The Characteristics of a Soln

1
The Characteristics of a Soln

• A homogenous mixture.
• A sample of matter containing two or more
  substances that has a uniform appearance and
  uniform properties throughout.
• Solid solution examples Steel, brass, bronze
• Liquid solution examples Alcohol in water,
  sugar water, coffee
• Gaseous solution examples An air sample, any
  mixture of gases

2
Solution Terminology

• Solute
• The substance present in a relatively small
  amount in a solution the solid or gas when a
  substance in that state is dissolved in a liquid
  to make a solution.
• Solvent
• The substance present in a relatively large
  amount in a solution the liquid when a solid or
  gas is dissolved to make a solution.

3
Microscopic View of a Soln
4
Solution Terminology

• Concentrated Solution
• Has a relatively large quantity of a specific
  solute
• per unit amount of solution.
• Dilute Solution
• Has a relatively small amount of a specific
  solute
• per unit amount of solution.

5
Solution Concentration
AgNO3 (aq) Cu0
Cu(NO3)2 (aq) Ag0
6
Solution Terminology

• Solubility
• A measure of how much solute will dissolve in a
  given
• amount of solvent at a given temperature.
• Saturated
• A solution whose concentration is at the
• solubility limit for a given temperature.
• Unsaturated
• A solution whose concentration is less than the
• solubility limit for a given temperature.

7
Solution Terminology
Supersaturated A solution whose concentration
is greater than the normal solubility limit.
8
Solution Terminology

• Miscible (Soluble) OR Immiscible (Insoluble)
• (usually in reference to solutions of liquids in
  liquids).

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The Formation of a Solution
10
The Formation of a Solution

• Development of Equilibrium in Forming a Saturated
  Solution
• Dissolving rate
• If temperature is held constant, the rate of
  dissolving
• per unit of solute surface is constant.
• Crystallization rate
• The rate per unit of surface area increases as
  the solution concentration at the surface
  increases.
• Dynamic equilibrium
• Dissolving rate is equal to crystallization rate.

11
The Formation of a Solution
Dynamic equilibrium
12
Determination of Solubility

• The extent to which a particular solute dissolves
  in a given solvent depends on three factors
• The strength of intermolecular forces within the
  solute, within the solvent, and between the
  solute and the solvent
• The partial pressure of a solute gas over a
  liquid solvent
• The temperature
• Golden Rule of Solubility LIKE DISSOLVES LIKE

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Determination of Solubility
Miscible- dipole forces and H-bonding
Immiscible- London forces only
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Pressure and Solubility
The solubility of a gaseous solute in a liquid is
directly proportional to the partial pressure of
the gas over the surface of the liquid.
15
Temperature and Solubility
The solubility of most solids increases with
rising temperature
16
Concentration Units

• Solution Concentration
• In general, concentration is

17
Percentage by Mass

• Example
• A solution is prepared by dissolving 1.23 g of
  sodium chloride in 500.0 mL of water. What is the
  percentage by mass?
• Solution The density of water is 1 g/mL.

18
Molarity
Molarity, M Moles of solute per liter of
solution
Molarity Calc Example 1 Calculate the
molarity of a solution prepared by dissolving
0.345 moles of NaCl in sufficient water to give
525 mL of solution. 0.345 mol / 0.525 L 0.657
M (be sure to convert mL to Liters)
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Molarity

• Molarity Calc Example 2
• Calculate the molarity of a solution made by
  dissolving 13.0 grams of sugar, C12H22O11, in
  enough water to make 4.00 ? 102 milliliters of
  solution.
• (convert grams to moles and mL to L)
• 0.0950 mol C12H22O11/L 0.0950 M C12H22O11

20
Molarity

• Example using molarity as a conversion factor
• How many moles of methanol are in 45.3 mL of
  0.550 M CH3OH?
• 0.0249 mol CH3OH

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Molarity

• To prepare a solution of a specified molarity
• Weigh the appropriate amount of solute.
• Add less than the total volume of solvent.
• Mix to completely dissolve the solute.
• Add additional solvent until the total solution
  volume is appropriate.

22
Molarity- Using Volumetric Glassware
23
Molality

• Molality, m
• The number of moles of solute dissolved
• in one kilogram of solvent
• Molarity (mol/L) is temperature dependent
• molality is temperature independent.

