1. Section 1 Types of Mixtures
Chapter 12
When you put sugar in water and stir it, why do the sugar crystals seem to disappear?

2. Section 1 Types of Mixtures
Chapter 12
Solutions
You know from experience that sugar dissolves in water. Sugar is described as “soluble in water.” By soluble we mean capable of being dissolved.
When sugar dissolves, all its molecules become uniformly distributed among the water molecules. The solid sugar is no longer visible.
Such a mixture is called a solution. A solution is a homogeneous (uniform throughout) mixture of two or more substances.

3. Chapter 12 Solutions, continued
Section 1 Types of Mixtures
Chapter 12
Solutions, continued
The dissolving medium in a solution is called the solvent, and the substance dissolved in a solution is called the solute.
Solutions may exist as gases, liquids, or solids. There are many possible solute-solvent combinations between gases, liquids, and solids.
example: Alloys are solid solutions in which the atoms of two or more metals are uniformly mixed.
Brass is made from zinc and copper.
Sterling silver is made from silver and copper.

4. Solutes, Solvents, and Solutions
Visual Concepts
Chapter 12
Solutes, Solvents, and Solutions
Click below to watch the Visual Concept.
Visual Concept

5. Solutes, Solvents, and Solutions
Visual Concepts
Chapter 12
Solutes, Solvents, and Solutions
In a solution of sugar and water, what is the solvent?
What is the solute?
Visual Concept

6. Visual Concepts
Chapter 12
Types of Solutions

7. Particle Models for Gold and Gold Alloy
Section 1 Types of Mixtures
Chapter 12
Particle Models for Gold and Gold Alloy

8. Section 1 Types of Mixtures
Chapter 12
Suspensions
If the particles in a solvent are so large that they settle out unless the mixture is constantly stirred or agitated, the mixture is called a suspension.
For example, a jar of muddy water consists of soil particles suspended in water. The soil particles will eventually all collect on the bottom of the jar, because the soil particles are denser than the solvent, water.
Particles over 1000 nm in diameter—1000 times as large as atoms, molecules or ions—form suspensions.

9. Chapter 12 Suspensions Visual Concepts -Demonstration
-Particles in a suspension can be removed by filtration.
Visual Concept

10. Section 1 Types of Mixtures
Chapter 12
Colloids
Particles that are intermediate in size between those in solutions and suspensions form mixtures known as colloidal dispersions, or simply colloids.
The particles in a colloid are small enough to be suspended throughout the solvent by the constant movement of the surrounding molecules.
example: Mayonnaise is a colloid.
It is an emulsion of oil droplets in water.

11. Chapter 12 Colloids, continued Tyndall Effect
Section 1 Types of Mixtures
Chapter 12
Colloids, continued
Tyndall Effect
Many colloids look similar to solutions because their particles cannot be seen.
The Tyndall effect occurs when light is scattered by colloidal particles dispersed in a transparent medium.
example: a headlight beam is visible from the side on a foggy night
The Tyndall effect can be used to distinguish between a solution and a colloid.

12. Visual Concepts
Chapter 12
Colloids

13. Chapter 12
Emulsions- a type of colloid; two liquids that don’t mix together (ex: oil and water)

14. Properties of Solutions, Colloids, and Suspensions
Section 1 Types of Mixtures
Chapter 12
Properties of Solutions, Colloids, and Suspensions

15. Section 2 The Solution Process
Chapter 12
Solubility
If you add spoonful after spoonful of sugar to tea, eventually no more sugar will dissolve.
This illustrates the fact that for every combination of solvent with a solid solute at a given temperature, there is a limit to the amount of solid that can be dissolved.
The point at which this limit is reached for any solute-solvent combination depends on the nature of the solute, the nature of the solvent, and the temperature.

16. Solubility of Common Compounds
Section 2 The Solution Process
Chapter 12
Solubility of Common Compounds

17. Solubility of Common Compounds
Section 2 The Solution Process
Chapter 12
Solubility of Common Compounds

18. Chapter 12 Concentration
Section 3 Concentration of Solutions
Chapter 12
Concentration
The concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution.
Concentration is a ratio: any amount of a given solution has the same concentration.
The opposite of concentrated is dilute.
These terms are unrelated to the degree to which a solution is saturated: a saturated solution of a solute that is not very soluble might be very dilute.

19. Section 3 Concentration of Solutions
Chapter 12
Concentration Units

20. Chapter 12 Concentration Visual Concepts
Click below to watch the Visual Concept.
Visual Concept

21. Section 3 Concentration of Solutions
Chapter 12
Molarity
Molarity is the number of moles of solute in one liter of solution.
For example, a “one molar” solution of sodium hydroxide contains one mole of NaOH in every liter of solution.
The symbol for molarity is M. The concentration of a one molar NaOH solution is written 1 M NaOH.

