Chapter 8 Solutions

1
Chapter 8 Solutions
2
Solutions Solute and Solvent

• Solutions
• Are homogeneous mixtures of two or more
  substances.
• Consist of a solvent and one or more solutes.

3
Nature of Solutes in Solutions

• Solutes
• Spread evenly throughout the solution.
• Cannot be separated by filtration.
• Can be separated by evaporation.
• Are not visible, but can give a color to the
  solution.

4
Examples of Solutions

• The solute and solvent in a solution can be a
  solid, liquid, and/or a gas.

5
Learning Check

• Identify the solute and the solvent in each.
• A. brass 20 g zinc 50 g copper
• solute 1) zinc 2) copper
• solvent 1) zinc 2) copper
• B. 100 g H2O 5 g KCl
• solute 1) KCl 2) H2O
• solvent 1) KCl 2) H2O

6
Solution

• Identify the solute and the solvent in each.
• A. brass 20 g zinc 50 g copper
• solute 1) zinc
• solvent 2) copper
• B. 100 g H2O 5 g KCl
• solute 1) KCl
• solvent 2) H2O

7
Water

• Water is the most common solvent.
• The water molecule is polar.
• Hydrogen bonds form between the hydrogen atom in
  one molecule and the oxygen atom in a different
  water molecule.

8
Water

9
Like Dissolves Like

• A solution forms when there is an attraction
  between the particles of the solute and solvent.
• A polar solvent such as water dissolves polar
  solutes such as sugar and ionic solutes such as
  NaCl.
• A nonpolar solvent such as hexane (C6H14)
  dissolves nonpolar solutes such as oil or grease.

10
Examples of Like Dissolves Like

• Solvents Solutes
• Water (polar) Ni(NO3)2
• (ionic)
• CH2Cl2 (nonpolar)
• I2 (nonpolar)

11
Learning Check

• Which of the following solutes will dissolve in
  water? Why?
• 1) Na2SO4
• 2) gasoline (nonpolar)
• 3) I2
• 4) HCl

12
Solution

• Which of the following solutes will dissolve in
  water? Why?
• 1) Na2SO4 Yes, ionic
• 2) gasoline No, nonnpolar
• 3) I2 No, nonpolar
• 4) HCl Yes, polar
• Most polar and ionic solutes dissolve in water
  because water is a polar solvent.

13
Formation of a Solution

• Na and Cl- ions on the surface of a NaCl crystal
  are attracted to polar water molecules.
• In solution, the ions are hydrated as several H2O
  molecules surround each.

14
Equations for Solution Formation

• When NaCl(s) dissolves in water, the reaction can
  be written as
• H2O
• NaCl(s) Na(aq) Cl- (aq)
• solid separation of
  ions

15
Learning Check

• Solid LiCl is added to water. It dissolves
  because
• A. The Li ions are attracted to the
• 1) oxygen atom (? -) of water.
• 2) hydrogen atom (?) of water.
• B. The Cl- ions are attracted to the
• 1) oxygen atom (? -) of water.
• 2) hydrogen atom (?) of water.

16
Solution

• Solid LiCl is added to water. It dissolves
  because
• A. The Li ions are attracted to the
• 1) oxygen atom (? -) of water.
• B. The Cl- ions are attracted to the
• 2) hydrogen atom (? ) of water.

17
Electrolytes

• Electrolytes
• Produce positive () and negative (-) ions when
  they dissolve in water.
• In water conduct an electric current.

