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Chapter 7: Solutions and Colloids Spencer L. Seager Michael R. Slabaugh www.cengage.com/chemistry/seager Jennifer P. Harris.

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Presentation on theme: "Chapter 7: Solutions and Colloids Spencer L. Seager Michael R. Slabaugh www.cengage.com/chemistry/seager Jennifer P. Harris."— Presentation transcript:

1 Chapter 7: Solutions and Colloids Spencer L. Seager Michael R. Slabaugh www.cengage.com/chemistry/seager Jennifer P. Harris

2 LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1. Classify mixtures as solutions or nonsolutions based on their appearance. (Section 7.1; Exercise 7.4) 2. Demonstrate your understanding of terms related to the solubility of solutes in solution. (Section 7.2; Exercises 7.6 and 7.12) 3. Predict in a general way the solubilities of solutes in solvents on the basis of molecular polarity. (Section 7.3; Exercise 7.16) 4. Calculate solution concentrations in units of molarity, weight/weight percent, weight/volume percent, and volume/volume percent. (Section 7.4; Exercises 7.22 b, 7.30 c, 7.34 a, and 7.38 c) 5. Describe how to prepare solutions of specific concentration using pure solutes and solvent, or solutions of greater concentration than the one desired. (Section 7.5; Exercises 7.46 and 7.48 b) 6. Do stoichiometric calculations based on solution concentrations. (Section 7.6; Exercise 7.56) 7. Understand colligative solution properties of boiling point, freezing point, and osmotic pressure and how to determine osmolarity; (Section 7.7; Exercises 7.64 a & c and 7.74) 8. Describe the characteristics of colloids. (Section 7.8; Exercise 7.82) 9. Describe the process of dialysis, and compare it to the process of osmosis. (Section 7.9; Exercise 7.84)

3 Mixtures Solutions (sometimes called true solutions) are homogeneous mixtures of two or more substances in which the components are present as atoms, molecules, or ions. Particles in these solutions are: too small to reflect light, are transparent (always clear but can be colored). in constant motion and not settled by the influence of gravity. Heterogeneous mixtures Particles in these “solutions” are: Large aggregates, reflect light (and are turbid) and do settle out due to gravity Colloidal solutions Particles in these solutions are: Large aggregates, reflect light (and are turbid) but do not settle out due to gravity (homogeneous mixture)

4 The dissolving process For true solutions

5 SOLUTION TERMINOLOGY Solutions can be solids, liquids or gases but we will deal primarily with liquid solutions. Solutions are composed of a solvent (main component) and one or more solutes (substance that is dissolved in the solvent) Solubility – the degree to which a solute dissolves (insoluble, slightly soluble, soluble, very soluble) Miscible/immiscible – terms that describe whether a liquid solute dissolves or not Saturated, unsaturated, supersaturated

6 SOLUBILITY The solubility of a solute is the maximum amount of the solute that can be dissolved in a specific amount of solvent under specific conditions of temperature and pressure.

7 EXAMPLES OF SOLUTE SOLUBILITES AT 0°C

8 EXAMPLES OF SOLUTE SOLUBILITES AT 0°C (continued)

9 EFFECT OF TEMPERATURE ON SOLUBILITY

10 THE SOLUTION PROCESS The solution process involves interactions between solvent molecules (often water) and the particles of solute. An example of the solution process for an ionic solute in water:

11 THE SOLUTION PROCESS (continued) An example of the solution process for a polar solute in water:

12 THE SOLUTION PROCESS (continued) A solute will not dissolve in a solvent if: the forces between solute particles are too strong to be overcome by interactions with solvent particles. the solvent particles are more strongly attracted to each other than to solute particles. A good rule of thumb for solubility is “like dissolves like.” Polar solvents dissolve polar or ionic solutes. Nonpolar solvents dissolve nonpolar or nonionic solutes.

13 INCREASING THE RATE OF DISSOLVING Crush or grind the solute. Small particles provide more surface area for solvent attack and dissolve more rapidly than larger particles. Heat the solvent. Solvent molecules move faster and have more frequent collisions with solute at higher temperatures. Stir or agitate the solution. Stirring removes locally saturated solution from the vicinity of the solute and allows unsaturated solvent to take its place.

