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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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Most popular questions from this chapter

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