Animated Solution for Chemistry - Ionic Equilibrium: The solubility of ${\text{CdSO}}_4$ in water is $8.0\times 10^{-4}{\text{ mol L}}^{-1}$. Its solubility in $0.01\text{ M }{\text{H}}_2{\text{SO}}_4$ solution is ……… $\times 10^{-6}{\text{ mol L}}^{-1}$. (Round off to the nearest integer) (Assume that, solubility is much less than $0.01\text{ M}$)

Imagine a beaker filled with pure water. When we add a sparingly soluble salt like cadmium sulfate (${\text{CdSO}}_4$), it establishes an equilibrium between the solid and its constituent ions in the solution.

The question provides the solubility of ${\text{CdSO}}_4$ in pure water as $S=8.0\times 10^{-4}{\text{ mol L}}^{-1}$. This is our starting point.

Solubility Product

In pure water, the dissociation of cadmium sulfate is straightforward:

Since one mole of the salt gives one mole of ${\text{Cd}}^{2+}$ and one mole of ${\text{SO}}_4^{2-}$, the concentration of both ions at equilibrium will be equal to the solubility, $S$. The solubility product constant, $K_{sp}$, is the product of these ion concentrations:

Let's substitute the given value of $S$ to find $K_{sp}$:

This $K_{sp}$ value is a constant at a given temperature and will not change even if we change the solvent.

Common Ion Effect

Now, the scenario changes. We are no longer dissolving the salt in pure water, but in a $0.01\text{ M}$ solution of sulfuric acid (${\text{H}}_2{\text{SO}}_4$). Sulfuric acid is a strong electrolyte and dissociates completely:

This means our solution already contains a significant amount of sulfate ions: $[{\text{SO}}_4^{2-}]=0.01\text{ M}$. When we try to dissolve ${\text{CdSO}}_4$ in this solution, it faces resistance. According to Le Chatelier's principle, the presence of the common ion (${\text{SO}}_4^{2-}$) pushes the equilibrium backward, suppressing the solubility of the salt. This phenomenon is known as the Common Ion Effect.

Let the new, reduced solubility of ${\text{CdSO}}_4$ in this acidic solution be $x$.

The concentration of ${\text{Cd}}^{2+}$ will be $x$.

The total concentration of ${\text{SO}}_4^{2-}$ will be the sum of what comes from the salt ($x$) and what is already present from the acid ($0.01\text{ M}$).

Now, we plug these into our $K_{sp}$ expression:

Solving a quadratic equation can be tedious. However, the question gives us a massive hint: "Assume that, solubility is much less than $0.01\text{ M}$".

Because $x$ is extremely small compared to $0.01$, adding it to $0.01$ barely changes the value. Therefore, we can safely approximate:

This simplifies our equation beautifully:

Now, it's just a matter of simple algebra to find $x$:

The question asks for the value to fill in the blank for $\cdots \times 10^{-6}{\text{ mol L}}^{-1}$. Comparing our result, the missing integer is 64.

This problem perfectly illustrates how the common ion effect drastically reduces solubility and how strategic approximations can save you valuable time during an exam!