14 HEAT AND SOLUTION FORMATION Endothermic: Exothermic: Solute + solvent → solution + heat Solute + solvent + heat→ solution

15 SOLUTION CONCENTRATIONS Solution concentrations express a quantitative relationship about the amount of solute contained in a specific amount of solution. Concentration units discussed include molarity and percentage.

16 MOLARITY The molarity of a solution expresses the number of moles of solute contained in one liter of solution. The mathematical calculation of the molarity of a solution involves the use of the following equation: In this equation, the number of moles of solute in a sample of solution is divided by the volume in liters of the same sample of solution.

17 PERCENT CONCENTRATIONS Percent concentrations express the amount of solute contained in 100 parts of solution. The parts of solution may be expressed in different units. Three variations exist for % concentrations. W/W, W/V or V/V generally expressed in grams and mL; if not then at least the same units. To make solutions of M, W/V% and V/V%, the amount of solute is measured out and solvent added to the volume needed. To make solutions of W/W%, the grams of solvent and solute are added to determine the weight of the solution.

18 CONCENTRATON CALCULATIONS Example 1: A 250-mL sample of solution contains 0.134 moles of solute. Calculate the molarity of the solution. 9.45 g of methyl alcohol, CH 3 OH, was dissolved in enough pure water to give 500 mL of solution. What was the molarity of the solution? Calculate the %(w/w) of a solution prepared by dissolving 15.0 grams of table sugar in 100 mL of water. The density of the water is 1.00 g/mL. Calculate the %(w/v) of a solution prepared by dissolving 8.95 grams of sodium chloride in enough water to give 50.0 mL of solution. A solution is made by dissolving 250 mL of glycerin in enough water to give 1.50 L of solution. Calculate the %(v/v) of the resulting solution

19 SOLUTION PREPARATION Solutions of known concentration are usually prepared in one of two ways. In one method, the necessary quantity of pure solute is measured using a balance or volumetric equipment. The solute is put into a container and solvent, usually water, is added until the desired volume of solution is obtained.

20 SOLUTION PREPARATION EXAMPLE Calculation example: Describe how to prepare 500 mL of 0.250 M NaCl solution. Solution: The mass of NaCl needed must first be determined. The volume and concentration of the desired solution are known, so the equation for molarity is rearranged to solve for the number of moles of solute needed. The result is: moles of solute = M x liters of solution = 0.250 M x 0.500 L = 0.125 mole Thus, 0.125 moles of NaCl is needed. NaCl has a formula weight of 58.4 u, so 0.125 moles has a mass of 0.125 x 58.4g or 7.30 grams. The solution is prepared by weighing a sample of NaCl with a mass of 7.30 grams. The sample is put into a 500 mL volumetric flask and pure water is added up to the mark on the flask.

21 SOLUTION PREPARATION (continued) In a second method, a quantity of solution with a concentration greater than the desired concentration is diluted with an appropriate amount of solvent to give a solution with a lower concentration. This type of problem is made simpler by using the following equation: (C c )(V c ) = (C d )(V d ) In this equation, C c is the concentration of the concentrated solution that is to be diluted, V c is the volume of concentrated solution that is needed, C d is the concentration of the dilute solution, and V d is the volume of dilute solution.

22 SOLUTION PREPARATION EXAMPLE Calculation example: Describe how to prepare 250 mL of 0.500 M HCl solution from a 1.50 M HCl solution. Solution: According to the definitions given above, C c = 1.50 M, C d = 0.500 M, and V d = 250 mL. The equation given above can be solved for V c, the volume of concentrated solution needed: The solution is prepared by measuring 83.3 mL of 1.50 M HCl and pouring it into a 250 mL volumetric flask. Pure water is then added up to the mark on the flask to give 250 mL of 0.500 M solution.

23 SOLUTION STOICHIOMETRY As shown earlier, the number of moles of solute in a volume of solution of known molarity can be obtained by multiplying together the known molarity and the solution volume in liters. Molarity is a ratio of moles of solute to liters of solution. This ratio can be written as two conversion factors: The conversion factor on the left is used to multiply by the molarity. It is selected to cancel the units of liters of solution and obtain the units of moles of solute. The conversion factor on the right is used to divide by the molarity. It is selected to cancel the units of moles of solute and obtain the units of liters of solution.

24 SOLUTION STOICHIOMETRY EXAMPLE Calculation example: Consider the balanced equation HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) How many mL of 0.100 M HCl solution would exactly react with 25.00 mL of 0.125 M NaOH solution?

