1. Chapter 8 Solutions
8.1 Solutions

2. Solutions: Solute and Solvent
are homogeneous mixtures of two or more substances
consist of a solvent and one or more solutes

3. Nature of Solutes in Solutions
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. Water Water is the most common solvent is a polar molecule
forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule

6. Combinations of Solutes and Solvents in Solutions

7. Formation of a Solution
Na+ and Cl– ions
on the surface of a NaCl crystal are attracted to polar water molecules
are hydrated in solution by many H2O molecules surrounding each ion

8. 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

9. Like Dissolves Like Two substances form a solution
when there is an attraction between the particles of the solute and solvent
when a polar solvent (such as water) dissolves polar solutes (such as sugar) and/or ionic solutes (such as NaCl)
when a nonpolar solvent such as hexane (C6H14) dissolves nonpolar solutes such as oil or grease

10. Water and a Polar Solute

11. Like Dissolves Like Solvents Solutes Water (polar) Ni(NO3)2
CH2Cl2 (nonpolar) (polar)
I2 (nonpolar)

12. 8.2 Electrolytes and Nonelectrolytes
Chapter 8 Solutions
8.2 Electrolytes and Nonelectrolytes

13. Solutes and Ionic Charge
In water,
strong electrolytes produce ions and conduct an electric current
weak electrolytes produce a few ions
nonelectrolytes do not produce ions

14. Strong Electrolytes Strong electrolytes
dissociate in water, producing positive and negative ions
dissolved in water will conduct an electric current
in equations show the formation of ions in aqueous (aq) solutions
H2O % ions
NaCl(s) Na+(aq) + Cl(aq)
H2O
CaBr2(s) Ca2+(aq) + 2Br(aq)

15. Weak Electrolytes A weak electrolyte
dissociates only slightly in water
in water forms a solution of a few ions and mostly undissociated molecules
HF(g) + H2O(l) H3O+(aq) + F(aq)
NH3(g) + H2O(l) NH4+(aq) + OH(aq)

16. Nonelectrolytes Nonelectrolytes dissolve as molecules in water
do not produce ions in water
do not conduct an electric current

17. Classification of Solutes in Aqueous Solutions

18. Equivalents
An equivalent (Eq) is the amount of an electrolyte or an ion that provides 1 mole of electrical charge (+ or –).
1 mole Na+ = 1 Eq
1 mole Cl− = 1 Eq
1 mole Ca2+ = 2 Eq
1 mole Fe3+ = 3 Eq

19. Electrolytes in IV Solutions

20. Electrolytes in Body Fluids
In replacement solutions for body fluids, the electrolytes are given in milliequivalents/L (mEq/L).
Ringer’s Solution
Cations Anion_____
Na mEq/L Cl− 155 mEq/L
K mEq/L
Ca mEq/L
155 mEq/L = mEq/L

21. Chapter 8 Solutions
8.3 Solubility

22. Solubility Solubility is
the maximum amount of solute that dissolves in a specific amount of solvent
expressed as grams of solute in 100 grams of solvent (usually water):
g of solute
100 g water

23. Unsaturated Solutions
contain less than the maximum amount of solute
can dissolve more solute
Dissolved solute

24. Saturated Solutions Saturated solutions contain
the maximum amount of solute that can dissolve
some undissolved solute at the bottom of the container
Dissolved
solute
Undissolved
solute

25. Effect of Temperature on Solubility
depends on temperature
of most solids increases as temperature increases
of gases decreases as temperature increases

26. Solubility and Pressure
Henry’s law states:
the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid
at higher pressures, more gas molecules dissolve in the liquid

27. Soluble and Insoluble Salts
Ionic compounds that
dissolve in water are soluble salts
do not dissolve in water are insoluble salts

28. Solubility Rules
Soluble salts typically contain at least one ion from Groups 1A(1), NO3−, or C2H3O2− (acetate).
Most other combinations are insoluble.