18
Strong Electrolytes

• Strong electrolytes ionize 100 in solution.
• Equations for the dissociation of strong
  electrolytes show the formation of ions in
  aqueous (aq) solutions.
• H2O 100 ions
• NaCl(s) Na(aq) Cl-(aq)
  H2O
• CaBr2(s) Ca2(aq) 2Br- (aq)

19
Learning Check

• Complete each of the following dissociation
  equations for strong electrolytes dissolved in
  water H2O
• A. CaCl2 (s) 1) CaCl2
• 2) Ca2 Cl2-
• 3) Ca2 2Cl-
• H2O
• B. K3PO4 (s) 1) 3K PO43-
• 2) K3PO4
• 3) K3 P3- O4-

20
Solution

• Complete each of the following dissociation
  equations for strong electrolytes dissolved in
  water
• H2O
• A. 3) CaCl2 (s) Ca2 2Cl-
• H2O
• B. 1) K3PO4 (s) 3K PO43-

21
Weak Electrolytes

• A weak electrolyte
• Dissolves mostly as molecules in solution.
• Produces only a few ions in aqueous solutions.
• Has an equilibrium that favors the reactants.
• HF H2O H3O(aq) F- (aq)
• NH3 H2O NH4(aq) OH- (aq)

22
Nonelectrolytes

• Nonelectrolytes
• Form only molecules in water.
• Do not produce ions in water.
• Do not conduct an electric current.

23
Equivalents

• An equivalent (Eq) is the amount of an ion that
  provides 1 mole of electrical charge ( or -).

24
Electrolytes inBody Fluids

• In replacement solutions for body fluids, the
  electrolyte amounts are given in milliequivalents
  per liter (mEq/L).

25
Examples of IV Solutions

• In intravenous solutions, the total mEq of
  positively charged ions is equal to the total mEq
  of negatively charged ions.

26
Learning Check

• A. In 1 mole of Fe3, there are
• 1) 1 Eq 2) 2 Eq 3) 3 Eq
• B. 0.5 equivalents of calcium is
• 1) 5 mEq 2) 50 mEq 3) 500 mEq
• C. If the Na in a NaCl IV solution is 34 mEq/L,
• the Cl- is
• 1) 34 mEq/L 2) 0 mEq/L 3) 68 mEq/L

27
Solution

• A. In 1 mole of Fe3, there are
• 3) 3 Eq
• B. 0.5 equivalents of calcium is
• 3) 500 mEq
• C. If the Na in a NaCl IV solution is 34
  mEq/L,
• the Cl- is
• 1) 34 mEq/L

28
Solubility

• Solubility states the maximum amount of solute
  that dissolves in a specific amount of solvent at
  a particular temperature.
• Typically, solubility is expressed as the grams
  of solute that dissolves in 100 g of solvent,
  usually water.
• g of solute
• 100 g water

29
Saturated Solutions

• A saturated solution
• Contains the maximum amount of solute that can
  dissolve.
• Has some undissolved solute at the bottom of the
  container.

30
Unsaturated Solutions

• An unsaturated solution
• Contains less than the maximum amount of solute.
• Can dissolve more solute.

31
Learning Check

• At 40?C, the solubility of KBr is 80 g/100 g H2O.
• Identify the following solutions as either
• (1) saturated or (2) unsaturated. Explain.
• A. 60 g KBr added to 100 g of water at 40?C.
• B. 200 g KBr added to 200 g of water at 40?C.
• C. 25 g KBr added to 50 g of water at 40?C.

32
Solution

• A. 2 Amount is less than the solubility.
• B. 1 In 100 g of water, 100 g KBr exceeds
  the
• solubility at 40?C.
• C. 2 This would be 50 g KBr in 100 g of
  water,
• which is less than the solubility at 40?C.

33
Solubility of Solids Changes with Temperature

• The solubility of most solids increases with an
  increase in temperature.

34
Solubility of Gases and Temperature

• The solubility of gases decreases with an
  increase in temperature.

35
Learning Check

• A. Why could a bottle of carbonated drink
  possibly burst (explode) when it is left out in
  the hot sun?
• B. Why do fish die in water that is too warm?

36
Solution

• A. The pressure in a bottle increases as the gas
  leaves solution as it becomes less soluble at
  high temperatures. As pressure increases, the
  bottle could burst.
• B. Because O2 gas is less soluble in warm water,
  fish cannot obtain the amount of O2 required for
  their survival.

37
Henrys Law

• According to Henrys Law, the solubility of a gas
  in a liquid is directly related to the pressure
  of that gas above the liquid.