25 SOLUTION PROPERTIES Absolutely pure water conducts electricity very poorly. Some solutes called electrolytes produce water solutions that conduct electricity well. Some solutes called nonelectrolytes produce water solutions that do not conduct electricity. A solution of a strong electrolyte conducts electricity well. A solution of a weak electrolyte conducts electricity poorly. A solution of a nonelectrolyte does not conduct electricity.

26 ELECTROLYTES STRONG ELECTROLYTES Strong electrolytes form solutions that conduct electricity because they dissociate completely into charged ions when they dissolve. WEAK ELECTROLYTES Weak electrolytes form weakly conducting solutions because they dissociate into ions only slightly when they dissolve. NONELECTROLYTES Nonelectrolytes form nonconducting solutions because they do not dissociate into ions at all when they dissolve.

27 COLLIGATIVE PROPERTIES OF SOLUTIONS Colligative solution properties are properties that depend only on the concentration of solute particles in the solution. Three colligative properties are boiling point, freezing point, and osmotic pressure. Experiments demonstrate that the vapor pressure of water (solvent) above a solution is lower than the vapor pressure of pure water.

28 SOLUTION BOILING POINT The boiling point of a solution is always higher than the boiling point of the pure solvent of the solution. Since BP elevation is a colligative property, dependent only on the concentration of particles, the following substances would have an equal impact on BP elevation (or any other colligative property) 0.1 M CaCl 2 0.15 M NaCl 0.3 M sugar (non-electrolyte)

29 SOLUTION BOILING POINT (continued) For example, the dissociation of calcium chloride is represented as: CaCl 2 Ca 2+ + 2 Cl - Thus, when 1 mole of CaCl 2 dissolves, 3 moles of particles (ions) are put into the solution.

30 The freezing point of a solution is always lower than the freezing point of the pure solvent of the solution. SOLUTION FREEZING POINT

31 OSMOTIC PRESSURE OF SOLUTIONS When solutions having different concentrations of solute are separated by a semipermeable membrane, solvent tends to flow through the membrane from the less concentrated solution into the more concentrated solution in a process called osmosis. When the more concentrated solution involved in osmosis is put under sufficient pressure, the net osmotic flow of solvent into the solution can be stopped. The pressure necessary to prevent the osmotic flow of solvent into a solution is called the osmotic pressure of the solution and can be calculated by using the following equation, which is similar to the ideal gas law given earlier: π = nMRT

32 OSMOTIC PRESSURE OF SOLUTIONS (continued) In this equation, π is the osmotic pressure, n is the number of moles of solute particles put into solution when 1 mole of solute dissolves, M is the molarity of the solution, R is the universal gas constant written as 62.4 L torr/K mol, and T is the solution temperature in Kelvin. The product of n and M is called the osmolarity of the solution.

33 COLLOIDS Colloids are homogeneous mixtures of two or more components called the dispersing medium and the dispersed phase. The dispersed phase substances in a colloid are in the form of particles larger than those found in solutions. DISPERSING MEDIUM OF A COLLOID The dispersing medium of a colloid is the substance present in the largest amount. It is analogous to the solvent of a solution. DISPERSED PHASE OF A COLLOID The dispersed phase of a colloid is the substance present in a smaller amount than the dispersing medium. It is analogous to the solute of a solution.

34 COLLOID PROPERTIES In colloids, the dispersed phase particles cannot be seen and do not settle under the influence of gravity. Colloids appear to be cloudy because the larger particles in the dispersed phase scatter light. Colloids demonstrate the Tyndall effect in which the path of the light through a colloid is visible because the light is scattered.

35 TYPES OF COLLOIDS

36 STABILIZING COLLOIDS Substances known as emulsifying agents or stabilizing agents are used to prevent some colloids from coalescing (e.g. egg yolk in oil and water to form mayonnaise, soap/ detergent ions forming a charged layer around nonpolar oils and greases).

37 STABILIZING COLLOIDS (continued)

38 DIALYSIS Dialysis can be used to separate small particles from colloids (e.g. cleaning the blood of people suffering from kidney malfunction).

39 DIALYSIS A dialyzing membrane is a semipermeable membrane with larger pores than osmotic membranes that allow solvent molecules, other small molecules, and hydrated ions to pass through. Dialysis is a process in which solvent molecules, other small molecules, and hydrated ions pass from a solution through a membrane.


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