29. Examples of Using the Solubility Rules

30. Formation of a Solid
When solutions of salts are mixed, a solid forms if ions of an insoluble salt are present
Example:
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

31. Equations for Forming Solids
A full equation shows the formulas of the compounds.
Pb(NO3)(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq)
An ionic equation shows the ions of the compounds.
Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq)
PbCl2(s) + 2Na+(aq) + 2NO3−(aq)
A net ionic equation shows only the ions that form a solid.
Pb2+(aq) + 2Cl−(aq) PbCl2(s)

32. Guide to Writing New Ionic Equations for an Insoluble Salt

33. Finding the Insoluble Salt
STEP 1 Write the ions of the reactants.
Ba2+(aq) + 2NO3−(aq) + 2Na+(aq) + CO32−(aq)
STEP 2 Write the new combinations of the ions and
determine if an insoluble salt forms.
BaCO3(s) + 2Na+(aq) + 2NO3−(aq)
STEP 3 Write the ionic equation, including the insoluble salt as a solid in the products.
Ba2+(aq) + 2NO3−(aq) + 2Na+(aq) + CO32 −(aq) BaCO3(s) + 2NO3−(aq) + 2Na+(aq)
STEP 4 Write the net ionic equation deleting spectator ions. Ba2+(aq) + CO32−(aq) BaCO3(s)

34. 8.4 Percent Concentration
Chapter 8 Solutions
8.4 Percent Concentration

35. Concentration The concentration of a solution
is the amount of solute dissolved in a specific amount of solution
concentration = amount of solute
amount of solution

36. Mass Percent Concentration
Mass percent (m/m) concentration is
the percent by mass of solute in a solution
calculated using the formula:
Mass = mass of solute(g) x 100%
percent mass of solute(g) + mass of solvent(g)
Mass percent = mass of solute(g) x 100%
mass of solution(g)

37. Mass of Solution

38. Calculating Mass Percent
The calculation of mass percent (m/m) requires the
grams of solute (g of KCl) and
grams of solution (g of KCl solution)
g of KCl = g
g of solvent (water) = g
g of KCl solution = g
8.00 g of KCl (solute) x 100% = 16.0% (m/m)
50.00 g of KCl solution

39. Guide to Calculating Solution Concentrations

40. Volume Percent The volume percent (v/v) is
percent by volume of solute (liquid) in a solution
volume % (v/v) = volume of solute x 100% volume of solution
volume solute (mL) in 100 mL of solution.
volume % (v/v) = mL of solute
100. mL of solution

41. Mass/Volume Percent The mass/volume percent (m/v) is
mass/volume % (m/v) = grams (g) of solute x 100%
volume(mL) of solution
mass of solute(g) in 100 mL solution = g of solute mL of solution

42. Preparation of a Solution: Mass/Volume Percent

43. Percent Conversion Factors
Two conversion factors can be written for each type of percent concentration.

44. Guide to Using Concentration to Calculate Mass or Volume

45. Example of Using Percent Concentration (m/m) Factors
How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution?
STEP 1 Given: 225 g of solution; 10.0% (m/m) NaCl
Need: g of NaCl (solute)
STEP 2 Plan: g of solution g of NaCl 45

46. Solution STEP 3 Write equalities and conversion factors:
10.0 g of NaCl = 100 g of NaCl solution
10.0 g NaCl and 100 g NaCl solution
100 g NaCl solution g NaCl
STEP 4 Set up problem to cancel grams of solution:
225 g NaCl solution x g NaCl = 22.5 g of NaCl g NaCl solution

47. Example Using Percent Concentration (m/v) Factors
How many mL of a 4.20% (m/v) will contain 3.15 g KCl?
STEP 1 Given: g of KCl (solute)
4.20% (m/v) KCl solution
Need: mL of KCl solution
STEP 2 Plan: g of KCl mL of KCl solution 47

48. Using Percent Concentration(m/v) Factors (continued)
STEP 3 Write equalities and conversion factors:
4.20 g KCl = 100 mL of KCl solution
4.20 g KCl and mL KCl solution
100 mL KCl solution g KCl
STEP 4 Set up the problem:
3.15 g KCl x 100 mL KCl solution = mL of KCl g KCl

