Animated Solution for Chemistry - Ionic Equilibrium: The molar solubility of ${\text{Cd(OH)}}_2$ is $1.84\times 10^{-5}\text{ m}$ in water. The expected solubility of ${\text{Cd(OH)}}_2$ in a buffer solution of $\text{pH}=12$ is

Overcoming the Common Ion Effect

Imagine you are a tiny cadmium ion, trying to break free from your solid crystal lattice and swim into the surrounding water. In pure water, it's a peaceful journey. But what happens when you try to dissolve into a solution that is already crowded with one of your own kind? This is the thrilling story of the Common Ion Effect.

Before we can predict how ${\text{Cd(OH)}}_2$ behaves in a hostile environment, we need to know its intrinsic strength—its Solubility Product Constant ($K_{sp}$). The problem generously gives us its solubility in pure water: $S=1.84\times 10^{-5}\text{ M}$.

When ${\text{Cd(OH)}}_2$ dissolves, it splits according to the equation:

For every $S$ moles of solid that dissolve, we get $S$ moles of ${\text{Cd}}^{2+}$ and $2S$ moles of ${\text{OH}}^{-}$. The equilibrium expression is:

Plugging in our known value of $S$:

This $K_{sp}$ is a fundamental constant at this temperature. It is the absolute limit of how much ${\text{Cd}}^{2+}$ and ${\text{OH}}^{-}$ can coexist in the solution before they start precipitating back into a solid.

Common Ion Effect

Now, we take our solid and drop it into a buffer solution with a $\text{pH}$ of 12. A $\text{pH}$ of 12 means the solution is highly basic. Let's find out exactly how basic it is by calculating the $\text{pOH}$:

This means the concentration of hydroxide ions already present in the buffer is:

Here is where the Common Ion Effect strikes! The buffer is already swarming with ${\text{OH}}^{-}$ ions. When ${\text{Cd(OH)}}_2$ tries to dissolve, it tries to add more ${\text{OH}}^{-}$ to the pool. However, the amount it can add ($2S^{\prime }$) is incredibly tiny compared to the $10^{-2}\text{ M}$ already there.

Therefore, we can safely approximate the total hydroxide concentration to just be the buffer's concentration:

Armed with our constant $K_{sp}$ and the overwhelming $[{\text{OH}}^{-}]$ from the buffer, we can find the new, suppressed solubility, $S^{\prime }$:

Solving for $S^{\prime }$:

To make it look elegant in standard scientific notation, we shift the decimal:

Notice the dramatic drop! In pure water, the solubility was on the order of $10^{-5}$. In the buffer, it plummeted to $10^{-10}$. The presence of the common ion effectively crushed the solid's ability to dissolve. This beautiful interplay of equilibrium is exactly why understanding $K_{sp}$ is so powerful in chemistry.