=4S^3$, is a universal truth for any salt that dissociates into three ions in a 1:2 ratio (like $MgC{l}_2$, $Ca{F}_2$, or our generic $A{B}_2$).

Crunching the Numbers

We are given the molar solubility of the salt in the problem:

All that's left is to plug this value into our simplified formula and crunch the numbers.

To calculate the cube of an exponential term, we simply multiply the exponent by 3. So, $(10^{-5}{)}^3$ becomes $10^{-15}$.

And there we have it! The solubility product constant for our sparingly soluble salt is $4\times 10^{-15}$. It's a straightforward calculation once you master the setup and avoid the classic stoichiometric traps.