49. Chapter 8 Solutions
8.5 Molarity and Dilution

50. Molarity (M) Molarity (M) is a concentration term for solutions
gives the moles of solute in 1 L solution
molarity (M) = moles of solute
liter of solution

51. Preparing a 1.0 Molar NaCl Solution
A 1.0 M NaCl solution is prepared
by weighing out 58.5 g of NaCl (1.00 mole) and
adding water to make 1.0 liter of solution

52. Guide to Calculating Molarity

53. Example of Calculating Molarity
What is the molarity of L NaOH solution if it
contains 6.00 g of NaOH?
STEP 1 Given: 6.00 g of NaOH in L solution
Need: molarity (mole/L) of NaOH solution
STEP 2 Plan:
g of NaOH moles of NaOH molarity

54. Example of Calculating Molarity (continued)
STEP 3 Write equalitites and conversion factors:
1 mole of NaOH = g of NaOH
1 mole NaOH and g NaOH
40.0 g NaOH mole NaOH
STEP 4 Set up problem to calculate molarity:
6.00 g NaOH x 1 mole NaOH = mole of NaOH
40.0 g NaOH
0.150 mole NaOH
0.500 L NaOH solution
= M NaOH

55. Molarity Conversion Factors
The units of molarity are used to write conversion factors for calculations with solutions.

56. Example of Calculations Using Molarity
How many grams of KCl are needed to prepare 125 mL
of a M KCl solution?
STEP 1 Given: 125 mL (0.125 L) of M KCl
Need: grams of KCl
STEP 2 Plan:
L of KCl moles of KCl g of KCl

57. Example of Calculations Using Molarity (continued)
STEP 3 Write equalities and conversion factors:
1 mole of KCl = g of KCl
1 mole KCl and g KCl
74.6 g KCl mole KCl
1 L of KCl = mole of KCl
1 L and mole KCl
0.720 mole KCl L
STEP 4 Set up problem to cancel mole KCl:
0.125 L x mole KCl x g KCl = 6.71 g of KCl
1 L mole KCl

58. Dilution In a dilution, water is added volume increases
concentration of solute decreases

59. Initial and Diluted Solutions
In the initial and diluted solution,
the moles of solute are the same
the concentrations and volumes are related by the following equations:
For percent concentration
C1V = C2V2
Concentrated Diluted
solution solution
For molarity
M1V = M2V2

60. Guide to Calculating Dilution Quantities

61. Example of Dilution Calculations Using Percent Concentration
What volume of a 2.00% (m/v) HCl solution can be
prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution?
STEP 1 Prepare a table:
C1 = 14.0% (m/v) V1 = mL
C2 = 2.00% (m/v) V2 = ?
STEP 2 Solve dilution expression for the unknown
C1V1 = C2V2 V2 = V1C1 C2
STEP 3 Set up the problem using known quantities:
V2 = V1C1 = (25.0 mL)(14.0%) = mL
C %

62. Example of Dilution Calculations Using Molarity
What is the molarity (M) of a solution prepared
by diluting L of M HNO3 to L?
STEP 1 Prepare a table:
M1 = M V1 = L
M2 = ? V2 = L
STEP 2 Solve the dilution expression for the unknown: M1V1 = M2V2
STEP 3 Set up the problem using known quantities:
M2 = M1V1 = (0.600 M)(0.180 L) = M
V L

63. Molarity in Chemical Reactions
In a chemical reaction,
the volume and molarity of a solution are used to determine the moles of a reactant or product
molarity ( mole ) x volume (L) = moles
1 L
if molarity (mole/L) and moles are given, the volume (L) can be determined
moles x L = volume (L)
moles

64. Guide to Calculations Involving Solutions in Chemical Reactions

65. Using Molarity of Reactants
How many mL of 3.00 M HCl are needed to completely
react with 4.85 g of CaCO3?
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
STEP 1 Given: M HCl; 4.85 g of CaCO3
Need: volume of HCl in mL
STEP 2 Write a plan:
g of CaCO moles of CaCO moles of HCl mL of HCl
STEP 3 Write equalities and conversion factors:
1 mole of CaCO3 = g
1 mole CaCO and g CaCO3
100.1 g CaCO mole CaCO3

