Question

Consider 1.0 L of an aqueous solution that contains 0.10 M sulfuric acid to which 0.30 mole of barium nitrate is added. Assuming no change in volume of the solution, determine the pH, the concentration of barium ions in the final solution, and the mass of solid formed.

Short answer

The pH of the solution is approximately 0.70, the concentration of barium ions in the final solution is 0.20 M, and the mass of solid formed is 23.34 g.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between sulfuric acid and barium nitrate is: H₂SO₄ + Ba(NO₃)₂ → BaSO₄ (s) + 2 HNO₃

Determine the initial moles of sulfuric acid

We know that the initial concentration of sulfuric acid is 0.10 M and the volume of the solution is 1.0 L. We can find the initial moles of sulfuric acid using the formula: moles = concentration × volume moles = 0.10 M × 1.0 L moles = 0.10 mol

Calculate the moles after the reaction

We have 0.10 moles of sulfuric acid and 0.30 moles of barium nitrate initially. The stoichiometry of the balanced equation shows that one mole of sulfuric acid reacts with one mole of barium nitrate to form one mole of barium sulfate. As there are fewer moles of sulfuric acid, it will be the limiting reagent. So, 0.10 moles of sulfuric acid will react with 0.10 moles of barium nitrate to form 0.10 moles of barium sulfate. The final moles of each species are: Sulfuric Acid: 0.10-0.10 = 0 mol Barium Nitrate: 0.30-0.10 = 0.20 mol Barium Sulfate: 0+0.10 = 0.10 mol

Calculate the concentration of barium ions in the final solution

After the reaction, there are 0.20 moles of barium nitrate left, which will dissolve in the 1.0 L of the solution. From the balanced equation, we can see that 1 mole of barium nitrate gives 1 mole of barium ions. Therefore, the concentration of barium ions in the final solution will be: Ba²⁺ concentration = moles of barium nitrate left / volume Ba²⁺ concentration = 0.20 mol / 1.0 L Ba²⁺ concentration = 0.20 M

Determine the mass of solid formed

We have calculated that 0.10 mol of barium sulfate is formed. To find the mass of the solid formed, we can use the molar mass of barium sulfate (233.43 g/mol): mass = moles × molar mass mass = 0.10 mol × 233.43 g/mol mass = 23.34 g

Calculate the pH of the solution

Since there is no sulfuric acid left after the reaction, the solution will contain nitric acid (HNO₃) formed during the reaction. The balanced equation shows that two moles of nitric acid are produced for every mole of sulfuric acid that reacts. As 0.10 moles of sulfuric acid react, 0.20 moles of nitric acid are produced. Therefore, the concentration of nitric acid in the 1.0 L solution is: HNO₃ concentration = moles of nitric acid / volume HNO₃ concentration = 0.20 mol / 1.0 L HNO₃ concentration = 0.20 M Nitric acid is a strong acid, and it will completely dissociate in the solution. So the concentration of hydronium ions (H₃O⁺) in the solution will be equal to the concentration of nitric acid: [H₃O⁺] = 0.20 M We can now calculate the pH of the solution using the formula: pH = -log₁₀([H₃O⁺]) pH = -log₁₀(0.20) pH ≈ 0.70 #Summary# 1. pH of the solution: ≈ 0.70 2. Concentration of barium ions in the final solution: 0.20 M 3. Mass of solid formed: 23.34 g

Key concepts

Stoichiometry

Stoichiometry is all about the quantitative relationships between the reactants and products in a chemical reaction. In our problem, we have sulfuric acid and barium nitrate reacting together. Understanding stoichiometry helps us determine how much of each substance is needed or produced. In a balanced equation, like the one in this problem—\( \text{H}_2\text{SO}_4 + \text{Ba(NO}_3\text{)}_2 \rightarrow \text{BaSO}_4 (s) + 2 \text{HNO}_3 \)—the coefficients tell us the molar ratios. Here, one mole of sulfuric acid reacts with one mole of barium nitrate.

The stoichiometry of this reaction indicates that sulfuric acid is the limiting reagent, which means it will be completely consumed first, controlling the amount of product formed. By knowing the initial moles of sulfuric acid (0.10 mol), we can predict the formation of barium sulfate (0.10 mol) based on a 1:1 stoichiometry. This is a key step in calculating the reactants' consumption and products' formation using stoichiometric principles.

Acid-Base Reactions

Acid-base reactions involve the transfer of protons (H⁺ ions) between reactants. In our scenario, sulfuric acid (\(\text{H}_2\text{SO}_4\)) is a strong acid that can donate protons easily, while barium nitrate is a salt that provides barium ions. The interaction doesn't usually fit a typical acid-base reaction, but when sulfuric acid reacts with the base part of barium nitrate, it forms nitrogen dioxide and water, which in turn generate nitric acid.

Because sulfuric acid is strong, it dissociates completely in water, increasing the concentration of hydrogen ions (H⁺) in the solution. This dissociation plays a crucial role in calculating the resulting pH after the reaction is complete. In this problem, we initially have 0.10 moles of the strong acid, leading to the complete consumption of sulfuric acid, ensuring only the resulting products (including the acidic \(\text{HNO}_3\)) influence the final pH.

Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. However, our exercise focuses on a reaction that proceeds to completion rather than equilibrium, since sulfuric acid and barium nitrate react fully due to the insolubility of barium sulfate.

Despite this, understanding equilibrium is valuable for similar reactions. Once a reaction reaches equilibrium, the concentrations of reactants and products remain constant. Our problem's complete reaction means no further changes in concentration, primarily involving an insoluble precipitate forming (\(\text{BaSO}_4\)), driving the reaction to completion and affecting final concentrations.

Solution Chemistry

Solution chemistry evaluates how substances dissolve and interact in solvents like water. In our problem, we initially have sulfuric acid dissolved in an aqueous system, creating a solution where it fully ionizes. Adding barium nitrate changes the chemistry—barium ions and leftover nitrate ions stay in solution unless they form a precipitate.

The reaction leads to the formation of barium sulfate, a solid that precipitates out. This event is crucial in solution chemistry because it affects both the composition and concentration within the solution. The remaining ions, particularly from the leftover barium nitrate, contribute to the final concentrations after precipitation. Calculating the barium ion concentration emphasizes understanding how residual ions continue to dissolve in the unchanged solution volume.