24
Molality

• Calculate the molality of a solution that
  contains 7.72 g of NaBr in 500 mL of water.
• First convert grams to moles
• Then calculate molality

25
Normality

• Normality, N
• The number of equivalents, eq, per liter of
  solution

Stop Here
26
Normality

• Equivalent, eq
• One equivalent of acid is the quantity that
  yields
• one mole of hydrogen ions in a chemical reaction.
• Once equivalent of base is the quantity that
  reacts
• with one mole of hydrogen ions.

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Normality

• eq acid/mol eq base/mol
• H3X NaOH NaH2X H2O 1 1
• H3X 2 NaOH Na2HX 2 H2O 2 1
• H3X 3 NaOH Na3X 3 H2O 3 1
• 2 HY Ba(OH)2 BaY2 2 H2O 1 2
• The number of equivalents of acid and base
• in each equation is the same.

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Solution Concentration

• General Form for Solution Concentration Ratios

29
Solution Concentration
30
Dilution of Solutions

• Concentrated solutions are diluted by
• adding more solvent particles.
• The number of solute particles remains the
• same before and after a dilution

31
Dilution of Solutions
32
Dilution of Solutions

• Example
• If 10.0 mL of a 16-M nitric acid solution is
  diluted to 1.00 L, what is the molar
  concentration of the dilute solution?
• Solution
• Solve with algebra.
• M1 16 M M2 ?
• V1 10.0 mL V2 1.00 L

33
Dilution of Solutions

• Example
• If 10.0 mL of a 16-M nitric acid solution is
  diluted to 1.00 L, what is the molar
  concentration of the dilute solution?
• M1 16 M M2 ?
• V1 10.0 mL V2 1.00 L

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Dilution of Solutions
35
Solution Stoichiometry

• For any reaction whose equation is known,
• the three steps for solving a stoichiometry
  problem are
• Convert the quantity of given species to moles.
• Convert the moles of given species to moles of
  wanted species.
• Convert the moles of wanted species to the
  quantity units required (usually volume, i.e.,
  mL).

36
Solution Stoichiometry
37
Solution Stoichiometry

• Example
• Aluminum shavings are dropped into 500.0 mL of
  0.77 M hydrochloric acid until the reaction is
  complete. How many grams of hydrogen are
  produced?
• Solution
• Solve with dimensional analysis.
• 2 Al 6 HCl 2 AlCl3 3 H2
• GIVEN 500.0 mL of 0.77 M HCl WANTED g H2

38
Solution Stoichiometry

• Aluminum shavings are dropped into 500.0 mL of
  0.77 M hydrochloric acid until the reaction is
  complete. How many grams of hydrogen are
  produced?
• 2 Al 6 HCl 2 AlCl3 3 H2
• GIVEN 500.0 mL of 0.77 M HCl WANTED g H2
• PER 0.77 mol HCl/1000 mL
• PATH mL
  mol HCl
• 6 mol HCl/3 mol H2 2.016 g H2/mol H2
• mol H2
  g H2

39
Solution Stoichiometry

• Aluminum shavings are dropped into 500.0 mL of
  0.77 M hydrochloric acid until the reaction is
  complete. How many grams of hydrogen are
  produced?
• 2 Al 6 HCl 2 AlCl3 3 H2
• GIVEN 500.0 mL of 0.77 M HCl WANTED g H2

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Titration Using Molarity

• Titration
• The very careful addition of one solution to
  another
• by means of a device that can measure
• delivered volume precisely, such as a buret.

41
Titration Using Molarity
42
Titration Using Molarity

• Buret
• A glass tube of uniform width calibrated to
  accurately measure volume of liquid delivered
  through an adjustable-flow stopcock at the bottom
  of the tube.
• Indicator
• A substance that changes from one color to
  another,
• used to signal the end of a titration.

43
Titration Using Molarity

• Standardize
• Determination of the concentration of a solution
  to be used in a titration by titrating it against
  a primary standard.
• Primary Standard
• A soluble solid of reasonable cost that is very
  stable and pure, preferably with a high molar
  mass, that can be weighed accurately for use in a
  titration.

