Question

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?

Short answer

(a) To begin precipitation, a concentration of \(1.1 \times 10^{-9} M\) SO42- is necessary for BaSO4, while \(3.2 \times 10^{-6} M\) is necessary for SrSO4. (b) Ba2+ precipitates first. (c) The concentration of SO42- when the second cation (Sr2+) begins to precipitate is \(3.2 \times 10^{-6} M\).

Step-by-step solution

Write the equilibrium expressions for both compounds

We will start by writing down the solubility product expressions for both BaSO4 and SrSO4: For BaSO4: \(K_{sp} = [Ba^{2+}][SO_4^{2-}]\) For SrSO4: \(K_{sp} = [Sr^{2+}][SO_4^{2-}]\)

Determine the initial concentrations of the ions

We know from the problem that the initial concentration of both Ba2+ and Sr2+ is 0.010M. We also know that the initial concentration of SO42- is zero. We now have: [Ba2+] = 0.010M [Sr2+] = 0.010M [SO42-] = x (initial concentration of SO42− is unknown)

Solve for x in terms of solubility products for both compounds

Now we will substitute these concentrations into the Ksp expressions: For BaSO4: \(1.1 \times 10^{-10} = (0.010)(x)\) For SrSO4: \(3.2 \times 10^{-7} = (0.010)(x)\) Now, solve for x to get the molar concentration of SO42− required for the beginning of precipitation for each cation: For BaSO4: \(x = \dfrac{1.1 \times 10^{-10}}{0.010} = 1.1 \times 10^{-9} M\) For SrSO4: \(x = \dfrac{3.2 \times 10^{-7}}{0.010} = 3.2 \times 10^{-6} M\)

Determine which cation precipitates first

To determine which cation will precipitate first, we will compare the values obtained in step 3. The cation with the lower required concentration of SO42- will begin to precipitate first. In this case, the required concentration of SO42- for BaSO4 precipitation is lower (1.1 x 10^-9 M) than for SrSO4 precipitation (3.2 x 10^-6 M), so the Ba2+ ion will precipitate first. Thus, the answer to part (b) of the problem is Ba2+.

Determine the concentration of SO42- when the second cation begins to precipitate

Since Ba2+ will precipitate first, the Sr2+ ions will still be in solution when the precipitation of Ba2+ is complete. When this happens, the concentration of Sr2+ ions will still be 0.010 M. Therefore, the concentration of SO42- necessary for SrSO4 precipitation can now be obtained using the following equation: \(3.2 \times 10^{-7} = (0.010)(x)\) Now, solve for x to get the concentration of SO42- when the second cation (Sr2+) begins to precipitate: \(x = \dfrac{3.2 \times 10^{-7}}{0.010} = 3.2 \times 10^{-6} M\) Now we found the answers to all three parts of the problem: (a) The concentration of SO42- necessary to begin precipitation is 1.1 x 10^-9 M for BaSO4 and 3.2 x 10^-6 M for SrSO4. (b) The cation precipitating first is Ba2+. (c) The concentration of SO42- when the second cation (Sr2+) begins to precipitate is 3.2 x 10^-6 M.

Key concepts

Chemical Equilibrium

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

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

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.