DISCLAIMER: these notes are provided to assist you in mastering the course material but they are not intended as a replacement of the
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conjunction with your own notes.
Chapter 6: Solutions
Objectives
 units of concentration
 solution formation
 reactions in solution – stoichiometry, titration
 acid – base reactions
 redox reactions
 colligative properties
 mixtures and distillation
 (colloids)
 solutions = mixtures of substances which are homogeneous, i.e.- intermingled at molecular
level
 components (may be more than one of each)
 solvent - major
 solute - minor
 say that solute is dissolved in solvent
 Examples of solutions
 can be gases, liquids, solids
 here focus on liquid solutions, especially with water as solvent (aqueous solutions)
6.1 Composition of Solutions - Units of Concentration
 qualitatively can describe solutions as dilute or concentrated
 quantitative expressions based on mass or on moles
1. (a) Mass %
% by mass of component =
mass of component in soln
x 100
total mass of soln
1. (b) Parts Per Million (ppm)
ppm of component =
2. (a) Mole Fraction (recall gas phase)
Chem 59-110 (’02)
mass of component in soln
x 106
total mass of soln
2
mole fraction of component =
moles of component
total moles of all components
 use symbol Xn where subscript n represents component of interest, eg. XHCl
2.(b) Molarity
moles of solute
liters of solution
-3
(SI would prefer mol m , but too large)
Molarity = M =
units: mol L-1
2.(c) Molality
Molality = m =
moles of solute
kg of solvent
 NB: these two easily confused
 Molarity in terms of volume of solution
 Molality in terms of mass of solvent
 do Examples 6.1 & 2, Problems 3-9, odd; read practical details (how to)
6.2 Nature of Dissolved Species
Solution Process (Dissolution) & Solubility
 an equilibrium:
dissolve
solute + solvent
solution
crystallize
 from left, continue adding solute until concentration of solvated (i.e.- solution) particles is
high enough that they frequently collide with and attach to the undissolved solute =
crystallization
 this reverse process also encouraged as solvating ability of solvent molecules is used up or
occupied
 eventually reach a point when the rates of dissolution and crystallization are equal, i.e.- at
equilibrium (terms in chapter 11)
Chem 59-110 (’02), ch. 6, Solutions
3
 solution in equilibrium with undissolved solute is a saturated solution
 concentration of solute in solution at that point is its solubility
 eg. solubility of NaCl = 37.5 g/100mL of water at 0 (less NaCl, undersaturated)
 possible to form supersaturated solution by preparing at elevated temperature, then cooling
carefully (eg. “heat packs”)
How and why do solutions form?
 recall, intermolecular forces that exist between molecules and ionic particles in pure
substances
 replace these forces with interactions between solvent and solute (the “how”)
 for liquids dissolving in liquids, terms: miscible, if solution forms; converse: immiscible
 many examples will be given of solution formation by ionic substances (i.e.- salts), but:
 solutions also formed by uncharged molecules (eg. fructose, Fig. 6.2)
 special case, colloids, at end of chapter
 water is not the only solvent, but it is commonly encountered and is of special relevance
to this class
 terms: solvation and hydration
 eg. gas solubilities in water, (previous chapter)
 increase with size due to dispersion forces
 for solids and liquids - “like dissolves like”
 polar and ionic solutes in polar solvents
 non-polar solutes in non-polar solvents
O
 eg. acetone, H3C C CH3 , and smaller alcohols are completely miscible with water due
to hydrogen bonding
 increase the size of hydrocarbon (non-polar) fragments of ketones and alcohols, solubility
drops off
 fat- and water-soluble vitamins
 eg. salt dissolution, eg. NaCl, K2SO4 (Fig. 6.3) - replace charge-charge interactions with ion dipole interactions in aqueous solution (strong electrolytes)
 solvation shell

