Animated Solution for Chemistry - Ionic Equilibrium: The molar solubility (in ${\text{mol L}}^{-1}$) of a sparingly soluble salt ${\text{MX}}_4$ is $S$. The corresponding solubility product is $K_{sp}$. $S$ is given in terms of $K_{sp}$ by the relation.

Imagine dropping a pinch of a sparingly soluble salt, ${\text{MX}}_4$, into a beaker of pure water. It doesn't vanish completely like table salt. Instead, a delicate equilibrium is established. A tiny fraction of the solid dissolves, releasing ions into the water, while simultaneously, those ions collide and reform the solid.

This dynamic balance is governed by a fundamental principle of ionic equilibrium: the Solubility Product Constant, denoted as $K_{sp}$. But how exactly does this constant relate to the actual amount of salt that dissolves, which we call the molar solubility, $S$?

The first step to unraveling this relationship is to write down the balanced chemical equation for the dissociation of the salt. When one molecule of ${\text{MX}}_4$ dissolves, it breaks apart into one ${\text{M}}^{4+}$ ion and four ${\text{X}}^{-}$ ions:

Now, let's define our terms. Let the molar solubility of the salt be $S$ ${\text{mol L}}^{-1}$. This means that at equilibrium, $S$ moles of the solid ${\text{MX}}_4$ have dissolved in every liter of solution.

Looking at the stoichiometry of our balanced equation, for every $S$ moles of ${\text{MX}}_4$ that dissolve, we get $S$ moles of ${\text{M}}^{4+}$ ions. However, because of the subscript '4', we get four times as many ${\text{X}}^{-}$ ions. Therefore, the equilibrium concentrations are:

The solubility product constant, $K_{sp}$, is defined as the product of the equilibrium concentrations of the dissolved ions, with each concentration raised to the power of its stoichiometric coefficient in the balanced equation. For our salt, the expression is:

This is where many students make a critical error. They forget that the concentration of ${\text{X}}^{-}$ is $4S$, AND that this entire value must be raised to the power of 4. Let's substitute our equilibrium concentrations into the $K_{sp}$ expression carefully:

Now, it's just a matter of algebraic expansion. We must apply the exponent of 4 to both the coefficient 4 and the variable $S$ inside the parentheses:

Substituting this back into our equation gives:

We have successfully expressed $K_{sp}$ in terms of $S$. But the question asks us to find $S$ in terms of $K_{sp}$. To isolate $S$, we first divide both sides by 256:

Finally, we take the fifth root of both sides to solve for $S$:

This elegant mathematical relationship allows chemists to predict exactly how much of a complex salt will dissolve in water just by knowing its solubility product constant. It's a beautiful example of how stoichiometry directly dictates the mathematical form of chemical equilibrium!