Problem 68 - Chemistry: The Central Science

Question

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ba}^{2+}\) and \(0.010 \mathrm{M}\) in \(\mathrm{Sr}^{2+}\). (a) What concentration of \(\mathrm{SO}_{4}{\underline{\phantom{xx}}}^{2-}\) is necessary to begin precipita- tion? (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 precipi- tates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}\) when the second cation begins to precipitate?

Step-by-Step Solution

Answer

Ba²⁺ ions precipitate first, and the concentration of SO₄²⁻ ions when the second cation, Sr²⁺, begins to precipitate is \(3.2 \times 10^{-5} M\).

Step 1: Write the solubility product expressions for each precipitate.

For the given compounds, the solubility product expressions are: For BaSO₄: \(K_{sp}=[\mathrm{Ba}^{2+}][\mathrm{SO}_{4}^{2-}] = 1.1 \times 10^{-10}\) For SrSO₄: \(K_{sp}=[\mathrm{Sr}^{2+}][\mathrm{SO}_{4}^{2-}] = 3.2 \times 10^{-7}\)

Step 2: Determine the concentration of SO₄²⁻ required to precipitate each cation.

We will use the Ksp expressions for both precipitates and the given concentrations of Ba²⁺ and Sr²⁺ ions. For BaSO₄: \([SO_{4}^{2-}] = \cfrac{K_{sp}}{[Ba^{2+}]} = \cfrac{1.1 \times 10^{-10}}{0.010} = 1.1 \times 10^{-8} \mathrm{M}\) For SrSO₄: \([SO_{4}^{2-}] = \cfrac{K_{sp}}{[Sr^{2+}]} = \cfrac{3.2 \times 10^{-7}}{0.010} = 3.2 \times 10^{-5} \mathrm{M}\)

Step 3: Determine which cation will precipitate first.

Comparing the concentrations of SO₄²⁻ required for precipitation: BaSO₄: \(1.1 \times 10^{-8} \mathrm{M}\) SrSO₄: \(3.2 \times 10^{-5} \mathrm{M}\) Since the required concentration of SO₄²⁻ is lower for BaSO₄, Ba²⁺ will precipitate first.

Step 4: Calculate the concentration of SO₄²⁻ when the second cation begins to precipitate.

Since Ba²⁺ precipitates first, we know that the concentration of SO₄²⁻ at this point is \(1.1 \times 10^{-8} \mathrm{M}\). In order for Sr²⁺ to precipitate, the concentration of SO₄²⁻ must reach \(3.2 \times 10^{-5} \mathrm{M}\). So, the concentration of SO₄²⁻ when the second cation begins to precipitate is \(3.2 \times 10^{-5} M\). In conclusion, Ba²⁺ ions precipitate first, and the concentration of SO₄²⁻ ions when the second cation, Sr²⁺, begins to precipitate is \(3.2 \times 10^{-5} M\).

Key Concepts

Precipitation Reactions

Precipitation reactions occur when two soluble ionic compounds in a solution form an insoluble solid called a precipitate. These reactions are guided by the concept of solubility product constant, or \( K_{sp} \), which helps determine when a compound will begin to precipitate from solution.
In these reactions, ions from each reactant are mixed in a solution and if the resulting product exceeds the solubility product constant, a precipitate forms. The concentration of ions and the specific \( K_{sp} \) values for the compounds involved govern when precipitation occurs.
A practical example is the gradual addition of a sodium sulfate (Na₂SO₄) solution to a mixture of barium ions \((\text{Ba}^{2+})\) and strontium ions \((\text{Sr}^{2+})\). Understanding which ion will precipitate first helps predict the sequence and properties of the formed precipitates.

Barium Sulfate

Barium sulfate (BaSO₄) is a compound known for its low solubility in water, which is reflected in its solubility product constant, \( K_{sp} = 1.1 \times 10^{-10} \). This low \( K_{sp} \) value indicates that BaSO₄ will precipitate out of a solution at a relatively low concentration of sulfate ions \((\text{SO}_{4}^{2-})\).
In the context of precipitation reactions, Ba²⁺ ions will combine with \(\text{SO}_{4}^{2-}\) ions as soon as their product exceeds the \( K_{sp} \) value. For example, when \([\text{SO}_{4}^{2-}]\) reaches \(1.1 \times 10^{-8} \text{ M}\), BaSO₄ starts precipitating out of solution.
This makes barium sulfate an ideal compound for studying the sequence of precipitation reactions, as its early precipitation can be easily predicted and observed.

Strontium Sulfate

Strontium sulfate (SrSO₄) has a relatively higher solubility than barium sulfate, with a \( K_{sp} = 3.2 \times 10^{-7} \). This means that it will precipitate from a solution only when the concentration of \( \text{SO}_{4}^{2-} \) reaches a higher threshold of \( 3.2 \times 10^{-5} \text{ M} \).
In mixtures containing both Ba²⁺ and Sr²⁺ ions, strontium sulfate precipitates after barium sulfate, due to its higher \( K_{sp} \). The delayed precipitation allows for the sequential removal of ions based on their solubility product constants.
This higher solubility can be leveraged in chemical processes to control the order of precipitation, making strontium sulfate particularly useful for studying the dynamics of ionic solubility and compound formation.

Cation Precipitation Sequence

Cation precipitation sequence refers to the order in which different metal ions precipitate from a solution. The sequence is determined primarily by the solubility products of the compounds they form with the anions present in the solution.
In cases where solutions contain multiple cations that could form precipitates with a common anion, like \( \text{SO}_{4}^{2-} \), the cation with the lower \( K_{sp} \) value will precipitate first. For the scenario with barium and strontium ions, Ba²⁺ precipitates first because BaSO₄ has a lower \( K_{sp} \) than SrSO₄.
Understanding this sequence is crucial in industrial processes and laboratory settings, as it aids in the selective separation and purification of specific ions based on their chemical properties.