SOLUTIONS AND THEIR BEHAVIOR

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SOLUTIONS AND THEIR BEHAVIOR

• CHAPTER 14

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CHAPTER OVERVIEW

• This chapter examines homogeneous mixtures called
  solutions, which are made up of a solute and a
  solvent.
• Concentrations of solutions can be expressed in a
  variety of units.
• Properties of solutions that depend only on the
  number of solute particles and not their type are
  called colligative properties.

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14.1 UNITS OF CONCENTRATION

• Molarity, M, moles solute per liter solution
• Molality, m, moles of solute per kilogram solvent
  
• Mole fraction, XA, moles A divided by moles
  total
• Weight percent (mass percent), wt. A,
  
• (mass A divided by mass total) x 100

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Molarity -vs- Molality
Each flask contains 19.4 g of K2CrO4
Water to the 1.00 L mark
Exactly 1.00 kg of water added
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UNITS OF CONCENTRATION

• parts per million, ppm, is calculated like
  percent, but multiply by 106
• Remember that the mass of the solution equals the
  mass of the solute plus solvent.
• Conversions between molarity and the other
  concentration units requires the density of the
  solution.

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Solutions
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• Why does a raw egg swell or shrink when placed in
  different solutions?

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Some Definitions
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• A solution is a HOMOGENEOUS mixture of 2 or more
  substances in a single phase.
• One constituent is usually regarded as the
  SOLVENT and the others as SOLUTES.

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14.2 THE SOLUTION PROCESS

• The key to understanding the solution process is
  intermolecular forces solvent - solvent
  solute - solute solute - solvent.

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THE SOLUTION PROCESS

• What prevents solubility is an energy barrier
  when the latter interaction is significantly
  weaker than the former interactions.
• Like dissolves like is a general rule, but not
  and explanation of the solubility phenomenon.

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• Liquids Dissolving in Liquids
• Miscible liquids are soluble in all proportions.
• Immiscible liquids do not mix, but form separate
  layers.
• Isopropanol is miscible with water but gasoline
  is not. Explain why.

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SOLUTIONS

• A saturated solution is one which has reached its
  equilibrium solubility at that temperature.
• An unsaturated solution is one that has not
  reached its equilibrium solubility.
• A supersaturated solution is one in which the
  equilibrium solubility has been temporarily
  exceeded.

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Definitions
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Solutions can be classified as unsaturated or
saturated.
A saturated solution contains the maximum
quantity of solute that dissolves at that
temperature.

• SUPERSATURATED SOLUTIONS contain more than is
  possible and are unstable.

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Energetics of the Solution Process
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• If the enthalpy of formation of the solution is
  more negative than that of the solvent and
  solute, the enthalpy of solution is negative.
• The solution process is exothermic!

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• Solids Dissolving in Liquids
• The same rules apply.
• Compare the intermolecular forces.
• I2 is quite soluble in CCl4, but not very soluble
  in water. Explain why?

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Supersaturated Sodium Acetate
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• One application of a supersaturated solution is
  the sodium acetate heat pack.
• Sodium acetate has an ENDOthermic heat of
  solution.

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Ionic Solutions

• The heat of solution for ionic compounds is the
  sum of the lattice energy (), bonds breaking,
  and the hydration energy (-), bonds forming.
• It may be positive (endo) or negative (exo)
  depending on the relative magnitudes of these
  energies.

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DHsoln can be calcd using Hess' Law.
821 kJ/mol 819 kJ/mol 2 kJ/mol (endo)
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DHsoln can be calcd using Hess' Law.
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Ion Size also determines Solubility
Remember Coulombs Law
-
(charge n)(charge n)
Force of Attraction
k
2
d
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Ionic Solutions

• Temperature has a significant effect on
  solubility for salts and is consistent with Le
  Chatelier's principle.

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The heat of solution for many salts is positive,
endothermic, as seen by the positive slope of the
graph.
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Supersaturated Sodium Acetate
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• Sodium acetate has an ENDOthermic heat of
  solution.
• NaCH3CO2 (s) heat ----gt Na(aq)
  CH3CO2-(aq)
• Therefore, formation of solid sodium acetate from
  its ions is EXOTHERMIC.
• Na(aq) CH3CO2-(aq) ---gt NaCH3CO2 (s)
  heat

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Dissolving Gases

• Gas solubility decreases with increasing
  temperature which means ?Hsolution lt 0 ,
  exothermic.
• Gas solubilities increase with increasing
  pressure.
• Write the general equation for the solubility of
  a gas showing that the process is exothermic and
  show how increasing the temperature decreases the
  solubility.

Gas Solvent ? Solution Heat
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Colligative Properties
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• On adding a solute to a solvent, the properties
  of the solvent are modified.
• Vapor pressure decreases
• Melting point decreases
• Boiling point increases
• Osmosis is possible (osmotic pressure)

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Colligative Properties
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• These changes are called COLLIGATIVE PROPERTIES.
• They depend only on the NUMBER of solute
  particles relative to solvent particles, not on
  the KIND of solute particles.

