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Chapter 5: Problem 13

Which of the following aqueous solutions has the highest concentration of \(\mathrm{K}^{+}\) ? (a) \(0.0850 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4};\) (b) a solution containing \(1.25 \mathrm{g} \mathrm{KBr} / 100 \mathrm{mL} ;\) (c) a solution having \(8.1 \mathrm{mg} \mathrm{K}^{+} / \mathrm{mL}\).

Short Answer

Expert verified
The solution having 8.1 mg K+ / mL has the highest concentration of K+.

Step by step solution

01

Calculate the Molar Concentration for Option (a)

For option a: Given that the solution is 0.0850 M K2SO4. Note that each molecule of K2SO4 has two molecules of K+. So, the concentration of K+ in the solution is \(0.0850 M \cdot 2 = 0.170 M\).
02

Calculate the Molar Concentration for Option (b)

For option b: Convert grams of KBr to moles using the molar mass. Given the volume of the solution is 100 mL, convert this to liters to calculate the molarity (moles per liter). We have \(1.25 g \mathrm{KBr} \times \frac{1 mol \mathrm{KBr}}{119.002 g} = 0.0105 mol \mathrm{KBr}\). Therefore, the concentration of KBr is \(0.0105 mol / 0.1 L = 0.105 M\). There is one mole of Potassium ion in KBr, hence the concentration of K+ is also 0.105 M.
03

Calculate the Molar Concentration for Option (c)

For option c: Given concentration is 8.1 mg K+ / mL. First, convert mg to g, and mL to L: \(8.1 mg \times \frac{1 g}{1000 mg} = 0.0081 g\) and \(1 mL = \frac{1}{1000} L\). Then convert grams to moles: \(0.0081 g \times \frac{1 mol}{39.098 g} = 0.000207 mol\). Therefore, the molar concentration is \(0.000207 mol / \frac{1}{1000} L = 0.207 M\) for K+.
04

Compare the Molar Concentrations

The molar concentrations for options (a), (b), and (c) are 0.170 M, 0.105 M, and 0.207 M, respectively. Thus, option c has the highest molar concentration of K+ ions.

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Most popular questions from this chapter

Assuming the volumes are additive, what is the \(\left[\mathrm{NO}_{3}^{-}\right]\) in a solution obtained by mixing \(275 \mathrm{mL}\) of \(0.283 \mathrm{M} \mathrm{KNO}_{3}, 328 \mathrm{mL}\) of \(0.421 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2},\) and \(784 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O} ?\)

The electrolyte in a lead storage battery must have a concentration between 4.8 and \(5.3 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) if the battery is to be most effective. A \(5.00 \mathrm{mL}\) sample of a battery acid requires \(49.74 \mathrm{mL}\) of \(0.935 \mathrm{M} \mathrm{NaOH}\) for its complete reaction (neutralization). Does the concentration of the battery acid fall within the desired range? [Hint: Keep in mind that the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) produces two \(\mathrm{H}^{+}\) ions per formula unit.]

Iron (Fe) is obtained from rock that is extracted from open pit mines and then crushed. The process used to obtain the pure metal from the crushed rock produces solid waste, called tailings, which are stored in disposal areas near the mines. The tailings pose a serious environmental risk because they contain sulfides, such as pyrite ( \(\mathrm{FeS}_{2}\) ), which oxidize in air to produce metal ions and \(\mathrm{H}^{+}\) ions that can enter into surface water or ground water. The oxidation of \(\mathrm{FeS}_{2}\) to \(\mathrm{Fe}^{3+}\) is described by the unbalanced chemical equation below. \(\mathrm{FeS}_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(\quad \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \quad(\text { not balanced })\) Thus, the oxidation of pyrite produces \(\mathrm{Fe}^{3+}\) and \(\mathrm{H}^{+}\) ions that can leach into surface or ground water. The leaching of \(\mathrm{H}^{+}\) ions causes the water to become very acidic. To prevent acidification of nearby ground or surface water, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is added to the tailings to neutralize the \(\mathrm{H}^{+}\) ions: \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \underset{\mathrm{Ca}^{2+}}{\longrightarrow}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Balance the equation above for the reaction of \(\mathrm{FeS}_{2}\) and \(\mathrm{O}_{2}\). [ Hint: Start with the half-equations \(\mathrm{FeS}_{2}(\mathrm{s}) \rightarrow\) \(\left.\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \text { and } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1) .\right]\) (b) What is the minimum amount of \(\mathrm{CaCO}_{3}(\mathrm{s})\) required, per kilogram of tailings, to prevent contamination if the tailings contain \(3 \%\) S by mass? Assume that all the sulfur in the tailings is in the form \(\mathrm{FeS}_{2}\).

Explain the important distinctions between (a) a strong electrolyte and strong acid; (b) an oxidizing agent and reducing agent; (c) precipitation reactions and neutralization reactions; (d) half-reaction and overall reaction.

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