reverse, precipitation reactions
eg. BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2 KCl(aq)
K+ and Cl- are spectator ions; could write net ionic reaction
Chem 59-110 (’02), ch. 6, Solutions
4
 solution formation occurs when the forces within pure components and their replacements
between solute and solvent are comparable, therefore, energy considerations
Energy Changes & Solution Formation (not in text)
 i.e.- the “why” solutions form
 from NaCl example, three energetic steps involved in solution formation
1. solute - solute interactions (must be overcome)
2. solvent - solvent interactions (some must be overcome)
3. solute - solvent interactions (must form)
 simplest approach (not foolproof), to consider heat (enthalpy) changes
 net change is algebraic sum of three:
H soln = H1 + H 2 + H 3
endothermic
H > 0
exothermic
H < 0
 calculation by two approaches, depending on data available, eg. for a salt in water:
 Hsoln = -lattice energy + enthalpy of hydration; (data in Tables)
 Hosoln = Ho f, soln - Ho f, solid; i.e.- from Hess’s Law (data in Tables)
 H2 term not evaluated independently; a part of hydration/solution enthalpy
 rough correlation of between enthalpy of solution and solubility
 Hsoln can be exothermic (eg. NaOH forms warm aqueous solution, Hsoln = -45 kJ/mol)
 Hsoln can be endothermic (eg. NH4NO3 forms chilled aqueous solution, Hsoln = 26 kJ/mol),
as long as it is not too endothermic (gain by H3 comparable to loss by (H1 + H2))
 hence, ionic or polar solutes (eg. NaCl) will not dissolve in non-polar solvents (eg. CCl4), i.e.H1 large, H3 small
 similarly, non-polar solutes (eg. CCl4) will not be solvated by polar solvents (eg. water), i.e.H2 large, H3 small
Solution Formation, Spontaneity & Disorder (not covered until chapter 8)
 enthalpy considerations, above, not the only ones
 why should endothermic process, eg. NH4NO3 dissolution, occur at all?
Chem 59-110 (’02), ch. 6, Solutions
5
 another surprise, two non-polar liquids like CCl4 and octane mix with very little energy
change yet they dissolve (in each other) spontaneously
 two factors must be considered with any chemical change, including dissolution:
1. “Energy” (“enthalpy” in the context of this chapter). Processes in which the energy
content of the system decreases tend to occur spontaneously. eg. falling book
2. Entropy (disorder, or randomness). Processes in which the disorder of the system
increases tend to occur spontaneously.
 the dissolution of NH4NO3 is endothermic, yet occurs spontaneously because the rigid order
of the crystal is totally disrupted as the ions are individually hydrated in solution
 hence, earlier statement that a substance will not dissolve if Hsoln too endothermic - it
is too large to be counteracted by an increase in disorder
 octane and CCl4 form solutions when mixed in any proportions because each of the molecules
has a bigger volume to “roam” around in the solution, i.e.- is more randomized
Factors Affecting Solubility
 nature of solute & solvent, i.e.- solute - solvent interactions (above)
 pressure, temp.
Pressure Effects
 solids and liquids little affected
 solubility of any gas in any solvent increases as pressure of the gas over the solvent increases
 quantitated by a form of Henry’s Law:
Sg = kHPg
Sg = solubility of the gas (M) in the solvent
Pg = partial pressure of the gas over the soln
kH = Henry’s Law constant (specific for solute - solvent pair and temp)
( a different form of Henry’s Law in section 6.6, below)
 eg. solubility of N2(g) in water at 25C and a partial pressure of 0.78 atm is 5.3 x 10-4 M
 if partial pressure doubled, what will solubility of N2(g) be?
 first calculate kH:
5.3 x 10-4 M
mol
kH =
= 6.8 x 10-4
0.78 atm
L - atm
 then, at elevated pressure:
mol
S N2 = k H PN2 = 6.8 x 10-4
x 1.56 atm
L - atm
= 1.06 x 10-3 mol / L
Temperature Effects
Chem 59-110 (’02), ch. 6, Solutions
6
 for most solid and liquid substances, solubility increases with temp.
 eg. for ionic substances in water
 solubility of gases in water generally decreases with increasing temp.
explained as increased kinetic energy, better able to escape
 can be explained by using Le Chatelier’s Principle, i.e.- treat enthalpy of reaction as a
component of the dissolution equilibrium:
 if exothermic, enthalpy a “product”, increased temp., decreases solubility
 if endothermic, enthalpy a “reactant”, increased temp., increases solubility
 problem: entropy ignored; wait until chapter 8
 eg. gas bubbles forming as glass of cold water warms to room temp.
 eg. lakes become anaerobic if subjected to thermal pollution
6.3 Stoichiometry of Reactions in Solution – Titration
 some “dissolution” processes involve chemical change, i.e.- “solvent” is also a reactant
 eg.
Ni(s) + 2 HCl(aq)  H 2 (g) + NiCl 2 (aq)
 not reversible; solute cannot be reclaimed by removing solvent

