Animated Solution for Chemistry - Ionic Equilibrium: If $K_{sp}$ of ${\text{Ag}}_2{\text{CO}}_3$ is $8\times 10^{-12}$, the molar solubility of ${\text{Ag}}_2{\text{CO}}_3$ in $0.1\text{ M}$ ${\text{AgNO}}_3$ is

A Crowded Solution

Imagine you are trying to dissolve a pinch of salt in a glass of water. Normally, it dissolves easily. But what if the water is already saturated with one of the ions present in your salt? This is the essence of the Common Ion Effect, a beautiful application of Le Chatelier's Principle.

In this problem, we are dealing with silver carbonate, ${\text{Ag}}_2{\text{CO}}_3$, which is a sparingly soluble salt. If we were dissolving it in pure water, it would establish its own quiet equilibrium. However, the beaker isn't filled with pure water; it contains a $0.1\text{ M}$ solution of silver nitrate, ${\text{AgNO}}_3$.

Since ${\text{AgNO}}_3$ is a strong electrolyte, it completely dissociates:

This means before our silver carbonate even starts to dissolve, the solution is already flooded with $0.1\text{ M}$ of ${\text{Ag}}^{+}$ ions. This pre-existing crowd of silver ions will heavily suppress the dissolution of ${\text{Ag}}_2{\text{CO}}_3$.

Solubility Product

Let's write down the equilibrium reaction for the dissolution of silver carbonate:

The equilibrium constant for this process is the solubility product, $K_{sp}$:

Notice the squared term! Because one molecule of ${\text{Ag}}_2{\text{CO}}_3$ produces two ${\text{Ag}}^{+}$ ions, the concentration of silver ions is raised to the power of 2 in the $K_{sp}$ expression.

Let the new molar solubility of the salt in this crowded solution be $S^{\prime }$. When $S^{\prime }$ moles of ${\text{Ag}}_2{\text{CO}}_3$ dissolve, they produce $2S^{\prime }$ moles of ${\text{Ag}}^{+}$ and $S^{\prime }$ moles of ${\text{CO}}_3^{2-}$.

But remember, the solution already had $0.1\text{ M}$ of ${\text{Ag}}^{+}$. Therefore, the total concentrations at equilibrium will be:

The Art of Neglecting

Now, we substitute these equilibrium concentrations into our $K_{sp}$ expression:

At first glance, this looks like a terrifying cubic equation. But in chemistry, we use logic to simplify math. The value of $K_{sp}$ is incredibly small ($8\times 10^{-12}$). This tells us that the reaction barely moves forward, meaning $S^{\prime }$ will be a microscopic number.

If $S^{\prime }$ is tiny, then $2S^{\prime }$ is also tiny. When you add a tiny number to a relatively large number like $0.1$, the tiny number is practically invisible. Therefore, we can safely make the approximation:

With our approximation, the complex cubic equation collapses into a simple linear one:

Squaring $0.1$ gives us $0.01$, or $10^{-2}$:

Solving for $S^{\prime }$:

And there we have it! The new solubility is $8\times 10^{-10}\text{ M}$. The common ion effect has done its job, drastically suppressing the solubility of the salt.