Wrap-Up

In this lesson we have studied a variety of things relating to concentrations of solutions. There are several different ways of expressing concentration: weight percent, volume percent, weight/volume percent, molarity, and normality. We have worked with calculations that relate M to moles of solute and volume of solution and also formula mass as well as color intensity. We looked at how, in reversible reactions, the concentrations of the chemicals in equilibrium are related to one another by the solubility product constant, K_sp, for the equilibrium.

Complete the Self Quiz now to test your knowledge.

Self Quiz

1. The formula mass of NaOH is 40.0 g/mol; what is the molarity of a solution made from 5.0 g of NaOH dissolved into enough water to make 200.0 mL of solution?

2. What mass percent CaCl₂ is formed by dissolving 20.0 g of CaCl₂ in 150.0 g of water?

3. When 40.0 mL of a 2.00 M FeBr₃ solution is diluted to a volume of 250.0 mL, what is the new concentration?

4. How many moles of glycerol are there in 50.0 mL of a 3.0 M solution?

5. In a 0.500 M solution of Na₂ SO₄ what are the ions present and what are their concentrations?

6. How many moles of chloride ion are present in 0.750 L of 0.005 M AlCl₃?

7. What units are typically used when computing concentration in colorimetry/spectrometry?

8. A 0.340 M solution of a compound has an absorbance of 0.180 at a wavelength of 558 nm. What is the concentration of the same compound that has an absorbance at 558 nm of 0.075? Assume 1 cm lengths.

9. Write an equation for the dissolution of Al₂S₃ and write an expression for its solubility product.

10. A 100 mL aqueous solution containing 74.5 g of CaCl₂ is exactly saturated. Determine the value of the solubility product constant for CaCl₂ from this data. Hint: first find the concentration of the compound, then the concentrations of the ions, and plug into the solubility product expression.

Self Quiz Answers

1. The formula mass of NaOH is 40.0 g/mol; what is the molarity of a solution made from 5.0 g of NaOH dissolved into enough water to make 200.0 mL of solution?

5.0 g → 0.125 mol

0.125 mol/0.2000 L = 0.625 M

2. What mass percent CaCl₂ is formed by dissolving 20.0 g of CaCl₂ in 150.0 g of water?

20.0 g/(20.0 g + 150.0 g) * 100% = 11.8%

3. When 40.0 mL of a 2.00 M FeBr₃ solution is diluted to a volume of 250.0 mL, what is the new concentration?

40.0 mL * 2.00 M = 250.0 mL * C

C = 0.32 M

4. How many moles of glycerol are there in 50.0 mL of a 3.0 M solution?

50.0 mL → 0.0500 L

3.0 mol/L * 0.0500 L = 0.15 mol

5. In a 0.500 M solution of Na₂ SO₄ what are the ions present and what are their concentrations?

Na⁺: 0.500 M * 2 = 1.00 M

SO₄^2-: 0.500 M * 1 = 0.500 M

6. How many moles of chloride ion are present in 0.750 L of 0.005 M AlCl₃?

0.005 M * 3 = 0.015 M

0.015 mol/L * 0.750 L = 0.011 mol

7. What units are typically used when computing concentration in colorimetry/spectrometry?

Molarity is typically used, though it is perfectly possible to use most concentration units

8. A 0.340 M solution of a compound has an absorbance of 0.180 at a wavelength of 558 nm. What is the concentration of the same compound that has an absorbance at 558 nm of 0.075? Assume 1 cm lengths.

0.180 = 0.340 M * 1 cm * ε

ε = 0.529 M^-1cm^-1

0.075 = C * 1 cm * 0.529 M^-1cm^-1

C = 0.142 M

Note: The wavelength (558 nm) is a red herring. That value is not used in your calculations, it just tells you what wavelength the colorimeter was set to measure. This is useful in the lab, but will never be used in these calculations.

9. Write an equation for the dissolution of Al₂S₃ and write an expression for its solubility product.

Al₂S₃ (s) ⇌ 2 Al³⁺ (aq) + 3 S^2- (aq)

K_sp = [Al³⁺]²*[S^2-]³

10. A 100 mL aqueous solution containing 74.5 g of CaCl₂ is exactly saturated. Determine the value of the solubility product constant for CaCl₂ from this data. Hint: first find the molar concentration of the compound, then the concentrations of the ions, and plug into the solubility product expression.

74.5 g → 0.671 mol

0.671 mol/0.100 L = 6.71 M

[Ca²⁺] = 6.71 M

[Cl^-] = 13.42 M

K_sp = [Ca²⁺]*[Cl^-]²

K_sp = 6.71*13.42²

K_sp = 1210