Animated Solution for Chemistry - Ionic Equilibrium: The $K_{sp}$ for ${\text{Cr(OH)}}_3$ is $1.6\times 10^{-30}$. The molar solubility of this compound in water is

The journey to understanding the solubility of sparingly soluble salts begins with a simple yet profound physical visualization. Imagine a beaker filled with pure water. When we drop solid chromium hydroxide into it, it doesn't just sit there inertly. A dynamic, invisible dance begins.

At the microscopic level, the solid lattice of chromium hydroxide is constantly exchanging ions with the surrounding water. For every single chromium ion that breaks free and enters the aqueous phase, exactly three hydroxide ions are released alongside it. This is not a one-way street; ions in the solution are also constantly colliding and reforming the solid. When the rate of dissolution perfectly matches the rate of precipitation, we reach a state of dynamic equilibrium.

Mathematically, we represent this delicate balance using the solubility product constant, $K_{sp}$. The law of mass action tells us that $K_{sp}$ is the product of the molar concentrations of the dissolved ions, with each concentration raised to the power of its stoichiometric coefficient from the balanced equation.

To solve for the molar solubility, which we will call $s$, we need to express the equilibrium concentrations of our ions in terms of this single variable. The molar solubility $s$ represents the number of moles of the solid salt that dissolve per liter of solution.

Because the stoichiometry of the dissolution is $1:3$, the concentration of chromium ions will simply be $s$. However, the concentration of hydroxide ions will be three times that amount, or $3s$.

Now comes the critical step where many students make a fatal error. We must substitute these expressions back into our $K_{sp}$ equation. It is absolutely vital to remember that the entire concentration of hydroxide, which is $3s$, must be cubed.

Let's expand this carefully. Cubing the $3$ gives us $27$, and cubing the $s$ gives us $s^3$. Multiplying this entire term by the $s$ from the chromium ion yields our simplified expression:

We are given the incredibly small value of $K_{sp}$ as $1.6\times 10^{-30}$. This tiny number confirms that chromium hydroxide is highly insoluble. We equate this given value to our derived expression:

To isolate $s^4$, we divide both sides by $27$:

Finally, to find the molar solubility $s$, we take the fourth root of the entire right side.

This elegant mathematical result perfectly matches option (c). It's a beautiful demonstration of how macroscopic properties like solubility are governed by strict, predictable microscopic rules!