38
Next Time

• We will continue with Chapter 8

39
Soluble and Insoluble Salts

• A soluble salt is an ionic compound that
  dissolves in water.
• An insoluble salt is an ionic compound that does
  not dissolve in water.

40
Solubility Rules

• A soluble salt dissolves in water.
• Insoluble salts do not dissolve in water.

41
Using the Solubility Rules

• The solubility rules predict whether a salt is
  soluble or insoluble in water.

42
Learning Check

• Indicate if each salt is (1) soluble or (2)
  insoluble.
• A. ______ Na2SO4
• B. ______ MgCO3
• C. ______ PbCl2
• D. ______ MgCl2

43
Solution

• Indicate if each salt is 1) soluble or 2)
  insoluble.
• A. 1 Na2SO4
• B. 2 MgCO3
• C. 2 PbCl2
• D. 1 MgCl2

44
Formation of a Solid

• When solutions of salts are mixed, a solid forms
  when ions of an insoluble salt combine.

45
Learning Check

• The formula of an insoluble salt in each mixture
  is
• A. BaCl2 Na2SO4
• 1) BaSO4 2) NaCl 3) Na2Cl2 4) none
• B. AgNO3 KCl
• 1) KNO3 2) AgK 3) AgCl 4) none
• C. KNO3 NaCl
• 1) KCl 2) NaNO3 3) ClNO3 4) none

46
Solution

• A. BaCl2 Na2SO4
• 1) BaSO4
• B. AgNO3 KCl
• 3) AgCl
• C. KNO3 NaCl
• 4) none all combinations are soluble.

47
Percent Concentration

• The concentration of a solution is the amount of
  solute dissolved in a specific amount of
  solution.
• amount of solute
• amount of solution
• The percent concentration describes the amount
  of solute that is dissolved in 100 parts of
  solution.
• amount of solute
• 100 parts solution

48
Mass Percent

• The mass percent (m/m)
• Concentration is the percent by mass of solute
  in a solution.
• mass percent g of solute x 100
  g of solution
• Is the g of solute in 100 g of solution.
• mass percent g of solute
  100 g of solution

49
Mass of Solution

• grams of solute grams of solvent
• 50.0 g KCl solution

50
Calculating Mass Percent

• Mass percent (m/m) is calculated from the grams
  of solute (g KCl) and the grams of solution (g
  KCl solution).
• g of KCl 8.0 g
• g of solvent (water) 42.0 g
• g of KCl solution 50.0 g
• 8.0 g KCl (solute) x 100 16 (m/m) KCl
• 50.0 g KCl solution

51
Learning Check

• A solution is prepared by mixing 15 g Na2CO3
  and 235 g of H2O. Calculate the mass percent
  (m/m) of the solution.
• 1) 15 (m/m) Na2CO3
• 2) 6.4 (m/m) Na2CO3
• 3) 6.0 (m/m) Na2CO3

52
Solution

• 3) 6.0 (m/m) Na2CO3
• mass solute 15 g Na2CO3
• mass solution 15 g 235 g 250 g
• mass (m/m) 15 g Na2CO3 x 100
  250 g solution
• 6.0 Na2CO3 solution

53
Mass/Volume Percent

• The mass/volume percent (m/v)
• Concentration is the ratio of the mass in grams
  (g) of solute in a volume (mL) of solution.
• mass/volume g of solute x 100
  mL of solution
• Is the g of solute in 100 mL of solution.
• mass/volume g of solute
  100 mL
  of solution

54
Preparing a Solution with a Mass/Volume
Concentration

• A percent mass/volume solution is prepared by
  weighing out the grams of solute (g) and adding
  water to give the final volume of the solution.

55
Calculation of Mass/Volume Percent

• Mass/volume percent (m/v) is calculated from
  the grams of solute (g KCl) and the volume of
  solution (mL KCl solution).
• g of KI 5.0 g KI
• mL of KI solution 250.0 mL
• 5.0 g KI (solute) x 100 2.0(m/v)
  KI
• 250.0 mL KI solution

56
Learning Check

• A 500. mL samples of an IV glucose solution
  contains 25 g glucose (C6H12O6) in water.
• What is the mass/volume (m/v) of glucose of
  the IV solution?1) 5.0 2) 20. 3) 50.

