Solutions
Chang 7th Edition Chapter 12
General Chemistry II
W hy does a raw egg swell or shrink when placed in different solutions?
Chapter Objectives
Define solution, solvent, solute and colligative properties and use them in calculations to solve for
molality, mole fraction, weight percent and ppm
Explain the differences between saturated, unsaturated and supersaturated solutions, miscible and
immiscible
Use lattice energy and enthalpy of hydration to explain enthalpy of solution
Use Henry's law and Le Chatelier's principle to explain solubility of gases
Solve problems using mole fraction, Raoult's law, and calculations predicting changes in colligative
properties, and solving for variables related to colligative properties
Define and calculate Osmotic pressure
Driving forces for solution formation
Some Definitions
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.
4 major topics
Ways to describe solution concentrations (amount of solute per unit of solution)
How and why solutions form
Colligative properties (those properties that depend on the number of solute particles per solvent
molecule - but not on the identity of the solute)
Colloids
Concentration Units
An IDEAL SOLUTION is one where the properties depend only on the concentration of solute.
Need conc. units to tell us the number of solute particles per solvent particle.
The unit “molarity” does not do this!
Concentration Units
MOLE FRACTION, X
For a mixture of A, B, and C
Concentration Units
MOLE FRACTION, X
For a mixture of A, B, and C
MOLALITY, m
Concentration Units
MOLE FRACTION, X
For a mixture of A, B, and C
MOLALITY, m
W EIGHT % = grams solute per 100 g solution
Calculating Concentrations
687320265
page 1 of 8
Dr. Myton, CH116
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mol fraction, molality, and weight
% of glycol.
Calculating Concentrations
250. g H2O = 13.9 mol
Calculating Concentrations
250. g H2O = 13.9 mol
X glycol = 0.0672
Calculating Concentrations
Calculate molality
Calculating Concentrations
Calculate molality
Calculate weight %
Problem Solving
12.33 A solution of ethyl alcohol, CH3CH2OH, in water has a concentration of 1.25 m. Calculate the
weight percent of ethyl alcohol
12.34 Calculate the molarity of an aqueous solution of NaCl with a concentration of 0.363 m and a density
of 1.0185 g/mL
Problem Solving
12.37 A solution of ammonia in water is at a concentration of 5.00% w/w. Its density is 0.9787 g/mL,
calculate the molarity and molality of the solution.
Problem Solving
Assume you add 1.2 kg of ethylene glycol HOCH2CH2OH as an antifreeze to 4.0 kg of water in the
radiator of your car. What are the mole fraction, molalaity, molarity, weight percent and ppm
concentrations of the solution?
Problem Solving
You dissolve 560 g of NaHSO4 in a swimming pool that contains 4.5e5 L of water at 25 ºC. What is the
sodium ion concentration in ppm?
Problem Solving
If you dissolve 10.0 g of sugar C12H2O11 (about one heaping teaspoon) in a cup of water (250. G) what are
the mole fraction, molarity, weight percent and ppm concentrations of the solution
Problem Solving
Sea water has a sodium ion concentration of 1.08e4 ppm. If the sodium is present in the form of
dissolved sodium chloride, how many grams of NaCl are in each liter of sea water? Assume the density
of sea water is 1.05 g/mL.
Definitions
Solutions can be classified as unsaturated or saturated.
Definitions
Solutions can be classified as unsaturated or saturated.
A saturated solution contains the maximum quantity of solute that dissolves at that temperature.
Definitions
Solutions can be classified as unsaturated or saturated.
687320265
page 2 of 8
Dr. Myton, CH116
A saturated solution contains the maximum quantity of solute that dissovles at that temperature.
SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.
Solutions and IM forces
Bond breaking is endothermic
Bond formation is exothermic
Types of bonds: solute-solute, solvent-solvent, and solute-solvent
An IDEAL SOLUTION the enthalpy of solution is zero
Energetics of the Solution Process
If the enthalpy of formation of the solution is more negative that that of the solvent and solute, the
enthalpy of solution is negative.
