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Chapter 16: Problem 111

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

Short Answer

Expert verified
The pH of the final solution is approximately 0.70, the concentration of barium ions in the final solution is \(0.20\,\mathrm{mol\,L^{-1}}\), and the mass of solid barium sulfate formed is 23.34 g.

Step by step solution

01

Write the balanced equation for the reaction between sulfuric acid and barium nitrate

The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and barium nitrate (Ba(NO3)2) is: \[H_2SO_4(aq) + Ba(NO_3)_2(aq) \rightarrow BaSO_4(s) + 2HNO_3(aq)\]
02

Determine the initial amount of sulfuric acid in the solution

As the solution contains 0.10 M sulfuric acid, we can find the initial amount by multiplying the molarity by the volume of the solution: \(0.10\,\mathrm{mol/L} \times 1.0\,\mathrm{L} = 0.10\,\mathrm{mol}\)
03

Calculate the limiting reactant and the amount of solid formed from the reaction

The 0.30 mol of barium nitrate is added to the solution, and the reaction has a 1:1 stoichiometry, as seen from the balanced equation. The limiting reactant is the one with the least amount of moles, which in this case is sulfuric acid (0.10 mol). Therefore, 0.10 mol of solid barium sulfate (BaSO4) will form.
04

Calculate the final amounts of sulfuric acid, barium ions, and nitrate ions in the solution

All of the sulfuric acid reacts, so the final amount is 0 moles. The barium ions react with sulfuric acid, so the remaining amount will be \(0.30\,\mathrm{mol} - 0.10\,\mathrm{mol} = 0.20\,\mathrm{mol}\). Since 0.20 mol of nitrate ions are produced from the reaction, the final amount of nitrate ions will be \(0.10\,\mathrm{mol} \times 2 + 0.20\,\mathrm{mol} = 0.40\,\mathrm{mol}\).
05

Determine the final concentration of each ion

Since the volume of the solution doesn't change (1.0 L), the concentration of each ion will be equal to the amount in moles. The final concentration of barium ions is: \(0.20\,\mathrm{mol\,L^{-1}}\). The final concentration of hydrogen ions in the solution is 0, as all the sulfuric acid reacts, and nitric acid is a strong acid that completely dissociates in water: \(2\times0.10\,\mathrm{mol\,L^{-1}} = 0.20\,\mathrm{mol\,L^{-1}}\).
06

Calculate the pH of the final solution

The pH of the solution can be calculated using the concentration of hydrogen ions: \[\mathrm{pH}=-\log\left[H^+\right]\] \[\mathrm{pH}=-\log\left(0.20\,\mathrm{mol\,L^{-1}}\right) \approx 0.70\]
07

Calculate the mass of the solid formed

To calculate the mass of barium sulfate formed from the reaction, we can use its molar mass (BaSO4: 233.43 g/mol): \(0.10\,\mathrm{mol} \times 233.43\,\mathrm{g/mol} = 23.34\,\mathrm{g}\) So, the pH of the final solution is approximately 0.70, the concentration of barium ions is 0.20 mol L-1, and the mass of solid formed is 23.34 g.

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