22. Chapter 12 Molarity, continued
Section 3 Concentration of Solutions
Chapter 12
Molarity, continued
To calculate molarity, you must know the amount of solute in moles and the volume of solution in liters.
When weighing out the solute, this means you will need to know the molar mass of the solute in order to convert mass to moles.
example: One mole of NaOH has a mass of g. If this quantity of NaOH is dissolved in enough water to make 1.00 L of solution, it is a 1.00 M solution.

23. Chapter 12 Molarity, continued
Section 3 Concentration of Solutions
Chapter 12
Molarity, continued
The molarity of any solution can be calculated by dividing the number of moles of solute by the number of liters of solution:
Note that a 1 M solution is not made by adding 1 mol of solute to 1 L of solvent. In such a case, the final total volume of the solution might not be 1 L.
Solvent must be added carefully while dissolving to ensure a final volume of 1 L.

24. Preparation of a Solution Based on Molarity
Visual Concepts
Chapter 12
Preparation of a Solution Based on Molarity
Click below to watch the Visual Concept.
Visual Concept

25. Chapter 12 Molarity, continued Sample Problem A
Section 3 Concentration of Solutions
Chapter 12
Molarity, continued
Sample Problem A
You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?

26. Chapter 12 Molarity, continued Sample Problem A Solution
Section 3 Concentration of Solutions
Chapter 12
Molarity, continued
Sample Problem A Solution
Given: solute mass = 90.0 g NaCl
solution volume = 3.50 L
Unknown: molarity of NaCl solution
Solution:

27. Chapter 12 Molarity, continued Sample Problem B
Section 3 Concentration of Solutions
Chapter 12
Molarity, continued
Sample Problem B
You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

28. Chapter 12 Molarity, continued Sample Problem B Solution
Section 3 Concentration of Solutions
Chapter 12
Molarity, continued
Sample Problem B Solution
Given: volume of solution = 0.8 L
concentration of solution = 0.5 M HCl
Unknown: moles of HCl in a given volume
Solution:

29. Factors Affecting the Rate of Dissolution
Section 2 The Solution Process
Chapter 12
Factors Affecting the Rate of Dissolution
Because the dissolution process occurs at the surface of the solute, it can be speeded up if the surface area of the solute is increased.
Stirring or shaking helps to disperse solute particles and increase contact between the solvent and solute surface. This speeds up the dissolving process.
At higher temperatures, collisions between solvent molecules and solvent are more frequent and of higher energy. This helps to disperse solute molecules among the solvent molecules, and speed up the dissolving process.

30. Factors Affecting the Rate of Dissolution
Visual Concepts
Chapter 12
Factors Affecting the Rate of Dissolution
Click below to watch the Visual Concept.
Visual Concept

31. Section 2 The Solution Process
Chapter 12
Solubility
If you add spoonful after spoonful of sugar to tea, eventually no more sugar will dissolve.
This illustrates the fact that for every combination of solvent with a solid solute at a given temperature, there is a limit to the amount of solid that can be dissolved.
The point at which this limit is reached for any solute-solvent combination depends on the nature of the solute, the nature of the solvent, and the temperature.

32. Particle Model for Soluble and Insoluble Substances
Section 2 The Solution Process
Chapter 12
Particle Model for Soluble and Insoluble Substances

33. Particle Model for Soluble and Insoluble Substances
Section 2 The Solution Process
Chapter 12
Particle Model for Soluble and Insoluble Substances

34. Chapter 12 Solubility, continued
Section 2 The Solution Process
Chapter 12
Solubility, continued
When a solute is first added to a solvent, solute molecules leave the solid surface and move about at random in the solvent.
As more solute is added, more collisions occur between dissolved solute particles. Some of the solute molecules return to the crystal.
When maximum solubility is reached, molecules are returning to the solid form at the same rate at which they are going into solution.

35. Chapter 12 Solubility, continued
Section 2 The Solution Process
Chapter 12
Solubility, continued
Solution equilibrium is the physical state in which the opposing processes of dissolution and crystallization of a solute occur at the same rates.

36. Chapter 12 Solubility, continued
Section 2 The Solution Process
Chapter 12
Solubility, continued
Saturated Versus Unsaturated Solutions
A solution that contains the maximum amount of dissolved solute is described as a saturated solution.
If more solute is added to a saturated solution, it falls to the bottom of the container and does not dissolve.
This is because an equilibrium has been established between ions leaving and entering the solid phase.
A solution that contains less solute than a saturated solution under the same conditions is an unsaturated solution.

