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Chapter 4: Problem 42

A stock solution containing \(\mathrm{Mn}^{2+}\) ions was prepared by dissolving \(1.584\) g pure manganese metal in nitric acid and diluting to a final volume of \(1.000 \mathrm{~L}\). The following solutions were then prepared by dilution: For solution \(A, 50.00 \mathrm{~mL}\) of stock solution was diluted to \(1000.0 \mathrm{~mL}\). For solution \(B, 10.00 \mathrm{~mL}\) of solution \(A\) was diluted to \(250.0 \mathrm{~mL}\). For solution \(C, 10.00 \mathrm{~mL}\) of solution \(B\) was diluted to \(500.0 \mathrm{~mL}\). Calculate the concentrations of the stock solution and solutions \(A, B\), and \(C .\)

Short Answer

Expert verified
The concentrations are: stock solution = \(\frac{1.584}{54.94(1.000)}\mathrm{\ M}\), solution A = \(\frac{C1(0.05000)}{1.000}\mathrm{\ M}\), solution B = \(\frac{C1(0.01000)}{0.250}\mathrm{\ M}\), and solution C = \(\frac{C1(0.01000)}{0.500}\mathrm{\ M}\), where \(C1\) is the concentration of the previous solution.

Step by step solution

01

Calculate the moles of manganese metal

To find the number of moles of manganese in the stock solution, we'll use the formula: Moles = mass / molar mass Given: Mass of manganese = 1.584 g Molar mass of manganese = 54.94 g/mol Moles of manganese = \(\frac{1.584\mathrm{\ g}}{54.94\mathrm{\ g/mol}}\)
02

Calculate the concentration of the stock solution

Now we'll find the concentration of Mn2+ in the stock solution: Concentration = moles / volume Given: Moles of manganese = \(\frac{1.584}{54.94}\mathrm{\ mol}\) Volume = 1.000 L Concentration (stock solution) = \(\frac{1.584}{54.94(1.000)}\mathrm{\ M}\)
03

Calculate the concentration of solution A

Now we'll use the dilution equation to find the concentration of solution A: \(C1V1=C2V2\) Given: \(C1\) = stock solution concentration \(V1\) = 50.00 mL \(V2\) = 1000.0 mL To make the units consistent, convert mL to L: \(V1 = 50.00\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.05000\mathrm{\ L}\) \(V2 = 1000.0\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 1.000\mathrm{\ L}\) Solve for \(C2\) (concentration of solution A): \(C2 = \frac{C1V1}{V2}\)
04

Calculate the concentration of solution B

Now we'll repeat the process to find the concentration of solution B, using solution A as our starting solution: Given: \(C1\) = solution A concentration \(V1\) = 10.00 mL \(V2\) = 250.0 mL Convert mL to L: \(V1 = 10.00\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.01000\mathrm{\ L}\) \(V2 = 250.0\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.250\mathrm{\ L}\) Solve for \(C2\)(concentration of solution B): \(C2 = \frac{C1V1}{V2}\)
05

Calculate the concentration of solution C

Finally, we'll find the concentration of solution C, using solution B as our starting solution: Given: \(C1\) = solution B concentration \(V1\) = 10.00 mL \(V2\) = 500.0 mL Convert mL to L: \(V1 = 10.00\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.01000\mathrm{\ L}\) \(V2 = 500.0\mathrm{\ mL} \times \frac{1\mathrm{\ L}}{1000\mathrm{\ mL}} = 0.500\mathrm{\ L}\) Solve for \(C2\)(concentration of solution C): \(C2 = \frac{C1V1}{V2}\)

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Most popular questions from this chapter

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) of \(0.100 M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) b. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow\)

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\). A \(0.205-g\) sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The \(\mathrm{BaSO}_{4}\) that is formed is filtered, dried, and weighed. Its mass is \(0.298 \mathrm{~g}\). What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}{ }^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

What volume of each of the following acids will react completely with \(50.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH}\) ? a. \(0.100 \mathrm{M} \mathrm{HCl}\) b. \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) c. \(0.200 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) ( 1 acidic hydrogen)

Balance each of the following oxidation-reduction reactions by using the oxidation states method. a. \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a a)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Pb}(s) \rightarrow \mathrm{Pb}(\mathrm{OH})_{2}(s)\) c. \(\mathrm{H}^{+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)
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