Question

Four solutions are prepared and mixed in the following order: (1) Start with \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{BaCl}_{2}\) (2) Add \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) (3) Add \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) (4) Add \(250.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NH}_{3}\) Write an equation for any reaction that occurs after each step, and calculate the concentrations of \(\mathrm{Ba}^{2+}, \mathrm{Cl}^{-}, \mathrm{NO}_{3}^{-}, \mathrm{NH}_{3}\), and \(\mathrm{NH}_{4}^{+}\) in the final solution, assuming that all reactions go to completion.

Short answer

Final concentrations are: [Cl⁻] = 0.0333 M, [NO₃⁻] = 0.0111 M, [NH₃] = 0.0556 M. Ba²⁺ precipitated as BaSO₄.

Step-by-step solution

Analyze the Initial Mixture

We begin with 100.0 mL of 0.100 M BaClequiv 2, thus you have BaCl₂. The initial number of moles of each ion is determined by multiplying the volume (in L) by the molarity: \[ \text{moles of } \text{Ba}^{2+} = 0.100 \, \text{L} \times 0.100 \, \text{mol/L} = 0.0100 \, \text{mol} \]\[ \text{moles of } \text{Cl}^- = 2 \times 0.0100 \, \text{mol} = 0.0200 \, \text{mol} \]

Reaction with AgNO3

Add 50.0 mL of 0.100 M AgNO₃. AgNO₃ reacts with BaCl₂ to form AgCl, which precipitates:\[ \text{Ag}^{+} + \text{Cl}^{-} \rightarrow \text{AgCl(s)} \]Determine moles of Ag⁺ added:\[ \text{moles of } \text{Ag}^{+} = 0.0500 \, \text{L} \times 0.100 \, \text{mol/L} = 0.00500 \, \text{mol} \]Ag⁺ reacts completely with Cl⁻, producing 0.00500 mol of AgCl and leaving 0.0150 mol Cl⁻ unreacted.

Reaction with H2SO4

Add 50.0 mL of 0.100 M H₂SO₄. BaCl₂ reacts with H₂SO₄ to form BaSO₄, a precipitate:\[ \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4(\text{s}) \]Calculate moles of \(\text{SO}_4^{2-}\):\[ \text{moles of } \text{SO}_4^{2-} = 0.0500 \, \text{L} \times 0.100 \, \text{mol/L} = 0.00500 \, \text{mol} \]All \(\text{Ba}^{2+}\) (0.0100 mol) reacts with \(\text{SO}_4^{2-}\) to form BaSO₄, leaving 0 mol \(\text{Ba}^{2+}\).

Final Addition of NH3

Add 250.0 mL of 0.100 M NH₃. \(\text{NH}_3\) mainly acts as a base forming \(\text{NH}_4^+\):\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]Calculate the moles of \(\text{NH}_3\):\[ \text{moles of } \text{NH}_3 = 0.2500 \, \text{L} \times 0.100 \, \text{mol/L} = 0.0250 \, \text{mol} \]

Final Concentration Calculations

Calculate the total volume after mixing all solutions:\[ \text{Total Volume} = 100.0 \, \text{mL} + 50.0 \, \text{mL} + 50.0 \, \text{mL} + 250.0 \, \text{mL} = 450.0 \, \text{mL} = 0.450 \, \text{L} \]Determine concentrations for ions:- Remaining \(\text{Cl}^-\): \[ [\text{Cl}^-] = \frac{0.0150 \, \text{mol}}{0.450 \, \text{L}} = 0.0333 \, \text{M} \]- \(\text{NO}_3^-\) from AgNO₃ (no reaction occurred with it affecting): \[ [\text{NO}_3^-] = \frac{0.00500 \, \text{mol}}{0.450 \, \text{L}} = 0.0111 \, \text{M} \]- \(\text{NH}_3\) (not all turns to \(\text{NH}_4^+\)): \[ [\text{NH}_3] = \frac{0.0250 \, \text{mol}}{0.450 \, \text{L}} = 0.0556 \, \text{M} \]- No \(\text{Ba}^{2+}\) remains; formed \(\text{BaSO}_4\) precipitate.

Additional Consideration for NH4+

Given \(\text{NH}_3\) reacts to form \(\text{NH}_4^+\), assuming minor dissociation, \([\text{NH}_4^+]\) is minimal. Therefore, based on the primary reactions and concentrations, \(\text{NH}_4^+\) concentration from \(\text{NH}_3\) equilibrium is negligible.

Key concepts

Precipitation Reactions

Precipitation reactions occur when two soluble salts in aqueous solutions combine to form an insoluble salt, or a precipitate. When the ions of the soluble salts exchange partners, they create a new pair of ions that is not soluble in water. In our exercise, this concept is beautifully illustrated with the reactions of BaCl₂ and AgNO₃, as well as BaCl₂ and H₂SO₄.

This process can be summarized with general chemical reactions showing the formation of a precipitate. For example, when AgNO₃ is added to BaCl₂, Ag⁺ ions react with Cl⁻ ions to form AgCl, an insoluble compound that appears as a precipitate. This can be represented by the equation: \[ \text{Ag}^{+} + \text{Cl}^{-} \rightarrow \text{AgCl(s)} \]Similarly, when H₂SO₄ is added to a solution containing BaCl₂, a precipitation reaction occurs: \[ \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4(\text{s}) \]These reactions are crucial in studying the behavior of ions in solution and predicting which insoluble salts will form under specific conditions. Understanding precipitation reactions helps in various fields, including chemical manufacturing and wastewater treatment.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is foundational in determining the quantitative relationships between substances as they participate in a chemical reaction. Might sound daunting, but it's simpler once you break it down into manageable parts.

In the case of our problem, stoichiometry helps us figure out how much of each reactant is needed to produce a certain amount of product. For example, when we start with 0.100 M BaCl₂ and react it with 0.100 M AgNO₃, we use stoichiometry to calculate the moles of AgCl formed and how much Cl⁻ is left unreacted. Similarly, for BaSO₄ formation, the moles of Ba²⁺ available dictate how much BaSO₄ can precipitate from the reaction with H₂SO₄.

Generally, stoichiometry involves:

• Balancing the chemical equation.
• Converting volumes of reactants to moles using molarity.
• Using mole ratios from the balanced equation to calculate other substances in the reaction.

Understanding stoichiometry allows chemists to predict the quantitative outcomes of reactions, ensuring optimal amounts of reactants are used to avoid waste and ensure complete reactions.

Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution (mol/L). Calculating molarity is essential in experiments where precise outcomes depend on reactant concentrations.

In our given exercise, molarity calculations help us find the concentration of ions after each reaction. For instance, when 0.100 M of a solute is added, you calculate the moles by multiplying the volume (in L) by the molarity. Subsequently, you can determine new concentrations after they participate in chemical reactions.

To calculate molarity:

• Determine the number of moles of solute, as illustrated with the initial moles of Ba²⁺ and Cl⁻.
• Add together the volumes of all solutions to find the total volume.
• Divide the moles of remaining solute by this total volume to obtain the final molarity.

This importance of molarity calculations lies in their use to maintain control over chemical reactions, ensuring desired reactions occur effectively by maintaining ideal conditions in terms of reactant ratios.