37. Mass of Solute Added Versus Mass of Solute Dissolved
Section 2 The Solution Process
Chapter 12
Mass of Solute Added Versus Mass of Solute Dissolved

38. Chapter 12 Solubility, continued Supersaturated Solutions
Section 2 The Solution Process
Chapter 12
Solubility, continued
Supersaturated Solutions
When a saturated solution is cooled, the excess solute usually comes out of solution, leaving the solution saturated at the lower temperature.
But sometimes the excess solute does not separate, and a supersaturated solution is produced, which is a solution that contains more dissolved solute than a saturated solution contains under the same conditions.
A supersaturated solution will form crystals of solute if disturbed or more solute is added.

39. Chapter 12 Solubility, continued Solubility Values
Section 2 The Solution Process
Chapter 12
Solubility, continued
Solubility Values
The solubility of a substance is the amount of that substance required to form a saturated solution with a specific amount of solvent at a specified temperature.
example: The solubility of sugar is 204 g per 100 g of water at 20°C.
Solubilities vary widely, and must be determined experimentally.
They can be found in chemical handbooks and are usually given as grams of solute per 100 g of solvent at a given temperature.

40. Solubility of Common Compounds
Section 2 The Solution Process
Chapter 12
Solubility of Common Compounds

41. Solubility of Common Compounds
Section 2 The Solution Process
Chapter 12
Solubility of Common Compounds

42. Solutes: Electrolytes Versus Nonelectrolytes
Section 1 Types of Mixtures
Chapter 12
Solutes: Electrolytes Versus Nonelectrolytes
A substance that dissolves in water to give a solution that conducts electric current is called an electrolyte.
Any soluble ionic compound, such as sodium chloride, NaCl, is an electrolyte.
The positive and negative ions separate from each other in solution and are free to move, making it possible for an electric current to pass through the solution.

43. Solutes: Electrolytes Versus Nonelectrolytes, continued
Section 1 Types of Mixtures
Chapter 12
Solutes: Electrolytes Versus Nonelectrolytes, continued
A substance that dissolves in water to give a solution that does not conduct electric current is called a nonelectrolyte.
Sugar is an example of a nonelectrolyte.
Neutral solute molecules do not contain mobile charged particles, so a solution of a nonelectrolyte cannot conduct electric current.

44. HCl contains a highly polar bond.
Section 1 Compounds in Aqueous Solution
Chapter 13
Ionization
Ions are formed from solute molecules by the action of the solvent in a process called ionization.
When a molecular compound dissolves and ionizes in a polar solvent, ions are formed where none existed in the undissolved compound.
Hydrogen chloride, HCl, is a molecular compound that ionizes in aqueous solution.
HCl contains a highly polar bond. -

45. Section 1 Compounds in Aqueous Solution
Chapter 13
Dissociation
Dissociation is separation of ions that occurs when an ionic compound dissolves. -
1 mol 1 mol 1 mol -
1 mol 1 mol 2 mol

46. Dissociation of NaCl Section 1 Compounds in Aqueous Solution
Chapter 13
Dissociation of NaCl

47. Dissociation, continued
Section 1 Compounds in Aqueous Solution
Chapter 13
Dissociation, continued
Sample Problem A
Write the equation for the dissolution of aluminum sulfate, Al2(SO4)3 , in water. How many moles of aluminum ions and sulfate ions are produced by dissolving 1 mol of aluminum sulfate? What is the total number of moles of ions produced by dissolving 1 mol of aluminum sulfate?

48. Dissociation, continued
Section 1 Compounds in Aqueous Solution
Chapter 13
Dissociation, continued
Sample Problem A Solution
Given: amount of solute = 1 mol Al2(SO4)3
solvent identity = water
Unknown: a. moles of aluminum ions and sulfate ions
b. total number of moles of solute ions produced (i)
Solution:

49. Soluble and Insoluble Ionic Compounds
Section 1 Compounds in Aqueous Solution
Chapter 13
Soluble and Insoluble Ionic Compounds

50. Dissociation, continued Net Ionic Equations, continued
Section 1 Compounds in Aqueous Solution
Chapter 13
Dissociation, continued Net Ionic Equations, continued
Sample Problem B
Identify the precipitate that forms when aqueous solutions of zinc nitrate and ammonium sulfide are combined. Write the equation for the possible double-displacement reaction. Then write the insoluble products as compounds that precipitate and soluble products as separate “spectator” ions.

51. Dissociation, continued Net Ionic Equations, continued
Section 1 Compounds in Aqueous Solution
Chapter 13
Dissociation, continued Net Ionic Equations, continued
Sample Problem B Solution, continued
Solution:
a. equation for the possible double-displacement reaction
b. The solubility chart reveals that zinc sulfide is not a soluble sulfide and is therefore a precipitate. Ammonium nitrate is soluble according to the table.
c. Zn(NO3)2 (aq) + (NH4)2S (aq)  ZnS (s) + 2NH NO3 -1

52. Section 3 Concentration of Solutions
Chapter 12
Molality
Molality is the concentration of a solution expressed in moles of solute per kilogram of solvent.
A solution that contains 1 mol of solute dissolved in 1 kg of solvent is a “one molal” solution.
The symbol for molality is m, and the concentration of this solution is written as 1 m NaOH.