57
Solution

• 1) 5.0
• Mass/volume (m/v)
• 25 g glucose x 100
• 500. mL solution
• 5.0 (m/v) glucose solution

58
Volume Percent

• The volume percent (v/v)
• Concentration is the percent volume (mL) of
  solute (liquid) to volume (mL) of solution.
• volume (v/v) mL of solute x
  100 mL of solution
• Is the mL of solute in 100 mL of solution.
• volume (v/v) mL of solute
  100 mL
  of solution

59
Percent Conversion Factors

• Two conversion factors can be written for any
  type of value.

60
Learning Check

• Write two conversion factors for each
  solutions
• A. 8(m/v) NaOH
• B. 12(v/v) ethyl alcohol

61
Solution

• A. 8(m/v) NaOH
• 8 g NaOH and 100 mL solution
  
• 100 mL solution 8 g NaOH
• B. 12(v/v) ethyl alcohol
• 12 mL alcohol and 100 mL solution
• 100 mL solution 12 mL alcohol

62
Using Percent Factors

• How many grams of NaCl are needed to prepare
• 250 g of a 10.0 (m/m) NaCl solution?
• 1. Write the 10.0 (m/m) as conversion factors.
• 10.0 g NaCl and 100 g solution
  
• 100 g solution 10.0 g NaCl
• 2. Use the factor that cancels given (g
  solution).
• 250 g solution x 10.0 g NaCl 25 g
  NaCl 100 g solution

63
Learning Check

• How many grams of NaOH are needed to
• prepare 2.0 L of a 12(m/v) NaOH solution?
• 1) 24 g NaOH
• 2) 240 g NaOH
• 3) 2400 g NaOH

64
Solution

• 2) 240 g NaOH
• 2.0 L x 1000 mL 2000 mL
• 1 L
• 2000 mL x 12 g NaOH 240 g NaOH
• 100 mL
• 12 (m/v) factor

65
Learning Check

• How many milliliters of 5 (m/v) glucose
  solution are given if a patient receives 150 g of
  glucose?
• 1) 30 mL
• 2) 3000 mL
• 3) 7500 mL

66
Solution

• 2) 3000 mL
• 150 g glucose x 100 mL 3000 mL
• 5 g glucose
• 5 m/v factor (inverted)

67
Molarity (M)

• Molarity is a concentration unit for the moles of
  solute in the liters (L) of solution.
• Molarity (M) moles of solute moles
• liter of solution
  L
• Examples
• 2.0 M HCl 2.0 moles HCl
• 1 L
• 6.0 M HCl 6.0 moles HCl
• 1 L

68
Preparing a 1.0 Molar Solution

• A 1.0 M NaCl solution is prepared by weighing out
  58.5 g NaCl ( 1.0 mole) and adding water to make
  1.0 liter of solution.

69
Calculation of Molarity

• What is the molarity of a NaOH solution prepared
  by adding 4.0 g of solid NaOH to water to make
  0.50 L of solution ?
• 1. Determine the moles of solute.
• 4.0 g NaOH x 1 mole NaOH 0.10 mole
• 40.0 g NaOH
• 2. Calculate molarity.
• 0.10 mole 0.20 mole 0.20 M NaOH
• 0.50 L 1 L

70
Learning Check

• Calculate the molarity of an NaHCO3 solution
  prepared by dissolving 36 g of solid NaHCO3 in
  water to give a solution volume of 240 mL.
• 1) 0.43 M
• 2) 1.8 M
• 3) 15 M

71
Solution

• 2) 1.8 M
• 36 g x 1 mole NaHCO3 0.43 mole NaHCO3
• 84 g
• 0.43 mole NaHCO3 1.8 M NaHCO3
• 0.240 L

72
Learning Check

• A glucose solution with a volume of 2.0 L
  contains 72 g glucose (C6H12O6). If glucose has
  a molar mass of 180. g/mole, what is the molarity
  of the glucose solution?
• 1) 0.20 M
• 2) 5.0 M
• 3) 36 M

73
Solution

• 1) 0.20 M
• 72 g x 1 mole x 1 0.20
  moles
• 180. g 2.0 L 1 L
• 0.20 M

74
Molarity Conversion Factors

• The units in molarity can be used to write
  conversion factors.

