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Solutions

Solutions. Chapter 18. Recall…. Solutions are homogeneous mixtures that may be solid, liquid, or gas. Examples. Scuba Tank. Solution of oxygen and several other gases. Salt-Water Aquarium. In this solution aquarium salt is the solute, water is the solvent. 6.1.

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Solutions

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  1. Solutions Chapter 18

  2. Recall….. • Solutions are • homogeneous mixtures • that may be solid, liquid, or gas

  3. Examples Scuba Tank Solution of oxygen and several other gases. Salt-Water Aquarium In this solution aquarium salt is the solute, water is the solvent.

  4. 6.1 Solutions and Other Mixtures • Suspension – a mixture that looks uniform when stirred or shaken that separates into different layers when it is no longer agitated • Colloid – a mixture of very tiny particles of pure substances that are dispersed in another substance but do not settle out of the substance • Emulsion – any mixture of immiscible liquids in which the liquids are spread throughout one another

  5. Suspensions • The pulp in orange juice is spread throughout the mixture right after the juice is shaken – over time, the pulp does not stay mixed with the water molecules and settles to the bottom of the container.

  6. Examples of suspensions • Water and coffee • Oil and vinegar • Sand and water • Water and oil • Blood

  7. Examples of colloids • Egg whites • Paint • Whipped cream (dispersing gas in a gas) • Marshmallows (dispersing gas in a solid) • Fog (water dispersed in air) • Smoke (solid particles dispersed in air)

  8. Emulsions • Examples: • Ice cream • Milk • Mayonnaise • Emulsifying agents like soap and detergent allow the formation of colloidal dispersions between liquids that do not ordinarily mix

  9. Factors determining solubility • Temperature • Stirring (agitation) • Surface area

  10. Properties of Solubility • Miscible – two liquids are miscible IF they dissolve in each other • EXAMPLE: Water and alcohol • Immiscible – liquids that are insoluble in each other • EXAMPLE: Oil and vinegar

  11. Factors affecting solubility • Solubility of most solid substances increases as temperature of solvent increases • Solubility of gases is greater in cold water than hot water

  12. The Dissolving Process More solid will dissolve as temperature increases Less gas will dissolve as temperature increases

  13. Henry’s Law • …at a given temperature the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid • (simply put, as pressure of gas above the liquid increases, the solubility increases; similarly, as pressure of the gas decreases, the solubility of the gas decreases)

  14. Solubility Curve Graphs

  15. What mass of sodium nitrate will dissolve in 100cm3 of water at 50°C? About 112g

  16. How much oxygen will dissolve in 100g water at 80C°? 0.5 mg How much oxygen will dissolve in 100g water at 10C°? 1.3 mg

  17. Henry’s Law - Equation S1 S2 ------ = ------ P1 P2 Solubility # 1 divided by pressure # 1 is equal to Solubility # 2 divided by Pressure #2

  18. Henry’s Law - Example • If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure what is its solubility (in g/L) at 1.0 atm of pressure? (The temperature is held constant at 250C) • 1. ANALYZEList the knowns and the unknown. Knowns: Unknown: P1 = 3.5 atm S2 = ? g/L S1 = 0.77 g/L P2 = 1.0 atm Henry’s Law: S1/P1 = S2/P2

  19. Example - continued • 2. CalculateSolve for the unknown. • Solving Henry’s law for S2 yields S2 = S1 X P2 P1 Substituting the known values and calculating yields S2 = 0.77 g/L X 1.0 atm = 0.22 g/L 3.5 atm

  20. Example continued • 3. EvaluateDoes the result make sense? • The pressure is reduced (from 3.5 atm to 1.0 atm, so the solubility of the gas should decrease, which it is shown to do. Because the new pressure is approximately one-third of the original pressure, the new solubility should be approximately one-third of the original, which it is. The answer is correctly expressed to two significant figures.