44
Titration Using Molarity

• Example
• A student titrates 17.5 mL of 0.387 M nitric acid
  solution into a 25.0-mL sample of barium
  hydroxide solution. What is the molar
  concentration of the barium hydroxide solution?
• Solution
• A titration is a solution stoichiometry problem.
  Use dimensional analysis to find mol Ba(OH)2 and
  algebra to find the molarity.
• GIVEN 17.5 mL of 0.387 M HNO3 WANTED M Ba(OH)2
• 2 HNO3 Ba(OH)2 2 H2O Ba(NO3)2

45
Titration Using Molarity

• A student titrates 17.5 mL of 0.387 M nitric acid
  solution into a 25.0-mL sample of barium
  hydroxide solution. What is the molar
  concentration of the barium hydroxide solution?
• GIVEN 17.5 mL of 0.387 M HNO3 WANTED M Ba(OH)2
• 2 HNO3 Ba(OH)2 2 H2O Ba(NO3)2
• PER 0.387 mol HNO3/1000 mL
• PATH mL
  mol HNO3
• 1 mol
  Ba(OH)2/2 mol HNO3
• mol Ba(OH)2

46
Titration Using Molarity

• A student titrates 17.5 mL of 0.387 M nitric acid
  solution into a 25.0-mL sample of barium
  hydroxide solution. What is the molar
  concentration of the barium hydroxide solution?
• PER 0.387 mol HNO3/1000 mL
• PATH mL
  mol HNO3
• 1 mol
  Ba(OH)2/2 mol HNO3
• mol Ba(OH)2
• 0.00339 mol Ba(OH)2

47
Titration Using Molarity

• A student titrates 17.5 mL of 0.387 M nitric acid
  solution into a 25.0-mL sample of barium
  hydroxide solution. What is the molar
  concentration of the barium hydroxide solution?
• 0.00339 mol Ba(OH)2
• 0.136 mol Ba(OH)2/L 0.136 M Ba(OH)2

48
Titration Using Normality

• Goal 17
• Given the volume of a solution that reacts with a
  known mass of a primary standard and the equation
  for the reaction, calculate the normality of the
  solution.
• Goal 18
• Given the volumes of two solutions that react
  with each other in a titration and the normality
  of one solution, calculate the normality of the
  second solution.

49
Titration Using Normality

• The number of equivalents of all species
• in a reaction is the same.
• For an acidbase reaction,
• equivalents of acid equivalents of base

50
Titration Using Normality

• Example
• What is the normality of a sodium hydroxide
  solution if 33.16 mL of the solution reacts with
  2.88 g of KHC8H4O4 in the reaction NaOH
  KHC8H4O4 H2O NaKC8H4O4?
• Solution
• Find the number of equivalents with dimensional
  analysis, and then use algebra to determine the
  normality.
• GIVEN 2.88 g KHC8H4O4 WANTED eq

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Titration Using Normality

• What is the normality of a sodium hydroxide
  solution if 33.16 mL of the solution reacts with
  2.88 g of KHC8H4O4 in the reaction NaOH
  KHC8H4O4 H2O NaKC8H4O4?
• 0.425 eq/L 0.425 N

52
Titration Using Normality

• Example
• 23.91 mL of the 0.425 N sodium hydroxide solution
  from the previous example is used to titrate a
  30.0-mL sample of phosphoric acid in the reaction
  NaOH H3PO4 H2O NaH2PO4. What is the
  normality of the acid?
• Solution
• Since the number of equivalents of all species in
  a reaction is the same, V1N1 eq V2N2

53
Titration Using Normality

• Example
• 23.91 mL of the 0.425 N sodium hydroxide solution
  from the previous example is used to titrate a
  30.0-mL sample of phosphoric acid in the reaction
  NaOH H3PO4 H2O NaH2PO4. What is the
  normality of the acid?
• V1N1 eq V2N2

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Colligative Properties

• Goal 19
• Given (a) the molality of a solution, or data
  from which it may be found, (b) the normal
  freezing or boiling point of the solvent, and c
  the freezing- or boiling-point constant, find the
  freezing or boiling point of the solution.
• Goal 20
• Given the freezing-point depression or
  boiling-point elevation and the molality of a
  solution, or data from which they may be found,
  calculate the molal freezing-point constant or
  molal boiling-point constant.

55
Colligative Properties

• Goal 21
• Given (a) the mass of solute and solvent in a
  solution, (b) the freezing-point depression or
  boiling-point elevation, or data from which they
  may be found, and c the molal
  freezing/boiling-point constant of the solvent,
  find the approximate molar mass of the solute.

