Animated Solution for Chemistry - Ionic Equilibrium: At $25^{\circ }\text{C}$, the solubility product of ${\text{Mg(OH)}}_2$ is $1.0\times 10^{-11}$. At which pH, will ${\text{Mg}}^{2+}$ ions start precipitating in the form of ${\text{Mg(OH)}}_2$ from a solution of $0.001\text{ M }{\text{Mg}}^{2+}$ ions?

The phenomenon of precipitation is like a hidden magic trick in chemistry. You can have a perfectly clear, transparent solution, but the moment you cross a specific chemical threshold, solid particles suddenly materialize out of nowhere and fall to the bottom of the beaker. This threshold is governed by a beautiful balance between the ions in the solution, known as the Solubility Product.

In this problem, we are exploring exactly when this magic trick happens for magnesium hydroxide, ${\text{Mg(OH)}}_2$. Let's dive into the mechanics of this ionic dance.

Imagine you are looking at a beaker filled with water. Dissolved in this water are magnesium ions, ${\text{Mg}}^{2+}$, floating around freely. We are told that the concentration of these magnesium ions is $0.001\text{ M}$, which we can write more conveniently as $10^{-3}\text{ M}$.

Right now, the solution is clear. But we want to force these magnesium ions to pair up with hydroxide ions, ${\text{OH}}^{-}$, to form a solid precipitate of ${\text{Mg(OH)}}_2$. To do this, we need to add hydroxide ions to the solution, which effectively means we are increasing the pH. The question is: at what exact pH will the solid just begin to appear?

To answer this, we need to consult the rulebook of solubility. For a sparingly soluble salt like ${\text{Mg(OH)}}_2$, the equilibrium reaction is:

Based on the stoichiometry of the reaction, the expression for the Ionic Product is:

Notice the square on the hydroxide concentration! This is a common trap. Because one molecule of ${\text{Mg(OH)}}_2$ releases two hydroxide ions, the concentration of ${\text{OH}}^{-}$ must be squared in the equilibrium expression.

So, our condition for the onset of precipitation is:

We know the $K_{sp}$ is $1.0\times 10^{-11}$, and we know our starting $[{\text{Mg}}^{2+}]$ is $10^{-3}\text{ M}$. Let's substitute these raw values into our master equation:

Now, we just need to isolate the hydroxide concentration. Dividing both sides by $10^{-3}$ gives us:

Using the laws of exponents, we subtract the denominator's exponent from the numerator's exponent ($-11-(-3)=-8$):

Taking the square root of both sides reveals the exact concentration of hydroxide ions required to trigger precipitation:

We have found the required hydroxide concentration, but the question asks for the pH. We need to build a bridge from $[{\text{OH}}^{-}]$ to pH. The first step across this bridge is calculating the pOH.

By definition, pOH is the negative base-10 logarithm of the hydroxide ion concentration:

Substituting our value:

The exponent $-4$ comes down to multiply with the negative sign, giving us a beautifully clean integer:

We are almost there. The final step is to convert pOH to pH. At standard room temperature ($25^{\circ }\text{C}$), the relationship between pH and pOH is rigidly fixed by the auto-ionization of water:

To find the pH, we simply subtract our calculated pOH from 14:

And there we have it! At a pH of exactly 10, the solution reaches its saturation point. If the pH is even slightly below 10, the solution remains clear. But the moment the pH hits 10 and tries to climb higher, the invisible magnesium ions will grab the abundant hydroxide ions, and a cloudy white precipitate of ${\text{Mg(OH)}}_2$ will begin to fall.