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Concentration Units
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• An IDEAL SOLUTION is one where the properties
  depend only on the concentration of solute.
• Need concentration units to tell us the number of
  solute particles per solvent particle.
• The unit molarity does not do this!

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Concentration Units
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MOLE FRACTION, X For a mixture of A, B, and C
mol A
X
mol fraction A



A
mol A

mol B

mol C
MOLALITY, m
mol solute
m of solute


kilograms solvent

• WEIGHT grams solute per 100 g solution

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Calculating Concentrations

• Dissolve 62.1 g (1.00 mol) of ethylene glycol in
  250. g of H2O.
• Calculate mole fraction, molality, and weight
  of glycol.

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Calculating Concentrations
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Dissolve 62.1 g (1.00 mole) of ethylene glycol in
250. g of H2O. Calculate X, m, and of glycol.

• 250. g H2O 13.9 mole

1.00 mol glycol
X


glycol
1.00 mol glycol

13.9 mol H
O
2
X glycol 0.0672
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Calculating Concentrations
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Dissolve 62.1 g (1.00 mol) of ethylene glycol in
250. g of H2O. Calculate X, m, and of glycol.

• Calculate molality

1.00 mol glycol
conc
(molality)


4.00 molal

0.250 kg H
O
2
Calculate weight
62.1 g

glycol



x 100

19.9
62.1 g

250. g
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Dissolving Gases Henrys Law
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• Gas solubility (M) kH Pgas
• kH for O2 1.66 x 10-6 M/mmHg
• When Pgas drops, solubility drops.

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Lake Nyos, Cameroon
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COLLIGATIVE PROPERTIES

• Changes in Vapor Pressure Raoult's Law
• The presence of a solute in the solvent lowers
  the vapor pressure of the solvent.
• Psolvent Xsolvent Posolvent
• If the solute is also volatile, a similar
  equation applies to the solute. Psolute
  Xsolute Posolute
• The total pressure for the solution is given
  by Ptotal Psolvent Psolute

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COLLIGATIVE PROPERTIES

• If the solute is nonvolatile, the total pressure
  is just the pressure of the solvent and is lower
  than that of the pure solvent.  
• Study examples and exercises.

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Understanding Colligative Properties
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• To understand colligative properties, study the
  LIQUID-VAPOR EQUILIBRIUM for a solution.

H
HO
surface
H
HO
HO
HO
HO
H
H
H
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Understanding Colligative Properties
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• To understand colligative properties, study the
  LIQUID-VAPOR EQUILIBRIUM for a solution.

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Understanding Colligative Properties Raoultss
Law

• VP of H2O over a solution depends on the number
  of H2O molecules per solute molecule.
• Psolvent proportional to Xsolvent
• OR
• Psolvent Xsolvent Posolvent
• VP of solvent over solution
• (Mol frac solvent)(VP pure solvent)
• RAOULTS LAW

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Raoults Law
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• An ideal solution is one that obeys Raoults law.
• PA XA PoA
• Because mole fraction of solvent, XA, is always
  less than 1, then PA is always less than PoA.
• The vapor pressure of solvent over a solution is
  always LOWERED!

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Raoults Law
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• Assume the solution containing 62.1 g of glycol
  in 250. g of water is ideal.
• What is the vapor pressure of water over the
  solution at 30 oC?
• (The VP of pure H2O is 31.8 mm Hg see App.)

Solution Xglycol 0.0672 and so Xwater
? Because Xglycol Xwater 1 Xwater 1.000 -
0.0672 0.9328 Pwater Xwater Powater
(0.9382)(31.8 mm Hg) Pwater 29.7 mm Hg
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Raoults Law

• Or (see next slide)
• ?PA VP lowering XBPoA
• VP lowering is proportional to mole fraction of
  the solute!
• For very dilute solutions,
• ?PA KmolalityB
• where K is a proportionality constant.
• This helps explain changes in melting and boiling
  points.

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See Exercise 14.6, p. 575
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Changes in Freezing and Boiling Points of Solvent
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• See Figure 14.13

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Boiling Point Elevation

• If a solute is added to the pure solvent at its
  normal boiling point, the equilibrium vapor
  pressure will decrease and the liquid will no
  longer boil.
• To reach the new boiling point the temperature
  must be increased, thus boiling point elevation.
  
• ?tbp Kbp msolute

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Figure 14.13
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The boiling point of a solution is higher than
that of the pure solvent.
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Elevation of Boiling Point
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• Elevation in BP DtBP KBP m
• (where KBP is characteristic of solvent)

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Change in Boiling Point
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• Dissolve 62.1 g of glycol (1.00 mol) in 250. g of
  water. What is the BP of the solution?
• KBP 0.512 oC/molal for water (Table 14.4).
• Solution
• 1. Calculate solution molality 4.00 m
• 2. DtBP KBP m
• DtBP 0.512 oC/molal (4.00 molal)
• DtBP 2.05 oC
• BP 102.05 oC

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Freezing Point Depression

• The freezing point is lowered by the presence of
  a solute since these particles cannot form the
  solid and some of them are occupying the low
  energy slots needed to form the solid solvent.
• ?tfp Kfp msolute
• Note the K is a negative value.