56
Colligative Properties

• A pure solvent has distinct physical properties.
• Introducing a solute into the solvent affects
  these properties.
• In dilute solutions of certain solutes, the
  change in
• some of these properties is proportional
• to the molal concentration of the solute
  particles.
• Colligative Property
• Solution property that is determined only by the
  number of solute particles dissolved in a fixed
  quantity of solvent and not by the identity of
  the solute particles.

57
Colligative Properties

• Boiling Point Elevation
• The boiling point of a solution is higher
• than the boiling point of the pure solvent.
• Example
• The normal boiling point of water is 100.0C.
• The normal boiling point of a 1 m
• solution of sugar water is 100.5C.

58
Colligative Properties

• Freezing Point Depression
• The freezing point of a solution is lower
• than the freezing point of the pure solvent.
• Example
• The normal freezing point of water is 0.0C.
• The normal boiling point of a 1 m
• solution of sugar water is 1.9C.

59
Colligative Properties
60
Colligative Properties

• Boiling point elevation and freezing point
  depression
• are colligative properties
• They depend on the number of solute
• particles but not their identity.
• ?Tb change in boiling temperature
• ?Tf change in freezing temperature

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Colligative Properties

• ?Tb ? m ?Tf ? m
• The proportionality constants are
• Kb molal boiling-point elevation constant
• Kf molal freezing-point depression constant
• ?Tb Kb ? m ?Tf Kf ? m

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Colligative Properties

• Molal Boiling-Point Elevation Constant Values
• Substance Boiling Point (C) Kb (C/m)
• Benzene 80 2.5
• Carbon disulfide 46 2.4
• Carbon tetrachloride 77 5.0
• Water 100 0.52
• Molal Freezing-Point Depression Constant Values
• Substance Freezing Point (C) Kf (C/m)
• Benzene 6 5.1
• Carbon disulfide 112 3.8
• Carbon tetrachloride 23 30
• Water 0 1.86

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Colligative Properties

• Example
• What is the freezing point of a solution made by
  adding 10.0 g of glucose, C6H12O6, to 7.50 ? 102
  grams of water?
• Solution
• Determine the solution molality, and then apply
  ?Tf Kf ? m.
• GIVEN 10.0 g C6H12O6, 7.50 ? 102 g H2O
• WANTED m C6H12O6
• PER 180.16 g C6H12O6/mol
  C6H12O6
• PATH g C6H12O6
  mol C6H12O6

64
Colligative Properties

• Example
• What is the freezing point of a solution made by
  adding 10.0 g of glucose, C6H12O6, to 7.50 ? 102
  grams of water?
• PER 180.16 g C6H12O6/mol
  C6H12O6
• PATH g C6H12O6
  mol C6H12O6

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Colligative Properties

• Tf 0C 0.138C 0.138C

66
Colligative Properties

• Procedure
• How to Calculate the Molar Mass of a Solute from
• Freezing-Point Depression or Boiling-Point
  Elevation Data
• Calculate molality from m ?Tf/Kf or m ?Tb/Kb.
  Express as mol solute/kg solvent.
• Using molality as a conversion factor between
  moles of solute and kilograms of solvent, find
  the number of moles of solute.
• Use the defining equation for molar mass, MM
  g/mol, to calculate the molar mass of the solute.

67
Colligative Properties

• Example
• When 31.5 g of a solute is dissolved in 4.00 ?
  102 g of water, the freezing point of the
  resulting solution is 1.22C. What is the molar
  mass of the solute?
• Solution
• Follow the three-step procedure.
• Step 1 Calculate molality ?Tf Kf ? m

68
Colligative Properties

• Example
• When 31.5 g of a solute is dissolved in 4.00 ?
  102 g of water, the freezing point of the
  resulting solution is 1.22C. What is the molar
  mass of the solute?
• Step 1 Calculate molality 0.656 m 0.656 mol
  solute/kg solvent
• Step 2 Determine the moles of solute
• 0.262 mol solute

69
Colligative Properties

• Example
• When 31.5 g of a solute is dissolved in 4.00 ?
  102 g of water, the freezing point of the
  resulting solution is 1.22C. What is the molar
  mass of the solute?
• Step 2 Determine the moles of solute 0.262 mol
  solute
• Step 3 Calculate molar mass by using its
  definition