53. Chapter 12 Molality, continued
Section 3 Concentration of Solutions
Chapter 12
Molality, continued
The molality of any solution can be calculated by dividing the number of moles of solute by the number of kilograms of solvent:
Unlike molarity, which is a ratio of which the denominator is liters of solution, molality is per kilograms of solvent.
Molality is used when studying properties of solutions related to vapor pressure and temperature changes, because molality does not change with temperature.

54. Making a Molal Solution
Section 3 Concentration of Solutions
Chapter 12
Making a Molal Solution

55. Chapter 12 Molality, continued Sample Problem D
Section 3 Concentration of Solutions
Chapter 12
Molality, continued
Sample Problem D
A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.

56. Chapter 12 Molality, continued Sample Problem D Solution
Section 3 Concentration of Solutions
Chapter 12
Molality, continued
Sample Problem D Solution
Given: solute mass = 17.1 C12H22O11
solvent mass = 125 g H2O
Unknown: molal concentration
Solution: First, convert grams of solute to moles and grams of solvent to kilograms.

57. Chapter 12 Molality, continued Sample Problem D Solution, continued
Section 3 Concentration of Solutions
Chapter 12
Molality, continued
Sample Problem D Solution, continued
Then, divide moles of solute by kilograms of solvent.

58. Chapter 12 Molality, continued Sample Problem E
Section 3 Concentration of Solutions
Chapter 12
Molality, continued
Sample Problem E
A solution of iodine, I2, in carbon tetrachloride, CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a m solution of iodine in CCl4 if g of CCl4 is used?

59. Chapter 12 Molality, continued Sample Problem E Solution
Section 3 Concentration of Solutions
Chapter 12
Molality, continued
Sample Problem E Solution
Given: molality of solution = m I2
mass of solvent = g CCl4
Unknown: mass of solute
Solution: First, convert grams of solvent to kilograms.

60. Chapter 12 Molality, continued Sample Problem E Solution, continued
Section 3 Concentration of Solutions
Chapter 12
Molality, continued
Sample Problem E Solution, continued
Solution, continued: Then, use the equation for molality to solve for moles of solute.
Finally, convert moles of solute to grams of solute.

61. Freezing-Point Depression Boiling-Point Elevation
Section 2 Colligative Properties of Solutions
Chapter 13
Colligative Properties of Solutions
Properties that depend on the concentration of solute particles but not on their identity are called colligative properties.
Freezing-Point Depression
Boiling-Point Elevation

62. The freezing-point depression, ∆Tf , is the difference between the freezing points of the pure solvent and a solution of a nonelectrolyte in that solvent, and it is directly proportional to the molal concentration of the solution. ∆Tf = Kf * m * i The boiling-point elevation, ∆Tb, is the difference between the boiling points of the pure solvent and a nonelectrolyte solution of that solvent, and it is directly proportional to the molal concentration of the solution. ∆Tb= Kb * m * i Kf and Kb are constants; m is molality, i = moles of ions formed per mole of compound dissolved

63. Freezing-Point Depression, continued
Section 2 Colligative Properties of Solutions
Chapter 13
Freezing-Point Depression, continued
Sample Problem C
Problem: What is the freezing-point depression of water in a solution of 17.1 g of sucrose, C12H22O11, in 200. g of water? What is the actual freezing point of the solution? Sucrose is a nonelectrolyte so i = 1
Given: solute g C12H22O11
solvent 200 g water
Unknown: a. freezing-point depression
b. freezing point of the solution

64. Freezing-Point Depression, continued
Section 2 Colligative Properties of Solutions
Chapter 13
Freezing-Point Depression, continued
Sample Problem C Solution, continued
Solution:

65. Freezing-Point Depression, continued
Section 2 Colligative Properties of Solutions
Chapter 13
Freezing-Point Depression, continued
Solution:
∆Tf = Kf * m * i
= °C/m * m * 1
= °C
New freezing point = 0°C + ( °C) = °C

66. mol of solute particles (i) 1
Section 2 Colligative Properties of Solutions
Chapter 13
Electrolytes and Colligative Properties
Electrolytes depress the freezing point and elevate the boiling point of a solvent more than expected.
The differences between the expected and calculated values are caused by the attractive forces that exist between dissociated ions in aqueous solution.
Electrolytes produce more than 1 mol of solute particles for each mole of compound dissolved.
mol of solute particles (i) 3 2

67. Boiling Point Elevation Example with Electrolyte Solute