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Chapter 4: Problem 28

Calculate each of the following quantities: (a) Molarity of a solution prepared by diluting \(37.00 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) potassium chloride to \(150.00 \mathrm{~mL}\) (b) Molarity of a solution prepared by diluting \(25.71 \mathrm{~mL}\) of \(0.0706 \mathrm{M}\) ammonium sulfate to \(500.00 \mathrm{~mL}\) (c) Molarity of sodium ion in a solution made by mixing \(3.58 \mathrm{~mL}\) of \(0.348 M\) sodium chloride with \(500 . \mathrm{mL}\) of \(6.81 \times 10^{-2} M\) sodium sulfate (assume volumes are additive)

Short Answer

Expert verified
Part (a): 0.0617 M. Part (b): 0.00363 M. Part (c): 0.1376 M.

Step by step solution

01

- Calculate the moles of potassium chloride in the initial solution (Part a)

Use the formula for molarity: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \). Rearrange to find moles: \( \text{moles} = M \times V \). For potassium chloride: \( 0.250 \text{ M} \times 0.037 \text{ L} = 0.00925 \text{ moles} \).
02

- Calculate the final molarity of the diluted solution (Part a)

Using the moles of solute (0.00925 moles) and the new volume (0.150 L), the final molarity is: \( M = \frac{0.00925 \text{ moles}}{0.150 \text{ L}} = 0.0617 \text{ M} \).
03

- Calculate the moles of ammonium sulfate in the initial solution (Part b)

Moles of ammonium sulfate can be calculated similarly: \( 0.0706 \text{ M} \times 0.02571 \text{ L} = 0.001815426 \text{ moles} \).
04

- Calculate the final molarity of the diluted ammonium sulfate solution (Part b)

Using the calculated moles of ammonium sulfate (0.001815426 moles) and the new volume (0.500 L), the final molarity is: \( M = \frac{0.001815426 \text{ moles}}{0.500 \text{ L}} = 0.00363 \text{ M} \).
05

- Calculate the moles of sodium chloride in the initial solution (Part c)

Calculate the moles of sodium chloride: \( 0.348 \text{ M} \times 0.00358 \text{ L} = 0.00124584 \text{ moles} \).
06

- Calculate the moles of sodium sulfate in the initial solution (Part c)

Calculate the moles of sodium sulfate: \( 6.81 \times 10^{-2} \text{ M} \times 0.500 \text{ L} = 0.03405 \text{ moles} \).
07

- Convert moles of sodium sulfate to moles of sodium ions (Part c)

Since each molecule of sodium sulfate (\text{Na}_2\text{SO}_4) dissociates into 2 \text{Na}^{+} ions, the moles of sodium ions from sodium sulfate are: \( 2 \times 0.03405 \text{ moles} = 0.0681 \text{ moles} \).
08

- Calculate total moles of sodium ions (Part c)

Sum the moles of sodium ions from both sources: \( 0.00124584 \text{ moles (NaCl)} + 0.0681 \text{ moles (Na}_2\text{SO}_4\text{)} = 0.06934584 \text{ moles} \).
09

- Calculate final molarity of sodium ions in the mixed solution (Part c)

The final volume of the mixed solution is: \( 0.00358 \text{ L} + 0.500 \text{ L} = 0.50358 \text{ L} \). The final molarity is: \( M = \frac{0.06934584 \text{ moles}}{0.50358 \text{ L}} = 0.1376 \text{ M} \).

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Key Concepts

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

molarity
Molarity is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute per liter of solution, symbolized by (M). To calculate molarity, use the formula: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \). For example, in Part (a) of the exercise, we calculated the molarity of a diluted potassium chloride solution. We first found the moles by multiplying the initial molarity by the initial volume. Once we had the moles, we divided by the final volume to get the new molarity.
dilution
Dilution is the process of decreasing the concentration of a solute in a solution, usually by adding more solvent. In the exercise, we performed dilutions for both potassium chloride and ammonium sulfate. The key point in dilution is that the number of moles of the solute remains the same before and after dilution; only the volume of the solution changes. The formula for dilution is: \( M_1V_1 = M_2V_2 \), where \( M_1 \) and \( V_1 \) are the initial molarity and volume, and \( M_2 \) and \( V_2 \) are the final molarity and volume. For instance, after diluting 37.00 mL of 0.250 M potassium chloride to 150.00 mL, we calculated the final molarity to be 0.0617 M.
chemical concentration
Chemical concentration refers to the amount of solute present in a given quantity of solvent or solution. It's crucial in reactions, as it dictates how much solute interacts with other substances. Concentration can be expressed in various ways, such as molarity (M), which we used in the exercise. For example, we calculated the chemical concentration of sodium ions in Part (c) after mixing two solutions. By knowing the initial concentrations and volumes, we derived the final concentration of sodium ions.
moles
A mole is a fundamental unit in chemistry used to quantify the amount of substance. One mole contains Avogadro's number (approximately \(6.022 \times 10^{23}\)) of particles (atoms, molecules, etc.). To find the moles of a solute in a solution, use the rearranged molarity formula: \( \text{moles} = M \times V \). For instance, in Part (b) of the exercise, we calculated the moles of ammonium sulfate by multiplying its molarity (0.0706 M) by its initial volume (0.02571 L), resulting in 0.001815426 moles.
solution composition
Solution composition involves understanding what components (solute and solvent) make up the solution and in what proportions. In Part (c) of the exercise, we dealt with the composition of a solution containing sodium ions from both sodium chloride and sodium sulfate. We calculated moles from each source and considered the total moles to find the final molarity. Understanding solution composition helps predict the behavior of the solution in different chemical contexts.

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