Problem 117 - Chemistry The Molecular Science

Question

A \(1.00-\mathrm{L}\) solution contains \(0.010-\mathrm{M} \mathrm{F}^{-}\) and \(0.010-\mathrm{M}\) \(\mathrm{SO}_{4}^{2-} .\) Solid barium nitrate is slowly added to the solution. (a) Calculate the \(\left[\mathrm{Ba}^{2+}\right]\) when \(\mathrm{BaSO}_{4}\) begins to precipitate. (b) Calculate the \(\left[\mathrm{Ba}^{2+}\right]\) when \(\mathrm{BaF}_{2}\) starts to precipitate. Assume no volume change occurs. \(K_{\mathrm{sp}}\) values: \(\mathrm{BaSO}_{4}=\) \(1.1 \times 10^{-10} ; \mathrm{BaF}_{2}=1.0 \times 10^{-6}\)

Step-by-Step Solution

Answer

(a) \(1.1 \times 10^{-8}\, \text{M}\); (b) \(1.0 \times 10^{-2}\, \text{M}\).

Step 1: Understanding the Problem

We need to find the barium ion concentration \( [\text{Ba}^{2+}] \) at which \( \text{BaSO}_4 \) and \( \text{BaF}_2 \) begin to precipitate. This means using the solubility product constants (\( K_{\text{sp}} \)) for these compounds.

Step 2: Calculate [Ba²⁺] for BaSO₄ Precipitation

BaSO₄ will start precipitating when \([\text{Ba}^{2+}][\text{SO}_4^{2-}] = K_{\text{sp}}\) for \(\text{BaSO}_4\). We are given \([\text{SO}_4^{2-}] = 0.010\, \text{M}\) and \(K_{\text{sp}} = 1.1 \times 10^{-10}\).\[[\text{Ba}^{2+}] = \frac{1.1 \times 10^{-10}}{0.010} = 1.1 \times 10^{-8}\, \text{M}\]

Step 3: Calculate [Ba²⁺] for BaF₂ Precipitation

BaF₂ will start precipitating when \([\text{Ba}^{2+}][\text{F}^-]^2 = K_{\text{sp}}\) for \(\text{BaF}_2\). We are given \([\text{F}^-] = 0.010\, \text{M}\) and \(K_{\text{sp}} = 1.0 \times 10^{-6}\).\[[\text{Ba}^{2+}] = \frac{1.0 \times 10^{-6}}{(0.010)^2} = 1.0 \times 10^{-2}\, \text{M}\]

Step 4: Conclusion: Compare Precipitation Points

The conditions for precipitation show \(\text{BaSO}_4\) starts precipitating at a much lower \([\text{Ba}^{2+}]\) than \(\text{BaF}_2\). Therefore, \(\text{BaSO}_4\) will begin to precipitate first.

Key Concepts

Precipitation Reactions

Precipitation reactions occur when two soluble salts in aqueous solutions react to form one or more insoluble products, known as precipitates. These reactions are essential in chemistry for separating and identifying compounds. When two ionic compounds dissolve in water, their ions separate and interact in the mixture. In some cases, the product of the ion concentration exceeds the solubility product constant (Ksp), leading to the formation of a solid. This solid is the precipitate. Precipitation is often a reversible process, meaning the solid can sometimes dissolve back into the solution if conditions change.

Key aspects of precipitation reactions include:

• Identifying the possible combinations of ions that can form precipitates.
  
• Calculating ion concentrations and comparing them to the Ksp values.
  
• Understanding that the precipitate formation will start when the ion concentration product exceeds the Ksp.
  

In the problem presented, barium ions are added to a solution containing fluoride and sulfate ions. As barium ions are introduced, they interact with both sulfate and fluoride ions to potentially form barium sulfate (\(BaSO_4\)) and barium fluoride (\(BaF_2\)) precipitates.

Solubility Product Constant

The solubility product constant (\(K_{ ext{sp}}\)) is a crucial concept in understanding how and when precipitates form in a solution. It represents the maximum product of the ion concentrations in a saturated solution at equilibrium. Essentially, it is the equilibrium constant for a solid substance dissolving in an aqueous solution. Ksp is specific to each compound and varies with temperature.

To determine the onset of precipitation, one compares the ion product to the Ksp:

• If the ion product is less than the Ksp, the solution is unsaturated and no precipitation occurs.
  
• If the ion product equals the Ksp, the solution is at equilibrium and is saturated.
  
• If the ion product exceeds the Ksp, the solution is supersaturated and precipitation will occur.
  

In this particular exercise, the Ksp values are given for barium sulfate (\(1.1 \times 10^{-10}\)) and barium fluoride (\(1.0 \times 10^{-6}\)). By comparing the calculated ion product to these constants, we can predict when each compound will begin to precipitate.

Solution Chemistry

Solution chemistry is the study of the properties and behaviors of substances dissolved in solvents, primarily water for chemistry. It focuses on understanding how solutes interact in solutions and how reactions occur within these liquid environments. Key concepts in solution chemistry include solubility, concentration, and the behavior of ions.

In this context, solution chemistry explores how ionic interactions lead to precipitation.

• Solubility refers to how much solute can dissolve in a solvent at a given temperature. It influences reaction occurrences in the solution.
  
• Concentration is the measure of how much solute exists within a volume of solvent, impacting the likelihood of reactions and precipitate formation.
  
• The behavior of ions, especially their ability to form new compounds with other ions, is central to understanding precipitation reactions.
  

In our exercise, understanding how barium ions (\(Ba^{2+}\)) interact with fluoride (\(F^-\)) and sulfate (\(SO_4^{2-}\)) ions in solution chemistry makes it possible to predict the formation of precipitates. Solution chemistry principles help to determine when certain compounds will separate from the solution to form solids.