66. Using Molarity of Reactants (continued)
STEP 3 (continued)
1 mole of CaCO3 = 2 moles of HCl
1 mole of CaCO3 and 2 moles of HCl
2 moles of HCl mole of CaCO3
1000 mL of HCl = 3.00 moles of HCl
1000 mL of HCl and moles of HCl
3.00 moles of HCl mL of HCl
STEP 4 Set up problem to calculate mL of HCl:
4.85 g CaCO3 x 1 mole CaCO3 x 2 moles HCl x 1000 mL HCl
100.1 g CaCO mole CaCO moles HCl
= mL of HCl required

67. 8.6 Properties of Solutions
Chapter 8 Solutions
8.6 Properties of Solutions

68. Solutions Solutions contain small particles (ions or molecules)
are transparent
do not separate, even by filtration or through a semipermeable membrane
do not scatter light

69. Colloids Colloids have medium-sized particles
cannot be separated by filtration
can be separated by semipermeable membranes
scatter light

70. Examples of Colloids

71. Suspensions Suspensions have very large particles settle out
can be separated by filtration
must be stirred to stay suspended
Examples: blood platelets, muddy water, and calamine lotion

72. Solutions, Colloids, and Suspensions

73. Comparison of Solutions, Colloids, and Suspensions

74. Colligative Properties
are changes in the properties of a solvent when solute particles are added
depend on the number of solute particles in solution
involve the lowering of freezing point
involve an increase in the boiling point

75. Freezing Point Lowering and Boiling Point Elevation
When 1 mole of solute particles is added to 1000 g of
water, the
freezing point of water decreases by 1.86 °C (from 0 °C to –1.86 °C)
boiling point of water increases by 0.52 °C (from 100 °C to °C)

76. Moles of Particles
The number of moles of particles depends on the type of
solute:
nonelectrolytes dissolve as molecules
1 mole of nonelectrolyte = 1 mole of particles in water
strong electrolytes dissolve as ions
1 mole of electrolyte = 2 to 4 moles of particles in water

77. Effect of Solute Particles

78. Osmosis
In osmosis,
water (solvent) flows from the lower solute concentration into the higher solute concentration
the level of the solution with the higher solute concentration rises
the concentrations of the two solutions become equal with time

79. Example of Osmosis
A semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens?
semipermeable membrane
4% starch
10% starch
H2O

80. Example of Osmosis (continued)
The 10% starch solution is diluted by the flow of water out of the 4% solution, and its volume increases.
The 4% solution loses water, and its volume decreases.
Eventually, the water flow between the two becomes equal.
7% starch
7% starch
H2O

81. Osmotic Pressure Osmotic pressure
is produced by the solute particles dissolved in a solution
is the pressure that prevents the flow of additional water into the more concentrated solution
increases as the number of dissolved particles in the solution increases

82. Osmotic Pressure of the Blood
Red blood cells
have cell walls that are semipermeable membranes
maintain an osmotic pressure that cannot change without damage occuring
must maintain an equal flow of water between the red blood cell and its surrounding environment

83. Isotonic Solutions An isotonic solution
exerts the same osmotic pressure as red blood cells
is known as a “physiological solution”
of 5.0% (m/v) glucose or 0.9% (m/v) NaCl is used medically because each has a solute concentration equal to the osmotic pressure equal to red blood cells

84. Hypotonic Solutions A hypotonic solution
has a lower osmotic pressure than red blood cells (RBCs)
has a lower concentration than physiological solutions
causes water to flow into RBCs
causes hemolysis (RBCs swell and may burst)

85. Hypertonic Solutions A hypertonic solution
has a higher osmotic pressure than RBCs
has a higher concentration than physiological solutions
causes water to flow out of RBCs
causes crenation (RBCs shrink in size)

86. Dialysis
In dialysis,
solvent and small solute particles pass through an artificial membrane
large particles are retained inside
waste particles such as urea from blood are removed using hemodialysis (artificial kidney)

87. Hemodialysis
When the kidneys fail, an artificial kidney uses hemodialysis to remove waste particles such as urea from blood.