Animated Solution for Chemistry - Ionic Equilibrium: If solubility product of $Z{r}_3(P{O}_4{)}_4$ is denoted by $K_{sp}$ and its molar solubility is denoted by $S$, then which of the following relation between $S$ and $K_{sp}$ is correct?

Imagine a beaker where solid Zirconium Phosphate, $Z{r}_3(P{O}_4{)}_4$, is dissolving in water. It establishes a dynamic equilibrium, breaking down into Zirconium and Phosphate ions. The balanced chemical equation for this dissolution process is the foundation of our problem:

If we define the molar solubility of the salt as $S$, it means that $S$ moles of the solid dissolve per liter of solution. According to the stoichiometry of the balanced equation, for every one mole of salt that dissolves, we get three moles of Zirconium ions and four moles of Phosphate ions. Therefore, at equilibrium, the concentrations of the ions will be:

Now, let's write the expression for the solubility product constant, $K_{sp}$. The $K_{sp}$ is defined as the product of the equilibrium concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the balanced dissolution equation.

For our salt, the expression is:

Let's substitute the equilibrium concentrations we found in terms of $S$ into our $K_{sp}$ expression:

Don't make a silly mistake here. We must carefully apply the exponents to both the coefficients and the variable $S$. We need to cube the $3$ to get $27$, and raise $4$ to the power of $4$ to get $256$. The powers of $S$ also multiply out:

Now, multiply the constants $27$ and $256$ together. This gives us $6912$. And $S^3$ multiplied by $S^4$ gives $S^7$ (since we add the exponents when multiplying the same base):

Finally, we need to isolate $S$ to find the correct relation. We divide $K_{sp}$ by $6912$ and then take the seventh root of both sides:

And there we have it! This matches option (b).

Pro Tip: You can actually use a direct shortcut formula for any sparingly soluble salt of the general form $A_x{B}_y$. The solubility product is simply given by $K_{sp}=x^x{y}^y{S}^{x+y}$. Try applying this to $Z{r}_3(P{O}_4{)}_4$ where $x=3$ and $y=4$, and you'll arrive at the same result instantly!