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Chapter 15: Problem 111

Consider \(1.0 \mathrm{L}\) 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 \(\mathrm{pH},\) 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 0.70, the concentration of barium ions in the final solution is \(0.20\,\mathrm{M}\), and the mass of solid barium sulfate formed is \(23.34\,\mathrm{g}\).

Step by step solution

01

Write the balanced reaction equation

The balanced reaction equation between sulfuric acid and barium nitrate is: \(\mathrm{H_2SO_4 + Ba(NO_3)_2 \rightarrow BaSO_4(s) + 2\,HNO_3}\) This tells us that 1 mole of sulfuric acid reacts with 1 mole of barium nitrate to produce 1 mole of barium sulfate and 2 moles of nitric acid.
02

Calculate the moles of sulfuric acid in the initial solution

First, let's find the moles of sulfuric acid (\(\mathrm{H_2SO_4}\)) in the 1.0 L solution: Moles = Molarity × Volume = \(0.10\,\mathrm{M} \times 1.0\,\mathrm{L} = 0.10\,\mathrm{mol}\)
03

Determine the limiting reactant and calculate the moles of products

We are given 0.30 mole of barium nitrate (\(\mathrm{Ba(NO_3)_2}\)), and calculated 0.10 mole of sulfuric acid. The balanced reaction equation tells us that 1 mole of sulfuric acid reacts with 1 mole of barium nitrate, so sulfuric acid is the limiting reactant. The moles of barium sulfate (\(\mathrm{BaSO_4}\)) that are formed will be equal to the moles of limiting reactant (sulfuric acid): Moles of \(\mathrm{BaSO_4}\) = 0.10 mol The moles of nitric acid (\(\mathrm{HNO_3}\)) produced will be double the moles of limiting reactant: Moles of \(\mathrm{HNO_3}\) = 2 × 0.10 mol = 0.20 mol
04

Calculate the final concentrations of hydronium ions (\(\mathrm{H^+}\)) and barium ions (\(\mathrm{Ba^{2+}}\))

Since there is no change in the volume of the solution, the concentration of hydronium ions from nitric acid will be equal to the concentration of nitric acid (the acidity of the solution is now only due to nitric acid and the sulfuric acid has reacted completely): \[ [\mathrm{H^+}] = \frac{0.20\,\mathrm{mol}}{1.0\,\mathrm{L}} = 0.20\,\mathrm{M} \] The concentration of barium ions in the final solution will be based on the initial amount of barium nitrate added and subtracting the moles that reacted with sulfuric acid: \[ [\mathrm{Ba^{2+}}] = \frac{0.30\,\mathrm{mol} - 0.10\,\mathrm{mol}}{1.0\,\mathrm{L}} = 0.20\,\mathrm{M} \]
05

Calculate the pH of the final solution

Now we can calculate the pH of the solution using the concentration of hydronium ions: \[\mathrm{pH} = -\log([\mathrm{H^+}]) = -\log(0.20\,\mathrm{M}) = 0.70\]
06

Calculate the mass of solid barium sulfate formed

To find the mass of solid barium sulfate formed, we multiply the moles of barium sulfate by its molar mass: Mass = Moles × Molar mass = \(0.10\,\mathrm{mol} \times 233.4\, \frac{\mathrm{g}}{\mathrm{mol}} = 23.34\,\mathrm{g}\) #Conclusion#: The pH of the final solution is 0.70, the concentration of barium ions in the final solution is 0.20 M, and the mass of solid barium sulfate formed is 23.34 g.

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Most popular questions from this chapter

You are browsing through the Handbook of Hypothetical Chemistry when you come across a solid that is reported to have a \(K_{\mathrm{sp}}\) value of zero in water at \(25^{\circ} \mathrm{C}\). What does this mean?

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) b. \(\mathrm{V}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}\)

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{FeC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.1 \times 10^{-7},\) or \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-7}\) b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\mathrm{sp}}=8.1 \times 10^{-12},\) or \(\mathrm{Mn}(\mathrm{OH})_{2}\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with HCl?What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10\) \(M\) \(\mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with \(\mathrm{HCl}\)?

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{S},\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are \(K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)\) \(=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},\) and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=\) \(1 \times 10^{-54}\).
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