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.

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

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

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

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

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

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