Find the molar solubility of \(\mathrm{SrCO}_{3}\left(K_{\mathrm{sp}}=5.4 \times 10^{-10}\right)\) in (a) pure water and (b) \(0.13 M \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}\).

The solubility product constant, or \(K_{sp}\), represents the equilibrium between a soluble ionic compound and its ions in solution. Essentially, it helps us understand how much of a solid can dissolve in a solution to form a saturated solution. It's given by the product of the molar concentrations of the ions each raised to the power of their coefficient in the dissociation equation.
For example, for \(\text{SrCO}_3\), which dissociates as follows: \(\text{SrCO}_3 (s) \rightleftharpoons \text{Sr}^{2+} (aq) + \text{CO}_3^{2-} (aq)\), \(K_{sp}\) is computed using the expression: \[K_{sp} = [\text{Sr}^{2+}] [\text{CO}_3^{2-}]\].
Understanding and calculating \(K_{sp}\) is key to determining molar solubility and predicting the extent to which a compound can dissolve in water or in solutions containing common ions.

When we talk about solubility equilibrium, we're referring to the dynamic balance between the solid phase of a solute and the dissolved ions in solution. This is crucial for understanding when a solution becomes saturated.
In the case of \(\text{SrCO}_3\), the system reaches equilibrium according to the dissociation equation: \(\text{SrCO}_3 (s) \rightleftharpoons \text{Sr}^{2+} (aq) + \text{CO}_3^{2-} (aq)\).
At equilibrium, the rate at which \(\text{SrCO}_3\) dissolves to form \(\text{Sr}^{2+}\) and \(\text{CO}_3^{2-}\) equals the rate at which these ions recombine to form the solid. The concentrations of these ions at this point determine the \(K_{sp}\).
This equilibrium state helps us understand that as more of the solute dissolves, the ions' concentration increases until it can't dissolve any more, thus maintaining a balance.

The common ion effect occurs when a compound's solubility decreases because of the addition of an ion common with the dissolved substance. This happens due to a shift in the solubility equilibrium according to Le Chatelier's principle.
For example, if we add \(\text{Sr(NO}_3)_2\) to a solution of \(\text{SrCO}_3\), the \(\text{Sr}^{2+}\) ions from \(\text{Sr(NO}_3)_2\) increase their concentration. This additional \(\text{Sr}^{2+}\) shifts the equilibrium to the left, reducing the solubility of \(\text{SrCO}_3\).
In calculating the molar solubility in a \(0.13 M\) \(\text{Sr(NO}_3)_2\) solution, the already existing \(\text{Sr}^{2+}\) ions reduce the amount of \(\text{SrCO}_3\) that can further dissolve. Hence, the solubility calculated is much lower compared to pure water.

The dissociation equation represents the process in which an ionic compound separates into its ions when dissolved in a solvent. It shows the components and stoichiometry of the dissociation.
For example, the dissociation equation for \(\text{SrCO}_3\) is: \[\text{SrCO}_3 (s) \rightleftharpoons \text{Sr}^{2+} (aq) + \text{CO}_3^{2-} (aq)\]
This equation tells us that one mole of solid \(\text{SrCO}_3\) dissociates into one mole of \(\text{Sr}^{2+}\) and one mole of \(\text{CO}_3^{2-}\).
Understanding the dissociation equation is essential for setting up the \(K_{sp}\) expression and determining the molar solubility. It provides a clear picture of the relationship and proportion between the compound and its ions in solution.