  21. Concentrations of Solutions • Concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent • A dilute solution is one that contains only a low concentration of solute • A concentrated solution contains a high concentration of solute

  22. Molarity (M) • THE most important unit of concentration in chemistry…… • Molarity (M) is the number of moles of a solute dissolved per liter of solution

  23. Molarity • Molarity = moles of solute or M = mol • liters of solution L For example: to calculate the molarity when 2 mol of glucose is dissolved in 5 L of solution, divide the number of moles by the volume in liters. 2 mol glucose 5 L of solution = 0.4 mol/L = 0.4 M mole

  24. Molarity Calculations • Rearrange the equation: Moles of solute = molarity (M) X liters of solution (V) Moles of solute = M1 X V1 M1 X V1 = M2 X V2 V1 = M2 X V2 / M1

  25. Colligative Properties • Colligative Properties – properties that depend only on the number of particles dissolved in a given mass of solvent. Three important ones: • Vapor-pressure lowering • Boiling-point elevation • Freezing-point depression

  26. Colligative properties • Vapor pressure – the pressure exerted by a vapor that is in dynamic equilibrium with its liquid in a closed system • Boiling-point Elevation – the difference in temperature between the boiling point of a solution and that of the pure solvent • Freezing-point depression – the difference in temperature between the freezing point of a solution and that of the pure solvent

  27. Calculations involving Colligative Properties • Molality (m) – is the number of moles of solute dissolved per kilogram (1000g) of solvent. (Also known as molal concentration) • Molality = moles of solute kilogram of solvent moles of solute = moles of solute kilogram of solvent 1000 g of solvent

  28. Percent composition • Percent by volume (% v/v) = Volume of solute Volume of solution X 100 Percent (mass/volume) (% m/v) = mass of solute (g) volume of solution (mL) X 100

  29. Assignment • Complete calculations…. • Practice Problems # 1,2 pg 507 • Practice Problems # 8,9,10,11 pg 511 • Practice Problems # 12,13 pg 513 • Practice Problems # 14,15 pg 514 • Practice Problems # 16,17 pg 515 • Practice Problems # 28,29 pg 521

  30. Problems page 507 • 1. The solubility of a gas in water is 0.16g/L at 104 kPa of pressure. What is the solubility when the pressure of the gas is increased to 288 kPa? (Assume the temperature is constant) • 2. A gas has a solubility in water at 0 C of 3.6 g/L at a pressure of 1.0 atm. What pressure is needed to produce an aqueous solution containing 9.5 g/L of the same gas at 0 C?

  31. Problems page 511 • 8. A solution has a volume of 2.0 L and contains 36.0 g of glucose. If the molar mass of glucose is 180 g/mol, what is the molarity of the solution? • 9. A solution has a volume of 250and contains 0.70 mol NaCl. What is its molarity?

  32. Problems page 511 • 10. How many moles of ammonium nitrate are in 335 mL of 0.425M NH4NO3? • 11. How many moles of solute are in 250 mL of 2.0 M CaCl2? How many grams of CaCl2 is this?

  33. Problems page 513 • 12. How many milliliters of a stock solution of 4.00M KI would you need to prepare 250.0 mL of 0.760M KI? • 13. Suppose you need 250 mL of 0.20M NaCl, but the only supply of sodium chloride you have is a solution of 1.0M NaCl. How do you prepare the required solution? Assume that you have the appropriate volume-measuring devices on hand.

  34. Problems page 514 • 14. If 10 mL of pure acetone is diluted with water to a total solution volume of 200 mL, what is the percent by volume of acetone of this solution? • 15. A bottle of hydrogen peroxide antiseptic is labeled 3.0% (v/v). How many mL H2O2 are in a 400.0 mL bottle of this solution?

  35. Problems page 515 • 16. Calculate the grams of solute required to make 250 mL of 0.10% MgSO4 (m/v). • 17. A solution contains 2.7g CuSO4 in 75 mL of solution. What is the percent (mass/volume) of the solution?

  36. Problems page 521 • 28. How many grams of sodium fluoride are needed to prepare a 0.400m NaCl solution that contains 750.0g of water? • 29. Calculate the molality of a solution prepared by dissolving 10.0g NaCl in 600g of water.

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