SOLUTIONS
CHAPTER 12
INTRODUCTION
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How do substances dissolve?
Why do substances dissolve?
What factors affect solubility?
How do dissolved substances affect
properties of the solution?
(Colligative Properties)
I. Solution Terminology
1. Solution: homogeneous mixture of two or more
substances
Solvent: substance present in greatest amount
substance doing the dissolving
Solute: substance present in lesser amount
substance which is dissolved
2. Solubility: amount of solute that will dissolve in
a given volume of solvent at equilibrium.
Soluble: a substance which dissolves in a
solvent.
Insoluble: a substance which does not
dissolve in a solvent
Miscible: when two liquids dissolve in each
other in any proportion
Immiscible: liquids will not dissolve in each
other
3. Dissolution: the process of dissolving solute
molecules in a solvent
Solvation: the process of solvent molecules
surrounding the solute molecules
Hydration: solvation in water. Water
molecules surround the solvent
molecules.
4. Hydrophilic: “water loving”, polar part of a
molecule attracted to water.
5. Hydrophobic: “water fearing”, nonpolar part of a
molecule not attracted to water.
6. Saturated Solution: when the concentration of
the solute equals its solubility. A saturated
solution contains as much of a solute as the
solubility allows. “A Dynamic Equilibrium”
See next Figure
7. Unsaturated Solution: when concentration of the
solute is less than its solubility.
8. Supersaturated Solution: when concentration of
solute exceeds its solubility.
Temporary state, unstable state
Formation of a Saturated Solution
Solid begins
to dissolve.
Eventually, the rates of dissolving
and of crystallization are equal; no
more solute appears to dissolve.
As solid dissolves,
some dissolved solute
begins to crystallize.
Longer standing does
not change the amount
of dissolved solute.
II. Explanation of Solubility
A. General Rule of Solubility
“Likes Dissolve Likes”
1. For something to dissolve, the solute and
solvent must have similar types of
intermolecular forces.
Polar or ionic compounds tend to be soluble
in polar solvents.
Nonpolar compounds tend to be soluble in
nonpolar solvents.
B. Energies Affecting Solubility
1. Attractive forces between solute molecules
must be broken. (energy requiring or
unfavorable).
2. Attractive forces between solvent
molecules must be broken. (energy
requiring or unfavorable).
3. Attractive forces between solute and
solvent molecules must be formed. (energy
producing or favorable). Must be equal
to or greater than unfavorable steps
(above) for solvation to occur.
4. Entropy (degree of energy randomness or
disorder) generally increases in solvation
process (favorable). Especially important for
gases.
C. Examples of Dissolving Liquids in Liquids
1. Water and Ethyl Alcohol are “Misible”
Explain
Diagram of interactions
Energetics of Process and ΔHsolution
See following diagram
2. Hydrocarbons are “Miscible”
Oil and Gasoline
Benzene and Octane
3. Water and Hydrocarbons are “Immiscible”
Water and Benzene
Water and Gasoline
D. Examples of Dissolving Solids in Liquids
1. Ionic Compounds in Water
In order to become solvated, the ionic interactions
between the ions in the crystal structure must be
broken. (unfavorable).
The energy which holds the crystal together is
known as the lattice energy.
How is this energy overcome?
1. Water molecules surround ions and “hydrate”
them. (favorable)
2. Usually an increase in entropy. (favorable)
Diagram of NaCl Dissolving in Water
See next Diagram
See Energy Diagram
Why are some ionic compounds( like AgCl)
not soluble in water?
2. Sugar (Glucose) Soluble in Water
(Glucose, C6H12O6, is a polar covalent
compound).
E. Factors Affecting Solubility
1. Temperature
Effect of temp. on solubility can best be
explained using Le Chatelier’s Principle.
When any change in concentration, temperature,
pressure, or volume is imposed on a system at
equilibrium, the system responds by attaning a new
equilibrium condition that minimizes the impact of
the imposed change.
Equilibrium will shift to the left or right!!!
a. Gas Dissolving in Liquid: Exothermic
Gas + solvent  saturated solution ΔH = - _
What is effect on solubility of increasing
temp?
b. Solids Dissolving in Liquid:
May be Exothermic or Endothermic?