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Change in Freezing Point
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Pure water
Ethylene glycol/water solution

• The freezing point of a solution is LOWER than
  that of the pure solvent.
• FP depression ?tFP KFPm

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Freezing Point Depression
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• Consider equilibrium at melting point
• Liquid solvent lt------gt Solid solvent
• Rate at which molecules go from S to L depends
  only on the nature of the solid.
• BUT rate for L ---gt S depends on how much is
  dissolved. This rate is SLOWED for the same
  reason VP is lowered.
• Therefore, to bring S ---gt L and L ---gt S rates
  into equilibrium for a solution, T must be
  lowered.
• Thus, FP for solution lt FP for solvent
• FP depression ?tFP KFPm

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Freezing Point Depression
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• Calculate the FP of a 4.00 molal glycol/water
  solution.
• KFP -1.86 oC/molal (Table 14.4, p. 577)
• Solution
• ?tFP KFP m
• (-1.86 oC/molal)(4.00 m)
• ?tFP -7.44 oC

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Colligative Properties Of Ionic Solutions

• Ionic compounds dissociate completely into ions
  in water.
• All calculations involving water and an ionic
  solute must account for the total number of
  particles present.
• This factor is called the van't Hoff factor, i.

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Freezing Point Depression
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• How much NaCl must be dissolved in 4.00 kg
  of water to lower FP to -10.00 oC?.
• Solution
• Calculate the required molality.
• ?tFP KFP m
• -10.00 oC (-1.86 oC/molal) Molality
• Concentration 5.38 molal

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Freezing Point Depression
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• How much NaCl must be dissolved in 4.00 kg of
  water to lower FP to -10.00 oC?.
• Solution
• Concentration required 5.38 molal

This means we need 5.38 mol of dissolved
particles per kg of solvent. Recognize that m
represents the total conc. of all dissolved
particles. Recall that 1 mol NaCl(aq) 1
mol Na(aq) 1 mol Cl-(aq)
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Freezing Point Depression
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• How much NaCl must be dissolved in 4.00 kg of
  water to lower FP to -10.00 oC?.
• Solution
• Concentration required 5.38 molal
• We need 5.38 mol of dissolved particles per kg of
  solvent.
• NaCl(aq) --gt Na(aq) Cl-(aq)

To get 5.38 mol/kg of particles we need 5.38
mol / 2 2.69 mol NaCl / kg 2.69 mol NaCl / kg
---gt 157 g NaCl / kg (157 g NaCl / kg)(4.00
kg) 629 g NaCl
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Boiling Point Elevation and Freezing Point
Depression
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• ?t K m i
• A generally useful equation
• i vant Hoff factor number of particles
  produced per formula unit.
• Compound Theoretical Value of i
• glycol 1
• NaCl 2
• CaCl2 3

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Osmosis
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Salt water
Pure water
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• Osmosis occurs when a molecule moves from a
  region of high concentration to lower
  concentration through a semipermeable membrane.

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Osmotic pressure is defined by p cRT
where c is the molarity of the solute and R is
the gas constant.
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Osmosis
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Semipermeable membrane

• The semipermeable membrane should allow only the
  movement of solvent molecules.
• Therefore, solvent molecules move from pure
  solvent to solution.

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Osmosis
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• The semipermeable membrane should allow only the
  movement of solvent molecules.
• Therefore, solvent molecules move from pure
  solvent to solution.

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Osmosis

• Equilibrium is reached when pressure produced by
  extra solution
• the OSMOTIC PRESSURE, p
• p cRT (where c is conc. in mol/L)
• counterbalances pressure of solvent molecules
  moving thru the membrane.

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Osmosis
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Osmosis
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• Osmosis of solvent from one solution to another
  can continue until the solutions are ISOTONIC
  they have the same concentration.

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Osmosis Calculating a Molar Mass
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• Dissolve 35.0 g of hemoglobin in enough water to
  make 1.00 L of solution. p measured to be 10.0
  mm Hg at 25 C. Calculate molar mass of
  hemoglobin.
• Solution
• (a) Calculate p in atmospheres
• p 10.0 mmHg (1 atm / 760 mmHg)
• 0.0132 atm
• (b) Calculate the concentration

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Calculating a Molar Mass
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Dissolve 35.0 g of hemoglobin in enough water to
make 1.00 L of solution. p measured to be 10.0 mm
Hg at 25 C. Calculate molar mass of
hemoglobin. Solution (b) Calculate concentration
from p cRT
0.0132 atm
Conc



(0.0821 L

atm/K

mol)(298
K)

• Concentration 5.39 x 10-4 mol/L
• (c) Calculate the molar mass
• Molar mass 35.0 g / 5.39 x 10-4 mol/L
• Molar mass 65,100 g/mol

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Reverse Osmosis
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COLLOIDS

• Colloids are a suspension of very small particles
  that do not settle out.
• (milk, jello, ..)

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Hydrophilic and hydrophobic colloids exist and
emulsions make use of molecules that contain both.
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Soap and Surfactants
H2O
H2O
H2O
H2O
Dirt
H2O
H2O
H2O
H2O
H2O
H2O
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Soap and Surfactants
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Detergent Fabric Softener