The solution process is exothermic!
Supersaturated
Sodium Acetate
One application of a supersaturated solution is the sodium acetate “heat pack.”
Sodium acetate has an ENDOthermic heat of solution.
Supersaturated Sodium Acetate
Sodium acetate has an ENDOthermic heat of solution.
NaCH3CO2 (s) + heat ---->
Na+(aq) + CH3CO2-(aq)
Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC.
Na+(aq) + CH3CO2-(aq) --->
NaCH3CO2 (s) + heat
Enthalpy of Solution
Determine the heat of solution for ammonium nitrate (used in cold packs)
Calculate the enthalpy of solution for sodium hydroxide (lye)
Enthalpy
Dissolving Gases & Henry’s Law
Gas solubility Sg = kH • Pgas
kH for O2 = 1.66 x 10-6 M/mmHg
W hen Pgas drops, solubility drops.
Henry’s Law Constants
At 25 ºC
Gas
kH (M/ mm Hg)
N2
8.42e-7
O2
1.66e-6
CO2
4.48e-5
Problem Solving
What is the concentration of oxygen in a fresh water stream in equilibrium with the atmosphere at 25 ºC
and 1.00 atm? Express your answer in ppm.
What is the concentration of carbon dioxide in equilibrium with an atmospheric partial pressure of 0.33
atm at 25 ºC?
Problem Solving
At 740 torr and 20C, nitrogen has a solubility in water of 0.018 g/L. At 620 torr and 20C, its
solubility is 0.015 g/L. Does nitrogen obey Henry’s law?
Temperature dependence of solubility
Colligative Properties
On adding a solute to a solvent, the props. of the solvent are modified.
Vapor pressure decreases
Melting point
decreases
Boiling point
increases
Osmosis is possible (osmotic pressure)
These changes are called COLLIGATIVE PROPERTIES.
687320265
page 3 of 8
Dr. Myton, CH116
They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of
solute particles.
Releasing pressure changes the solubility of a gas in solution
Lake Nyos, Cameroon
Courtesy of George Kling, page 656-657
Understanding
Colligative Properties
To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.
Understanding
Colligative Properties
To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.
Adding a solute lowers vapor pressure
Understanding
Colligative Properties
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)
RAOULT’S LAW
Raoult’s Law
An ideal solution is one that obeys Raoult’s 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 LOW ERED!
Raoult’s Law
Assume the solution containing 62.1 g of glycol in 250. g of water is ideal. W hat is the vapor pressure of
water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg; see App. G)
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
Problem Solving
Ethylene glycol, HOCH2CH2OH is a common ingredient in automobile antifreeze. If 651 g of ethylene
glycol is dissolved in 1.50 kg of water (represents a typical 30.2% solution), what is the vapor pressure of
water over the solution at 90 ºC?
The vapor pressure of water at 90 ºC is 535.8 mm Hg, assume ideal behavior
Problem Solving
Assume you dissolve 10.0 g of sugar (C12H22O12 in 225 mL of water and warm the water to 60 ºC. What
is the vapor pressure of the water over this solution at equilibrium?
Psolvent = -Xsolute Pºsolvent
Problem solving
687320265
page 4 of 8
Dr. Myton, CH116
12.43 The vapor pressure of water at 20C is 17.5 torr. A 20% by weight solution of ethylene glycol
HOCH2CH2OH in water is prepared. Assuming that the solute is nonvolatile, do a calculation to estimate
the vapor pressure of the solution
Problem Solving
Example 12.5 CCl4 has a vapor pressure of 100 torr at 23 degrees Celsius. Assume candle wax has a
molecular weight of 331 and calculate the vapor pressure of carbon tetrachloride in a solution made of
10.0 g of the nonvolatile wax in 40.0 g of carbon tetrachloride
Problem Solving
12.45 Benzene and toluene help get good engine performance from lead-free gasoline. At 40C the vapor
pressure of benzene is 180 torr and that of toluene is 60 torr. To prepare a solution of these that will have
a total vapor pressure of 96 torr at 40C requires what mole percent concentration of each?