75
Learning Check

• Stomach acid is 0.10 M HCl solution. How
  many moles of HCl are present in 1500 mL of
  stomach acid?
• 1) 15 moles HCl
• 2) 1.5 moles HCl
• 3) 0.15 mole HCl

76
Solution

• 3) 0.15 mole HCl
• 1500 mL x 1 L 1.5 L
• 1000 mL
• 1.5 L x 0.10 mole HCl 0.15 mole HCl
• 1 L
• Molarity factor

77
Learning Check

• Calculate the grams of KCl that must be
  dissolved in water to prepare 0.25 L of a 2.0 M
  KCl solution.
• 1) 150 g KCl
• 2) 37 g KCl
• 3) 19 g KCl

78
Solution

• 3) 37 g KCl
• Determine the number of moles of KCl.
• 0.25 L x 2.0 mole KCl 0.50 moles KCl
  1 L
• Convert the moles to grams of KCl.
• 0.50 moles KCl x 74.6 g KCl 37 g KCl
• 1 mole KCl
• molar mass of KCl

79
Learning Check

• How many milliliters of 6.0 M HNO3 contain
• 0.15 mole of HNO3?
• 1) 25 mL
• 2) 90 mL
• 3) 400 mL

80
Solution

• 1) 25 mL
• 0.15 mole HNO3 x 1 L x
  1000 mL
• 6.0 moles HNO3 1 L
• Molarity factor inverted
• 25 mL HNO3

81
Next Time

• We complete Chapter 8
• Review for Exam 3

82
Solutions

• Solutions are mixtures that
• Contain small solute particles (ions or
  molecules).
• Are transparent.
• Cannot be separated by filters.

83
Colloids

• Colloidal dispersions are
• mixtures that
• Contain medium-sized particles called colloids.
• Cannot be separated by filters.
• Are separated by semipermeable membranes.
• Scatter light (Tyndall effect).

84
Tyndall Effect

• A beam of light going through a colloid is
  visible because the light is scattered by the
  large solute particles.
• The Tyndall effect does not occur with solutions.

85
Examples of Colloids
86
Suspensions

• Suspensions are mixtures that
• Contain very large particles that are visible.
• Settle out rapidly.
• Are separated by filters.

87
Comparing the Properties of Solutions, Colloids,
and Suspensions

• (a) Suspensions settle.
• (b) Filters separate suspensions, but not
  solutions or colloids.
• (c) Only solution particles go through
  semipermeable membranes.

88
Comparing Solutions, Colloids, and Suspensions
89
Learning Check

• A mixture with solute particles that do not
  settle, but are too large to pass through a
  semipermeable membrane is called a
• 1) solution
• 2) colloid
• 3) suspension

90
Solution

• A mixture with solute particles that do not
  settle, but are too large to pass through a
  semipermeable membrane is called a
• 2) colloid

91
Osmosis

• In osmosis,
• Water moves through a semipermeable membrane that
  separates two solutions with different
  concentrations.
• Water flows out of the solution with the lower
  solute concentration and into the solution with
  the higher solute concentration.
• The concentrations of the two solutions become
  equal.

92
Osmosis

• As water flows into the sucrose solution, the
  volume of the sucrose solution increases.
• The concentration of the sucrose solution
  decreases.

93
Osmosis

• During osmosis, water flows across the
  semipermeable membrane from the 4 starch
  solution into the 10 solution.
• semipermeable
• membrane

4 starch
10 starch
H2O
94
Equilibrium

• Eventually, the flow of water across the
  semipermeable membrane becomes equal in both
  directions.

7 starch
7 starch
H2O