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Chapter 4: Problem 42

A stock solution containing \(\mathrm{Mn}^{2+}\) ions was prepared by dissolving \(1.584 \mathrm{~g}\) 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 of the stock solution and solutions A, B, and C are as follows: Stock solution concentration: \(\frac{1.584}{54.93}\) mol/L Solution A concentration: \(\frac{(1.584/54.93) * 0.050}{1.000}\) mol/L Solution B concentration: \(\frac{(C1 * 0.010)}{0.250}\) mol/L, where C1 is the concentration of solution A. Solution C concentration: \(\frac{(C2 * 0.010)}{0.500}\) mol/L, where C2 is the concentration of solution B.

Step by step solution

01

Calculate the moles of Mn\(^{2+}\) ions in the stock solution

First, we need to determine the moles of Mn\(^{2+}\) ions in the stock solution. We'll use the molar mass of manganese metal which is 54.93 g/mol. Number of moles of Mn = (mass of Mn) / (molar mass of Mn) Number of moles of Mn = (1.584 g) / (54.93 g/mol)
02

Concentration of the stock solution

Now that we have the moles of Mn\(^{2+}\) ions, we can calculate the concentration of the stock solution in mol/L. Concentration of stock solution = (number of moles of Mn\(^{2+}\) ions) / (volume of stock solution in L) Concentration of stock solution = (1.584/54.93) / 1.000 L
03

Concentration of solution A

Now we'll determine the concentration of solution A. Remember that dilution follows the equation: C1V1 = C2V2 Where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the concentration and volume of the diluted solution. We know: C1 = Concentration of stock solution = (1.584 / 54.93) mol/L V1 = Volume of stock solution used = 50.00 mL = 0.050 L V2 = Volume of solution A = 1000.0 mL = 1.000 L Now we can solve for C2, the concentration of solution A: C2 = (C1 * V1) / V2
04

Concentration of solution B

Similarly, we'll determine the concentration of solution B using the concentration of solution A: C1 = Concentration of solution A (from Step 3) V1 = Volume of solution A used = 10.00 mL = 0.010 L V2 = Volume of solution B = 250.0 mL = 0.250 L Again, we can find C2 (concentration of solution B) using the dilution equation: C2 = (C1 * V1) / V2
05

Concentration of solution C

Lastly, we'll determine the concentration of solution C using the concentration of solution B: C1 = Concentration of solution B (from Step 4) V1 = Volume of solution B used = 10.00 mL = 0.010 L V2 = Volume of solution C = 500.0 mL = 0.500 L Again, we can find C2 (concentration of solution C) using the dilution equation: C2 = (C1 * V1) / V2 By following these steps, you will be able to find the concentration of the stock solution and solutions A, B, and C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity Calculation
Molarity is a way to express the concentration of a solution, specifically in terms of moles of solute per liter of solution (mol/L). To calculate molarity, you first need to know the number of moles of the substance you're interested in and the total volume of the solution. In this exercise, the stock solution is prepared by dissolving manganese metal. To find out how many moles of manganese you have, divide the mass of the manganese by its molar mass. In this case, the mass is 1.584 grams and the molar mass is 54.93 g/mol. This gives the moles of manganese ions, allowing us to proceed to calculate the concentration. Once you have the moles, simply divide by the volume of the solution in liters. The stock solution in this scenario has a volume of 1 liter, simplifying our calculation significantly. This resulting value gives the molarity or concentration of the stock solution, an essential piece of information for subsequent dilution calculations.
Dilution Principle
The dilution principle is widely used in chemistry to prepare solutions of desired concentrations from more concentrated stock solutions. The relationship between the concentrations and volumes of a solution before and after dilution is represented by the equation: \[ C_1V_1 = C_2V_2 \]This equation shows that the product of the initial concentration \(C_1\) and volume \(V_1\) is equal to the product of the final concentration \(C_2\) and volume \(V_2\). It is fundamental in understanding how to scale solutions to different concentrations without changing the amount of solute present.When you perform a dilution, say of the stock solution to make solution A, you keep the moles of solute constant while only the volume changes. By knowing how solutions A, B, and C are prepared through dilution, you can apply this equation in each step. Each dilution further lowers the concentration as the solution is diluted further鈥攁chieving a weaker solution with each progressive step.
Chemical Solutions
Chemical solutions are homogenous mixtures composed of solutes dissolved in solvents. In this exercise, the solute is manganese \(\mathrm{Mn}^{2+}\) ions, and it's initially dissolved in nitric acid. Preparing chemical solutions often requires understanding the relationships between solute, solvent, concentration, and volume.The concentration of a chemical solution is a critical factor, influencing various properties such as reactivity and conductivity. Solutions are used in experiments, industrial processes, and everyday life, such as saltwater or sugar in coffee.In laboratory settings, preparing precise concentrations involves weighing solutes, dissolving them in solvents, and adjusting volumes accurately. The fundamental concept of solution and dilution ensures the correct chemical environment for reactions to proceed as planned, illustrating the vast utility and importance of solution chemistry.

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Most popular questions from this chapter

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The balanced equation is \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow\) $$ 4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) $$ This reaction is an oxidation-reduction reaction. What species is reduced, and what species is oxidized? How many electrons are transferred in the balanced equation above?

A solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in water is prepared by dissolving \(75.0 \mathrm{~mL}\) of ethanol (density \(=0.79 \mathrm{~g} / \mathrm{cm}^{3}\) ) in enough water to make \(250.0 \mathrm{~mL}\) of solution. What is the molarity of the ethanol in this solution?

What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)

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Many plants are poisonous because their stems and leaves contain oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) or sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\); when ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) in aqueous solution.
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