Animated Solution for Chemistry - Ionic Equilibrium: What is the molar solubility of ${\text{Al(OH)}}_3$ in $0.2\text{ M NaOH}$ solution? Given that, solubility product of ${\text{Al(OH)}}_3=2.4\times 10^{-24}$

Understanding the Common Ion Effect

Imagine you are trying to squeeze into a crowded subway train. If the train is already packed with people, it's going to be incredibly difficult for you to get in. This is exactly what happens at the molecular level when we try to dissolve a sparingly soluble salt into a solution that already contains one of its ions. This phenomenon is known as the Common Ion Effect, and it is the star of this problem.

In our scenario, we have a beaker filled with a $0.2\text{ M}$ solution of Sodium Hydroxide ($\text{NaOH}$). Sodium hydroxide is a strong base, meaning it completely dissociates in water:

Because it completely dissociates, the concentration of hydroxide ions (${\text{OH}}^{-}$) in the beaker is a robust $0.2\text{ M}$. Now, we introduce Aluminum Hydroxide (${\text{Al(OH)}}_3$), a sparingly soluble salt. It tries to establish its own delicate equilibrium:

Let's define the molar solubility of ${\text{Al(OH)}}_3$ as $S$. If it were dissolving in pure water, it would produce $S$ moles of ${\text{Al}}^{3+}$ and $3S$ moles of ${\text{OH}}^{-}$. However, the water is not pure; it's already teeming with hydroxide ions from the $\text{NaOH}$.

Therefore, the total concentration of hydroxide ions in the solution is the sum of the hydroxide from the strong base and the hydroxide from the dissolving salt:

Now, we turn to the governing law of this system: the Solubility Product Constant ($K_{sp}$). For ${\text{Al(OH)}}_3$, the expression is:

Substituting our known values and expressions into this equation, we get:

At first glance, expanding a cubic binomial like $(0.2+3S{)}^3$ seems like a mathematical nightmare. But here is where we must pause and look at the physical reality of the numbers. The $K_{sp}$ value is $2.4\times 10^{-24}$. This number is astronomically small! It tells us that ${\text{Al(OH)}}_3$ barely dissolves at all, even in pure water.

Because the solubility $S$ is going to be so minuscule, multiplying it by 3 and adding it to $0.2$ will barely make a dent. It's like adding a drop of water to the ocean. Therefore, we can safely make the following approximation:

This single, logical assumption transforms our terrifying cubic equation into a walk in the park.

With our approximation in place, the equation simplifies beautifully:

Calculating the cube of $0.2$ gives us $0.008$ (or $8\times 10^{-3}$):

Now, we simply isolate $S$:

Adjusting the scientific notation to standard form, we arrive at our final answer:

This incredibly small number proves our earlier assumption was absolutely correct. The common ion effect drastically suppressed the solubility of the aluminum hydroxide, demonstrating the elegant predictability of chemical equilibrium.