Animated Solution for Chemistry - Ionic Equilibrium: Two salts $A_{2}X$ and $MX$ have the same value of solubility product of $4.0\times 10^{-12}$. The ratio of their molar solubilities i.e $\frac{S(A_{2}X)}{S(MX)}=\dots \dots$ (Round off to the nearest integer)

Imagine you are handed two beakers, each containing a different sparingly soluble salt: $A_{2}X$ and $MX$. You are told that both salts share the exact same solubility product constant, $K_{sp}=4.0\times 10^{-12}$. Your intuition might scream that because their $K_{sp}$ values are identical, their molar solubilities must be identical too. But nature loves a good plot twist, and this is one of the classic traps in Ionic Equilibrium.

The $A_{2}X$ Salt

Let's break down the first salt, $A_{2}X$. When it dissolves in water, it doesn't just split into two pieces; it shatters into three ions according to its stoichiometry:

If we define its molar solubility as $S$, then for every mole of $A_{2}X$ that dissolves, it yields $2S$ moles of $A^{+}$ and $S$ moles of $X^{2-}$. The Law of Mass Action dictates that the solubility product is the product of these ion concentrations, raised to the power of their stoichiometric coefficients:

Substituting our terms, we get:

Solving for $S$, we divide by 4 to get $S^3=10^{-12}$. Taking the cube root is straightforward here: $S=10^{-4}\text{ M}$.

The $MX$ Salt

Now, let's shift our focus to the second salt, $MX$. This is a simpler 1:1 salt. Its dissociation is straightforward:

Let's call its molar solubility $S_1$. The concentration of both $M^{+}$ and $X^{-}$ will simply be $S_1$. The $K_{sp}$ expression is much simpler:

Equating this to our given $K_{sp}$:

Taking the square root gives us $S_1=2.0\times 10^{-6}\text{ M}$.

The Grand Reveal

We now have the solubilities of both salts. $S(A_{2}X)=10^{-4}\text{ M}$ and $S(MX)=2.0\times 10^{-6}\text{ M}$. Notice how $A_{2}X$ is significantly more soluble than $MX$, despite having the same $K_{sp}$! This is because the $S^3$ term in $A_{2}X$ allows it to reach the $K_{sp}$ threshold at a much higher concentration than the $S^2$ term in $MX$.

The question asks for the ratio of their molar solubilities:

To make the math visually cleaner, we can rewrite $10^{-4}$ as $100\times 10^{-6}$:

The ratio is exactly 50. This problem beautifully illustrates why you can never directly compare the $K_{sp}$ values of salts with different stoichiometries to determine which is more soluble. You must always calculate their individual molar solubilities first!