Chapter 7: Solutions and Colloids

1. Chapter 7:Solutions and Colloids

2. SOLUTIONS Solutions are homogeneous mixtures of two or more substances in which the components are present as atoms, molecules, or ions. Particles in liquid solutions are: too small to reflect light, thus solutions are transparent (clear). in constant motion. not settled by the influence of gravity. Some solutions are colored.

3. SOLVENT OF A SOLUTION The solvent of a solution is the substance present in the largest amount in the solution. SOLUTE OF A SOLUTION A solute of a solution is any substance present in an amount less than that of the solvent. A solution may contain more than one solute.

4. PHYSICAL STATES OF SOLUTIONS The physical state of a solution, solid, liquid, or gas, is usually the same as the physical state of the solvent.

5. 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.

6. SOLUBLE SUBSTANCE This is a term used to describe a substance that dissolves to a significant extent in a solvent without stating how much actually will dissolve. INSOLUBLE SUBSTANCE This is a term used to describe a substance that does not dissolve to a significant extent in a solvent. Sugar has limited solubility in water

7. IMMISCIBLE This is a term used to describe liquids that are not soluble in each other. The liquids don't mix together to form a solution. Click here to play Chemistry Interactive

8. SATURATED SOLUTION A saturated solution is a solution that contains the maximum amount possible of dissolved solute in a stable situation under the prevailing conditions of temperature and pressure. SUPERSATURATED SOLUTION An unstable solution that contains an amount of solute greater than the solute solubility under the prevailing conditions of temperature and pressure.

9. EXAMPLES OF SOLUTE SOLUBILITES AT 0°C

10. EXAMPLES OF SOLUTE SOLUBILITES AT 0°C cont.

11. EFFECT OF TEMPERATURE ON SOLUBILITY

12. 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: Click here to play Chemistry Interactive Click here to play Coached Problem

13. Click here to play Coached Problem An example of the solution process for a polar solute in water:

14. A solute will not dissolve in a solvent if: (1) the forces between solute particles are too strong to be overcome by interactions with solvent particles. (2) 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. Click here to play Chemistry Interactive

15. GENERAL SOLUBILITIES OF IONIC COMPOUNDS IN WATER Click here to play Coached Problem

16. 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.

17. HEAT AND SOLUTION FORMATION Endothermic: Exothermic:

18. 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.

19. 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.

20. EXAMPLES OF MOLARITY CALCULATIONS Example 1: A 250-mL sample of solution contains 0.134 moles of solute. Calculate the molarity of the solution. Solution: The solution sample volume in liters is 0.250 L. The number of moles of solute in the sample is 0.134 mol. These two quantities are substituted into the equation given earlier: Click here to play Coached Problem Click here to play Coached Problem

21. Example 2: 9.45 g of methyl alcohol, CH3OH, was dissolved in enough pure water to give 500 mL of solution. What was the molarity of the solution? Solution: The solution volume was 500 mL or 0.500L. The amount of solute was given in grams, but needs to be expressed in moles. The molecular weight of methyl alcohol is 32.0 u, so 1mol = 32.0 g. This fact can be used to convert the mass of methyl alcohol into moles: The number of moles of alcohol and the number of liters of solution can now be substituted into the equation for molarity:

22. 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.

23. WEIGHT/WEIGHT PERCENT Weight/weight percent, abbreviated %(w/w), is a concentration that expresses the mass of solute contained in 100 mass units of solution. Any mass units may be used, but the mass of solute and solution must be in the same units. The equation used for mathematical calculations is:

24. Calculation example: 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. Solution: The mass of the water used is 100 grams because according to the density each mL has a mass of 1.00 grams. The mass of solution is equal to the mass of water plus the mass of the sugar solute or 100 + 15.0 = 115 grams. The calculation is:

25. WEIGHT/VOLUME PERCENT Weight/volume percent, abbreviated %(w/v), is a concentration that expresses the number of grams of solute contained in 100 mL of solution. The solute mass must be expressed in grams, and the amount of solution must be expressed in mL. This percent concentration is normally used when the solute is a solid and the solvent and resulting solutions are liquids. The equation used for mathematical calculations is:

26. Calculation example: 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. Solution: The data are given in units that may be put directly into the equation. The calculation is:

27. VOLUME/VOLUME PERCENT Volume/volume percent, abbreviated %(v/v), is a concentration that expresses the volume of liquid solute contained in 100 volumes of solution. Any volume units may be used, but they must be the same for both the solute and the solution. The equation used for mathematical calculations is:

28. Calculation example: 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. Solution: The data give the solute volume and solution volume in different units. They must be the same. Either quantity could be changed, but the solute volume will be expressed in liters as 0.250 L. The calculation is:

29. 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.

