A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is 0.010\(M\) in \(\mathrm{Ba}^{2+}(a q)\) and 0.010\(M\) in \(\mathrm{Sr}^{2+}(a q) .\) (a) What concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. \(\mathrm{BaSO}_{4} : K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}\): \(K_{s p}=3.2 \times 10^{-7} . )\) (b) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) when the second cation begins to precipitate?

Chemical equilibrium plays a pivotal role in understanding precipitation reactions like the one described in the exercise with barium sulfate (BaSO₄) and strontium sulfate (SrSO₄). It refers to a state where the rate of the forward reaction matches the rate of the backward reaction, resulting in no net change in the concentrations of reactants and products over time. In this context, the equilibrium is captured by the solubility product constant (\(K_{sp}\)), which is a measure of the product of the molar concentrations of the ions in a saturated solution, raised to the power of their coefficients in the balanced equation.

When adding sodium sulfate (\(Na_2SO_4\)) to a solution with \(Ba^{2+}\) and \(Sr^{2+}\) ions, the \(K_{sp}\) values guide us in predicting which salt will precipitate first. The point at which precipitation begins is where the product of ion concentrations exceeds the \(K_{sp}\) – indicating the system has reached its equilibrium limit for solubility at that temperature and pressure.

Precipitation reactions occur when ions in solution combine to form an insoluble solid, known as a precipitate. The exercise illustrates a classic example with the formation of BaSO₄ and SrSO₄, where the solubility product principle determines the onset of precipitation. It's all about saturation: a solution holds a maximum amount of dissolved substance, and if this limit is surpassed - typically by changing the concentration of the ions - solid formation ensues.

In the process, if we were to neglect volume changes, we could approach this tactically by solving for the molar concentration of sulfate ions (\(SO_4^{2-}\)) needed to start the precipitation. Given the \(K_{sp}\) values, we can calculate the precipitate formation threshold and conclude that BaSO₄ will form first due to its lower solubility product, effectively dictating the sequence of precipitation in a solution containing both \(Ba^{2+}\) and \(Sr^{2+}\) ions.

Ion concentration is a quantitative measure of the amount of an ion in a solution, typically expressed in molarity (moles per liter). In the exercise, the concentration of sulfate ions required to initiate precipitation is calculated based on the known \(K_{sp}\) values and initial ion concentrations. The concept is crucial as it determines the point at which the solution becomes supersaturated and excess ions begin to form a precipitate.

When the concentration of \(SO_4^{2-}\) ions reaches a certain level, specifically 1.1 x 10^-9 M, it exceeds the solubility for BaSO₄, resulting in barium precipitating out first. With Ba²⁺ out of the solution, the concentration of \(SO_4^{2-}\) can increase further without any precipitation until it reaches 3.2 x 10^-6 M, at which point strontium begins to precipitate as SrSO₄. Understanding ion concentration is essential for predicting and controlling such reactions in a variety of chemical applications.