Animated Solution for Chemistry - Ionic Equilibrium: The solubility product of ${\text{Cr(OH)}}_3$ at $298\text{ K}$ is $6.0\times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of ${\text{Cr(OH)}}_3$ will be

Visualizing the Saturated Solution

Imagine a beaker containing a saturated solution of chromium(III) hydroxide, ${\text{Cr(OH)}}_3$. At equilibrium, the solid precipitate at the bottom is in a dynamic balance with its dissolved ions in the water.

When one molecule of ${\text{Cr(OH)}}_3$ dissolves, it breaks apart to yield one chromium ion, ${\text{Cr}}^{3+}$, and three hydroxide ions, ${\text{OH}}^{-}$. This stoichiometric ratio is the heartbeat of the entire problem. If we define the molar solubility of the salt as $S$, then the equilibrium concentrations of the ions will be:

Solubility Product Constant

The solubility product constant, $K_{sp}$, is the ultimate mathematical tool for sparingly soluble salts. It is defined as the product of the equilibrium ion concentrations, with each concentration raised to the power of its stoichiometric coefficient from the balanced equation.

For our salt, the expression is:

Stoichiometry in Concentration and Power

Here lies the most notorious trap in ionic equilibrium! Students often substitute $3S$ for the hydroxide concentration but forget to cube it, or they cube it but forget the $3$ inside. You must do both. The concentration itself is physically three times the solubility, and the law of mass action dictates that this entire concentration must be cubed.

Let's substitute our values carefully:

Expanding the cube, we get $3^3=27$ and $S^3$. Multiplying by the initial $S$ gives us:

Simplifying the Expression

We are given that $K_{sp}=6.0\times 10^{-31}$. Equating this to our derived expression, we can isolate $S$:

But wait! The question doesn't ask for the solubility $S$; it asks for the concentration of hydroxide ions, $[{\text{OH}}^{-}]$. Remember our initial setup? The hydroxide concentration is $3S$.

To match the options provided in the exam, we need to perform a neat algebraic trick. We can bring the coefficient $3$ inside the fourth root by raising it to the power of $4$. Since $3^4=81$, the expression becomes:

Now, the math becomes beautifully simple. Dividing $81$ by $27$ gives exactly $3$.

Multiplying $3$ by $6.0$ yields $18$. Thus, our final, elegant answer is:

This perfectly matches option (b). Always trust the algebra, and never rush the final simplification steps!