Raoult’s Law
For a 2-component s ystem where A is the solvent and B is the solute
DP
PA = VP lowering = XBPoA
VP lowering is proportional to mol frac solute!
For very dilute solutions, DP
PA = K•molalityB where K is a proportionality constant.
This helps explain changes in melting and boiling points.
Boiling Point Elevation
 Tbp = Kbp msolute
The boiling point of a solution is higher than that of the pure solvent.
BP/FP Elevation/Depression Constants
Solvent
BP
Kbp
ºC
ºC/m
FP
water
100
+0.5121
0.0
-1.86
Benzene
80.10
+2.53
5.50
-5.12
Kfp
ºC
ºC/m
Camphor
207.4
+5.611
179.75 -39.7
Change in Boiling Point
Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. W hat is the BP of the solution?
KBP = +0.512 oC/molal for water (see Table 14.3).
Solution
1.
Calculate solution molality = 4.00 m
2.
DttBP = KBP • m
DttBP = +0.512 oC/molal (4.00 molal)
DttBP = +2.05 oC
BP = 102.05 oC
Problem Solving
Eugenol, the active ingredient in cloves, has the formula C10H12O2. What is the boiling point of a solution
when 0.144 g of this compound is dissolved in 10.0 g of benzene
What quantity of ethylene glycol, HOCH 2CH2OH, must be added to 125 g of water to raise the boiling
point by 1.0 ºC?
Using BP for MW
A solution prepared from 1.25 g of oil of wintergreen (methyl salicylate) in 99.0 g of benzene has a boiling
point of 80.31 ºC. Determine the molar mass of methyl salicylate.
Change in Freezing Point
The freezing point of a solution is LOW ER than that of the pure solvent.
FP depression = DttFP = KFP•m
Freezing Point Depression
Consider equilibrium at melting point
Liquid solvent <------> Solid solvent
687320265
page 5 of 8
Dr. Myton, CH116
•
Rate at which molecules go from S to L depends only on the nature of the solid.
•
BUT — rate for L ---> S depends on how much is dissolved. This rate is SLOW ED for the same
reason VP is lowered.
•
Therefore, to bring S ---> L and L ---> S rates into equilibrium for a solution, T must be lowered.
Thus, FP for solution < FP for solvent
FP depression = DttFP = KFP•m
Freezing Point Depression
Calculate the FP of a 4.00 molal glycol/water solution.
KFP = -1.86 oC/molal (Table 14.4)
Solution
DttFP = KFP • m
= (-1.86 oC/molal)(4.00 m)
DttFP = -7.44 oC
Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?.
Solution
Calc. required molality
DttFP = KFP • m
-10.00 oC = (-1.86 oC/molal) • Conc
Conc = 5.38 molal
Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?.
Solution
Conc req’d = 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)
Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?.
Solution
Conc req’d = 5.38 molal
W e need 5.38 mol of dissolved particles per kg of solvent.
NaCl(aq) --> 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 ---> 157 g NaCl / kg
(157 g NaCl / kg)•(4.00 kg) = 629 g NaCl
Problem Solving
How many grams of ethylene glycol, HOCH2CH2OH, must be added to 5.50 kg of water to lower the
freezing point of the water from 0.0 ºC to -10.0 ºC?
Problem Solving
Some people have summer homes on a lake or in the woods. In Northern climates these homes may be
closed for winter and antifreeze added to the toilet tank to prevent damage from water freezing in the trap.
Will adding 525 g of ethylene glycol, HOCH2CH2OH, prevent freezing at -25 ºC?
Problem Solving
Glycerol C3H8O3 (molecular mass 92) is essentially a nonvolatile liquid that is very soluble in water. A
solution is made by dissolving 46.0 g of glycerol in 250 g of water. By calculations, estimate the following
the boiling point of the solution at 1 atm
its freezing point
its vapor pressure at 25C (at this temperature the vapor pressure of water is 23.8 torr)
Boiling Point Elevation and Freezing Point Depression
687320265
page 6 of 8
Dr. Myton, CH116
Dtt = K • m • i
A generally useful equation
i = van’t Hoff factor = number of particles produced per formula unit.
