Animated Solution for Chemistry - Ionic Equilibrium: The solubility product of a salt having general formula $M{X}_2$, in water is $4\times 10^{-12}$. The concentration of $M^{2+}$ ions in the aqueous solution of the salt is

Imagine you have a beaker filled with pure water. You drop a pinch of a sparingly soluble salt, let's call it $M{X}_2$, into it. It doesn't vanish completely like table salt; instead, it settles at the bottom. But don't be fooled by its stillness! A dynamic, invisible dance is happening right at the boundary between the solid and the water. The solid is constantly dissolving into ions, and those ions are constantly recombining to form the solid. When the rates of these two processes match, we reach a state of equilibrium.

To understand this equilibrium mathematically, we first need to write down the chemical equation for the dissociation. For our generic salt $M{X}_2$, the equation looks like this:

Notice the stoichiometry here. It's the secret key to the whole problem. For every one mole of solid $M{X}_2$ that dissolves, it releases exactly one mole of $M^{2+}$ ions and two moles of $X^{-}$ ions into the aqueous solution.

Now, let's introduce a variable to represent how much of the salt actually dissolves. We call this the molar solubility, denoted by $S$ (in mol/L). If $S$ moles of $M{X}_2$ dissolve per liter of water, then based on our balanced equation, the equilibrium concentration of the ions will be:

Next, we bring in the heavy hitter: the solubility product constant, $K_{sp}$. The $K_{sp}$ is defined as the product of the equilibrium concentrations of the dissolved ions, with each concentration raised to the power of its stoichiometric coefficient from the balanced equation.

Let's substitute our expressions involving $S$ into the $K_{sp}$ equation. This is where many students make a silly mistake, so pay close attention to the squaring!

Expanding the squared term gives us $4S^2$. Multiplying that by $S$ yields a beautifully simple expression:

The problem graciously provides us with the value of $K_{sp}$, which is $4\times 10^{-12}$. Let's plug that in and solve for $S$:

We can easily cancel the $4$ on both sides of the equation:

Taking the cube root of both sides, we find the molar solubility:

The question specifically asks for the concentration of $M^{2+}$ ions in the solution. Looking back at our initial setup, we established that $[M^{2+}]=S$. Therefore, the concentration of $M^{2+}$ is exactly $1.0\times 10^{-4}$ M.

Always remember: The stoichiometry of the salt dictates everything. If the question had asked for the concentration of $X^{-}$ ions, the answer would have been $2S$, or $2.0\times 10^{-4}$ M. Mastering this relationship will make any $K_{sp}$ problem a walk in the park!