Question

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ba}^{2+}(a q)\) and \(0.010 \mathrm{M}\) in \(\mathrm{Sr}^{2+}(a q)\). (a) What concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. BaSO \(_{4}: K_{i p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}\) \(\left.K_{s p}=3.2 \times 10^{-7} .\right)(\mathbf{b})\) 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) 1.1x10^-8 M; (b) Ba^2+; (c) 3.2x10^-5 M

Step-by-step solution

Determine Ion Product for Precipitation of BaSO4

The ion product of precipitation can be defined using the solubility product, \(K_{sp}\), of the compound. For \(\text{BaSO}_4\), \[\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \\text{K}_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]\]We need the concentration of \(\text{SO}_4^{2-}\) to begin precipitation.\[1.1 \times 10^{-10} = 0.010 \times [\text{SO}_4^{2-}]\]Solve for \([\text{SO}_4^{2-}]\):\[[\text{SO}_4^{2-}] = \frac{1.1 \times 10^{-10}}{0.010} = 1.1 \times 10^{-8}\, \text{M}\]

Determine Ion Product for Precipitation of SrSO4

Next, we determine the required \([\text{SO}_4^{2-}]\) for \(\text{SrSO}_4\) using its solubility product:\[\text{Sr}^{2+} + \text{SO}_4^{2-} \rightarrow \text{SrSO}_4 \\text{K}_{sp} = [\text{Sr}^{2+}][\text{SO}_4^{2-}]\]Given \(K_{sp} = 3.2 \times 10^{-7}\),\[3.2 \times 10^{-7} = 0.010 \times [\text{SO}_4^{2-}]\]Solve for \([\text{SO}_4^{2-}]\):\[[\text{SO}_4^{2-}] = \frac{3.2 \times 10^{-7}}{0.010} = 3.2 \times 10^{-5}\, \text{M}\]

Identify First Cation to Precipitate

Compare the required sulfate concentrations for precipitation of both compounds. \([\text{SO}_4^{2-}] = 1.1 \times 10^{-8}\, \text{M}\) for \(\text{BaSO}_4\) is less than \(3.2 \times 10^{-5}\, \text{M}\) for \(\text{SrSO}_4\).Thus, \(\text{Ba}^{2+}\) precipitates first as \(\text{BaSO}_4\).

Determine Sulfate Concentration for Second Cation Precipitation

Since \(\text{SrSO}_4\) starts to precipitate at a higher \([\text{SO}_4^{2-}]\) than \(\text{BaSO}_4\), the concentration at which \(\text{Sr}^{2+}\) begins to precipitate is when \([\text{SO}_4^{2-}] = 3.2 \times 10^{-5}\, \text{M}\).

Key concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, or \(K_{sp}\), is a fundamental concept in understanding precipitation reactions. It represents the equilibrium constant for a solid substance dissolving in an aqueous solution. The formula for \(K_{sp}\) is given by multiplying the concentrations of the ions formed in a saturated solution.
For a general salt \(AB\), dissolving into ions \(A^+\) and \(B^-\), the equation is:

• \(K_{sp} = [A^+][B^-]\)

In our example, we calculate the \(K_{sp}\) for two compounds, \(\text{BaSO}_4\) and \(\text{SrSO}_4\). Each has a specific \(K_{sp}\) value indicating how likely they are to stay dissolved in solution versus precipitating out as solid salts.
Here, \(K_{sp}\) helps us determine at what point these ions will form a precipitate. A small \(K_{sp}\) means a compound is less soluble. We find \( [\text{SO}_4^{2-}] \) necessary for precipitation by setting \(K_{sp}\) equal to the corresponding ion concentrations.

Selective Precipitation

Selective precipitation is a technique used to separate ions in a solution by adding a reagent that forms a precipitate with one of the ions, but not the others. This method relies on differences in solubility products.
In the example, the process is used to identify which cation will precipitate first as \(\text{BaSO}_4\) or \(\text{SrSO}_4\).

• The reagent used is \(\text{Na}_2\text{SO}_4\) which provides \(\text{SO}_4^{2-}\) ions.
• The ion that requires a lower concentration of \(\text{SO}_4^{2-}\) to begin precipitation will precipitate first.

By computing \([\text{SO}_4^{2-}]\) needed for forming each precipitate and comparing these values, we find that \(\text{Ba}^{2+}\) precipitates before \(\text{Sr}^{2+}\). This concept is crucial in separating ions based on differing \(K_{sp}\) values.

Ionic Equilibria

Ionic equilibria refer to the balance of species in a solution that contains ions. It encompasses all interactions and reactions that occur between ions in a solution, including dissolution and precipitation.
In precipitation reactions, ionic equilibria determine when a salt will start to precipitate from the solution as its solubility limit is exceeded. Equilibrium is reached when the rate of dissolution equals the rate of precipitation.

• The initial concentrations of ions are critical in these calculations.
• The equilibrium concentration needed for one ion can affect the concentration needed for another ion.

In this exercise, the equilibrium for \(\text{Ba}^{2+}\) and \(\text{Sr}^{2+}\) with \(\text{SO}_4^{2-}\) is established by determining at what concentration of \(\text{SO}_4^{2-}\) each cation begins to precipitate. Here, the ionic equilibria help identify that while \(\text{Ba}^{2+}\) reacts first, \(\text{Sr}^{2+}\) requires a much higher \(\text{SO}_4^{2-}\) concentration to reach equilibrium and precipitate. This understanding of equilibria is fundamental in chemists' toolkit for predicting and manipulating reactions in ionic solutions.