Compound
Theoretical Value of i
glycol
1
NaCl
2
CaCl2
3
Problem Solving
A 0.00200 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at -0.00732 ºC. How
many moles of ions does 1 mole of the salt give upon being dissolved?
Calculate the freezing point of 525 g of water containing 25.0 g of NaCl assuming a van’t Hoff factor of
1.85
Problem Solving
Calculate the percent ionization of a 1.00 m aqueous acetic acid solution based on the following
equilibrium:
HC2H3O2 (aq)  H+ (aq) + C2H3O2- (aq)
Kf = 1.86C/m. The solution actually freezes at –1.90 C
12.74 The van’t Hoff factor for the solute in 0.118 m LiCl is 1.89. Calculate the freezing point of the
solution. Why is this factor so much larger than that of NiSO 4?
Problem Solving
An experiment calls for the use of the dichromate ion in sulfuric acid as an oxidizing agent for
propyl alcohol. The chief product is acetone which forms according to the following reaction:
3C3H8O + Na2Cr2O7 + 4 H2SO4  3C3H6O + Cr2(SO4)3 + Na2SO4 + 7 H2O
the oxidizing agent is only available as sodium dichromate dihydrate. In theory how many grams
of sodium dichromate dihydrate are needed to oxidize 21.4 g of isopropyl alcohol according to the
balanced equation?
continued
The amount of acetone actually isolated was 12.4 g. Calculate the percentage yield of acetone
The reaction produces a volatile byproduct. When a sample of it with a mass of 8.654 mg was
burned in oxygen, it was converted into 22.368 mg of carbon dioxide and 10.655 mg of water.
Assume any unaccounted for material is oxygen. Calculate the percentage composition of the
byproduct and determine its empirical formula
continued
A solution prepared by dissolving 1.338 g of the byproduct in 115.0 g of benzene had a freezing point of
4.87C. Calculate the molecular mass of the byproduct and write its molecular formula.
Osmosis
Osmosis
The semipermeable membrane should allow only the movement of solvent molecules.
Therefore, solvent molecules move from pure solvent to solution.
Osmosis
The semipermeable membrane should allow only the movement of solvent molecules.
Therefore, solvent molecules move from pure solvent to solution.
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.
687320265
page 7 of 8
Dr. Myton, CH116
Osmosis
Osmosis
Osmotic Pressure
Adding a solute increases osmotic pressure
Osmosis
Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they
have the same concentration.
Osmotic pressure in living systems: FIGURE 14.16
Problem solving
12.64 What is the osmotic pressure in torr of a 0.010 M aqueous solution of a molecular
compound at 25C?
12.65 An aqueous solution of a compound with a very high molecular mass was prepared in a
concentration of 2.0 g/L at 298 K. Its osmotic pressure was 0.021 torr. Calculate the molecular
mass of the compound
Osmosis
Calculating a Molar Mass
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
C. Calc. molar mass of hemoglobin.
Solution
(a)
Calc. p in atmospheres
p = 10.0 mmHg • (1 atm / 760 mmHg)
= 0.0132 atm
(b)
Calc. concentration
Osmosis
Calculating a Molar Mass
Osmosis
Calculating a Molar Mass
Conc = 5.39 x 10-4 mol/L
(c )
Calc. molar mass
Molar mass = 35.0 g / 5.39 x 10-4 mol/L
Molar mass = 65,100 g/mol
Problem solving
A 1.40 g sample of polyethylene, a common plastic, is dissolved in enough benzene to give exactly 100
mL of solution. The measured osmotic pressure of the solution is 1.86 mm Hg at 25 ºC. What is the
average molecular mass of the polymer
Surfactants
687320265
page 8 of 8
Dr. Myton, CH116