30. 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. Click here to play Coached Problem Click here to play Coached Problem

31. 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: (Cc)(Vc) = (Cd)(Vd) In this equation, Cc is the concentration of the concentrated solution that is to be diluted, Vc is the volume of concentrated solution that is needed, Cd is the concentration of the dilute solution, and Vdis the volume of dilute solution.

32. 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, Cc = 1.50 M, Cd = 0.500 M, and Vd = 250 mL. The equation given above can be solved for Vc, 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. Click here to play Coached Problem Click here to play Coached Problem

33. 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. Click here to play Coached Problem

34. Calculation example: Consider the balanced equation HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) How many mL of 0.100 M HCl solution would exactly react with 25.00 mL of 0.125 M NaOH solution? Solution: The number of moles of NaOH in 25.00 mL of 0.125 M solution is calculated by multiplying the molarity and the solution volume in liters: moles of NaOH = 0.125 M x 0.02500 L = 0.00313 moles of NaOH According to the reaction, 1 mole of NaOH reacts with 1 mole of HCl, so 0.00313 moles of HCl will be needed.

35. The volume of 0.100 M HCl solution that contains 0.00313 can be calculated by dividing the needed number of moles by the solution molarity: Thus, it is seen that 0.0313 liters or 31.3 mL of 0.100 M HCl solution will be required to react with the 25.00 mL of 0.125 M NaOH solution. Click here to play Coached Problem Click here to play Coached Problem

36. 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.

37. 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. Click here to play Chemistry Interactive

38. 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.

39. SOLUTION BOILING POINT The boiling point of a solution is always higher than the boiling point of the pure solvent of the solution. The difference between the boiling point of the pure solvent and the solution is represented by the symbol ∆tb. The difference in boiling point between pure solvent and solution depends on the concentration of solute particles, and is calculated using the following equation: ∆tb = nKbM In this equation, ∆tb is the difference between the boiling point of the solution and the boiling point of the pure solvent.

40. The n in the equation is the number of moles of solute particles put into the solution when 1 mole of solute dissolves. For solutes that do not dissociate, n = 1. For solutes that are strong electrolytes, n varies depending on the number of ions formed when the solute dissolves. For example, the dissociation of sodium chloride is represented as NaCl Na+ + Cl-. Thus, when 1 mole of NaCl dissolves, 2 moles of particles (ions) are put into the solution and n = 2. Kb in the equation is a constant called the boiling point constant. Kb is characteristic of the solvent used to make the solution. M is the molarity concentration of solute in the solution. Click here to play Coached Problem

41. SOLUTION FREEZING POINT The freezing point of a solution is always lower than the freezing point of the pure solvent of the solution. The difference in freezing point of the solution and pure solvent is ∆tf, which is calculated by subtracting the freezing point of the solution from the freezing point of the pure solvent. The value of ∆tf is calculated by using the following equation: ∆tf = nKfM The symbols in this equation have meaning similar to those given for the boiling point elevation equation. Kf is a constant characteristic of the solvent used to form the solution. Click here to play Coached Problem

42. 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

43. In this equation, π is the osmotic pressure, n is the number 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. Click here to play Coached Problem Click here to play Chemistry Interactive

44. COLLOIDS Colloids are homogeneous mixtures of two or more components called the dispersing medium and the dispersed phase. The dispersed phase substances present in a colloid are present 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.

45. 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.

46. TYPES OF COLLOIDS

47. STABILIZING COLLOIDS Ions in a dispersing medium are attracted to colloid particles and stick on their surfaces. All colloid particles within a particular system will attract ions of only one charge or the other. The colloid particles all acquire the same charge and repel each other.

48. STABILIZING COLLOIDS The repulsion helps prevent the particles from coalescing into aggregates large enough to settle out. 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) Click here to play Chemistry Interactive

49. DESTRUCTION OF COLLOIDS Colloid solids are removed from gaseous smoke stack wastes before they are released into the atmosphere in the Cottrel precipitator. The precipitator contains a number of highly charged plates or electrodes. As smoke passes over the charged surfaces, the colloid particles lose their charges. The particles then coalesce into largerparticles that settle out and are collected for disposal. Dialysis can be used to separate small particles from colloids (e.g. cleaning the blood of people suffering from kidney malfunction).

50. 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.