KOH(s) → K+(aq) + OH-(aq) ΔHsoln = -57.6 kJ/mol
H2 O
NH4NO3(s) → NH4+(aq) + NO3-(aq) ΔHsoln = +25.7 kJ/mol
H2 O
F. Pressure Effects on Dissolving
Gases
(Remember to use “Likes Dissolve Likes” to
predict what gases will dissolve in a liquid).
Solubility of a gas in a liquid
increases as the pressure of the gas
increases.
Explain why. See Figure
Quantitative Relationship For Pressure
and Solubility of a Gas
Henry’s Law
Sg = kH Pg
S = solubility of gas
Pg = partial pressure of gas
kH = constant (different for each gas)
(see Table 12.4 for values)
Problem Solving:
Nitrogen comprises 78% of the atmosphere and
has a “kH” value of 8.42 x 10-7 M / mm Hg in
water.
a. How many grams of N2 will be present in
1.00 L of water that is in equilibrium with air
at atmospheric pressure, 760mm Hg?
b. How many grams of N2 would be present if the
atmospheric pressure is increased to 10
atmospheres?
III. Concentration Units
Other concentration terms useful for describing
properties of dilute solutions.
volume of of
solutesolute
volume
x100
volume of solution
A. Volume Percent =
x100
volume of solution
B.
Mass Percent =
mass of solute
x100
mass of solution
Example Problem:
What is the mass percent of methanol in a
solution made up by adding 27.5 mL of
methanol to 500 g of water?
(Density of methanol = 0.791 g / mL)
C. Mole Fraction =
A solution is prepared by adding 0.76 g
NaCl, 0.21 g KOH, and 54 g of water.
a. What is the mass percent of NaCl?
b. What is the mole fraction of NaCl?
c. What is the mole fraction of water?
d. What is the mole percent of water?
D. Parts Per Million =ppm
1 g solute / 1,000,000 g solution
1 mg solute / 1000 g solution
Parts per Billion = ppb
1 g solute / 1,000,000,000 g solution
1 mg solute / 1,000,000 g solution
The mass percent of NaCl = 0.043% in an
aqueous solution. What is the ppm NaCl in
the solution?
E. Molality
Recall Molarity = M = mol solute / L solution
Volume is affected by temperature changes,
and therefore another concentration term,
molality, will be used in problem solving with
temperature changes.
Molality = m =
molesof solute
kg of solvent
Example Problem:
What is the molality of a solution prepared by
dissolving 7.53 g of methanol (CH3OH, 32.0 g/mol)
in 200.0 g of water?
If the density of the solution is 0.987 g/mL, what is
the molarity?
IV. Colligative Properties
A colligative property of a solution depends only
on the concentration of the solute particles
(molecules or ions), but not on the identity of
the solute.
* All colligative properties are entropy driven!!
Explain
(A 2.0 molal NaCl solution has the same
colligative properties as a 4.0 molal glucose
nonelectrolyte) solution, or 1.0 molal Na3PO4
solution).
Molality
# Moles Particles
Per Formula Unit
# Moles Particles
Per Kg Solvent
2m NaCl
2
4
4m Glucose
1
4
1m Na3PO4
4
4
Examples of Colligative Properties
1. Vapor Pressure Lowering
2. Boiling Point Elevation
3. Melting Point Depression (Lowering)
4. Osmotic Pressure (Osmosis)
A. Vapor Pressure Lowering
1. The vapor pressure of any pure solvent will
be lowered by the addition of a nonvolatile
solute to the solvent.
2. Explanation
a. Pure solvent  vapor has greater increase in
entropy than solution  vapor. Solvent in
solution is in a more disordered state and thus
less increase in entropy when going to gas.
*** b. Consider ability of solvent molecules in a
solution to escape to gas.
3. Raoult’s Law
Simple relationship between the vapor pressure of
one component (solvent in this case) and its mole
fraction in solution.
Psolution = Xsolvent P0solvent
Psolution = vapor pressure of solvent gas over solution
Xsolvent = mole fraction of solvent in the solution
P0solvent = vapor pressure of pure solvent
Example Vapor Pressure Problem
Calculate the vapor pressure of water at 90oC in
a solution prepared by dissolving 5.00 g glucose
(C6H12O6) in 100 g water. The vapor pressure of
water at 900C is 525.8 mm Hg. (Glucose is a
nonelectrolyte, nonvolatile covalent compound.
