Find the molar solubility of \(\mathrm{BaCrO}_{4}\left(K_{\mathrm{sp}}=2.1 \times 10^{-10}\right) \mathrm{in}\) (a) pure water and (b) \(1.5 \times 10^{-3} M \mathrm{Na}_{2} \mathrm{CrO}_{4}\)

The solubility product constant, denoted as \(K_{\mathrm{sp}}\), is a special type of equilibrium constant. It represents the extent to which a sparingly soluble salt can dissociate in water. For a general salt \(AB\), which dissociates into \(A^+\) and \(B^-\), the expression for \(K_{\mathrm{sp}}\) is: \[ K_{\mathrm{sp}} = [A^+][B^-] \] This expression showcases the product of the molar concentrations of the ions produced. By knowing \(K_{\mathrm{sp}}\), one can determine how much of the salt dissolves in water, which is useful in various chemical processes and applications. It's the foundation for calculating molar solubility, which is the subject of the current exercise.

Keep in mind that \(K_{\mathrm{sp}}\) values are usually very small for sparingly soluble salts, indicating their limited solubility in water.

A dissociation equation shows how a compound separates into its ions in a solution. For the current exercise, the salt is \(\text{BaCrO}_{4}\), and its dissociation into water can be written as: \[ \text{BaCrO}_{4} (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{CrO}_{4}^{2-} (aq) \] This equation states that \(\text{BaCrO}_{4}\) dissociates into its constituent ions, \(\text{Ba}^{2+}\) and \(\text{CrO}_{4}^{2-}\), upon dissolving. This dissociation is crucial for setting up the \(K_{\mathrm{sp}}\) expression, as it tells us which ions are present and their stoichiometric relationship.

Understanding the dissociation process is key because it directly impacts the calculations we perform for finding molar solubility, both in pure water and in the presence of common ions.

The common ion effect refers to the suppression of the solubility of an ionic compound when a common ion is added to the solution. In simpler terms, if you add an ion that's already a part of the dissociation of the salt, it will reduce the amount of the salt that can dissolve. In the exercise, when \(\text{Na}_2\text{CrO}_4\) is added, it introduces additional \(\text{CrO}_{4}^{2-}\) ions into the solution.

This increases the concentration of \(\text{CrO}_{4}^{2-}\) from the start, leading to a shift in equilibrium according to Le Chatelier's principle. The system compensates by decreasing the dissociation of \(\text{BaCrO}_{4}\), resulting in lower solubility. Mathematically, this results in:
\[\text{BaCrO}_{4} (s) \rightleftharpoons \text{Ba}^{2+} (aq) + (\text{CrO}_{4}^{2-} + 1.5 \times 10^{-3} M) (aq)\] Thus, the molar solubility is significantly reduced when a common ion is present.

Molar solubility refers to the number of moles of a solute that can dissolve in one liter of solution until the solution becomes saturated. Understanding molar solubility is important for predicting how much of a salt can dissolve in a given volume of water.

To find the molar solubility of \(\text{BaCrO}_{4}\) in pure water or in the presence of a common ion, we set up expressions using \(K_{\mathrm{sp}}\) and solve for the solubility.

For instance, in pure water:
\[K_{\mathrm{sp}} = s^2 = 2.1 \times 10^{-10}\]
Solving for \(s\), the molar solubility is: \[s = \sqrt{2.1 \times 10^{-10}} = 1.45 \times 10^{-5} M\]

In the presence of the common ion \(\text{CrO}_4^{2-}\):
\[K_{\mathrm{sp}} = s \times (1.5 \times 10^{-3} + s)\] Given the initial concentration of \(\text{CrO}_4^{2-}\) is much larger than \(s\), it simplifies to: \[s = \frac{2.1 \times 10^{-10}}{1.5 \times 10^{-3}} = 1.4 \times 10^{-7} M\]This results demonstrate the drastic reduction in solubility due to the common ion effect.