B. Boiling Point Elevation
1. Why does addition of a nonvolatile solute to a
solvent elevate the boiling point?
2. Equation:
ΔTb = Kbmi
ΔTb = amount the boiling point is raised (elevated)
Kb = constant dependent on solvent only
m = molality of solute (mol solute / kg solvent)
i = van’t Hoff factor , (# particles per formula unit
of solute)
Formula Unit of Compound
CH3OH
NaCl
Na2SO4
K3PO4
i
1
2
3
4
What is the boiling point for the following aqueous
solutions? The Kb for water is 0.52oC kg mol-1.
a. Pure water
b. 1 molal CH3OH solution
c. 1 molal NaCl solution
3. Molar Mass From Boiling Point Elevation
A solution is prepared by dissolving 11.0 g of a
nonvolatile organic solute in 100 g of chloroform
solvent. The boiling point of the solution was
84.30oC. Given: B.P. of chloroform = 61.20oC.
Kb for chloroform = 3.63oC / m
What is the molar mass of the unknown solute?
(Det. ΔT, m, #mol solute, molar mass)
4. Significant of BP Elevation
C. Freezing Point Depression
1. Explanation For Lower Freezing Point
For pure solvent, the speed of the particles
decreases until they can occupy space in the
growing crystal clusters.
The presence of solute molecules gets in the way of
the solvent molecules joining the growing crystal
clusters.
2. Equation:
ΔTf = Kfmi
3. Problems
a. What is the freezing point of a 0.0222 m CaCl 2
aqueous solution?
Kf for water = 1.86oC/m.
b. A solution made by dissolving 3.46 g of a
nonelectrolyte solute in 85.0 g benzene froze at
4.13oC.
F.P. of benzene = 5.45oC.
Kf for benzene = 5.07 OC/m.
1) Calculate the molality of the solution.
2) Calculate the molar mass of the solute.
4. Significance
D. Osmotic Pressure
1. Definitions
Osmosis is defined as the passage of solvent
molecules through a semipermeable membrane
from a solution of lower solute concentration to a
solution of higher solute concentration.
Semipermeable membranes allow only certain
small molecules to pass through.
Osmotic pressure is the pressure that must be
applied to the solution to stop osmosis. Fig. 12.16
or the following:
3. Equation:
Π = M RTi
Π = osmotic pressure
M = concentration of solution in molarity (M)
R = gas constant (0.0821 L atm / K mol
T = temperature in Kelvin scale
i = van’t Hoff factor
** Note similarity to Ideal Gas Law
4. Problems
a. Calculate the osmotic pressure for a 3.75 x 10-4 M
aqueous solution of the polymer dextran at 25oC.
Dextran is a nonelectrolyte.
b. Calculate the molar mass of dextran if a solution of
5.0 g of dextran in 1.0 L of water gives an osmotic
pressure of 2.9 x 10-3 atm. at 25oC.
(Det. M, #mol solute, molar mass)
5. Significance / Practical Applications
Practical Applications of
Osmosis
Ordinarily a patient must be given intravenous fluids that are
isotonic—have the same osmotic pressure as blood.
External solution is
hypertonic; produces
osmotic pressure > πint.
Net flow of water out
of the cell.
Red blood cell in
isotonic solution
remains the same
size.
External solution is
hypotonic; produces
osmotic pressure <
πint. Net flow of water
into the cell.
Practical
Applications of
Osmosis (cont’d)
• Reverse osmosis (RO):
reversing the normal net flow
of solvent molecules through
a semipermeable membrane.
• Pressure that exceeds the
osmotic pressure is applied to
the solution.
• RO is used for water
purification.
Pressure greater than π is
applied here …
… water flows from the more
concentrated solution, through
the membrane.
V. Additional Reading
Responsible for Terms and Significance
1.
2.
3.
Reverse osmosis
Isotonic
Hypotonic
Hypertonic
Colloids and Tyndall Effect
(These are found in Sections 12.7 and 12.8)