or, mix two solutions:

2 Br-(aq) + Cl2(aq)  2 Cl-(aq) + Br2(aq)
 fixed volume of one, eg. 50 mL of 0.060 M NaBr(aq)
 how much of other to completely react?
 Concentration? Volume?
Do Example 6.5

Stages in the above reaction, a titration; at the stoichiometric ratio, the end point – signaled
by some indicator
 2 types here: acid – base, oxidation – reduction (redox)
Acid – Base Reactions
 water dissociates to a very small extent: H2O(l)  H+(aq) + OH-(aq)
 an acid produces a net increase in [H+] over that in water, eg. HCl, HNO3, H2SO4,
CH3COOH
 a base produces a net increase in [OH-] over that in water, eg. NaOH, NH3
 acid – base titration, Example 6.6 (a H+ transfer)
Chem 59-110 (’02), ch. 6, Solutions
7
50.0 mL of acetic acid solution titrated with 1.306 M NaOH; 31.66 mL of latter needed to
end point; conc’n of acetic acid?
Oxidation – Reduction (Redox) Reactions
 electron transfer reactions
 eg. Br- + Cl2 reaction, above

Br- oxidized, loses e
Cl2 reduced, gains e eg. Zn(s) in aqueous acid:
Zn(s)  2 H (aq.)  Zn 2  (aq.)  H2 (g)
oxidation state: (0)
(+1)
(+2)
(0)
reducing oxidizing
oxidized
reduced
agent,
agent,
reductant oxidant
NB: conjugates (compare, acid - base)
 not necessary to have ionic components, eg.:
2 PH 3 (g)  4O 2 (g)  P2 O 5 (s)  3H 2 O(g)
(-3)(+1)
(0)
(+5)(-2)
(+1)(-2)
red. agent
ox. agent
oxidized
reduced

another eg.: Zn strip in a CuSO4(aq) solution, over time blue colour of solution disappears,
copper “plates out” and Zn strip slowly disappears
overall: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
comprised of two half-reactions:
oxidation:
Zn(s)  Zn2+(aq) + 2 ereduction: Cu2+(aq) + 2 e-  Cu(s)
 note, overall equation above already balanced with respect to mass and charge
Balancing Redox Equations in aqueous solution
 2 methods to balance electrons: Oxidation Numbers and Half - Reactions (preferred)
Oxidation Number Method
 eg., unbalanced (NB: need reactants and products)
Chem 59-110 (’02), ch. 6, Solutions
8
Al(l)  MnO 2 (s)  Al 2 O3 (s)  Mn(l)
(0)
(+4)
(+3)
(0)
 for Al, change +3; for Mn, change -4
 to balance Al and Mn, take 4 Al and 3 Mn, then balance O by inspection:
 4 Al  3MnO 2
 2 Al 2 O 3
 3Mn
Method of Half - Reactions
 separate oxidation and reduction processes
Sn 2 (aq.)  2 Fe3 (aq.)  Sn 4 (aq.)  2 Fe2 (aq.)
oxidation: Sn 2 (aq.)  Sn 4 (aq.)  2e
reduction: 2Fe3 (aq.)  2e
 2Fe2 (aq.)
 (could also readily do this one by oxidation numbers)
 eg. decolorization of permanganate solution by oxalate in aqueous acid solution:
 unbalanced:
MnO4 (aq.)  C2O24 (aq.)  Mn2  CO2 (g)
 incomplete half-reactions:
oxidant: MnO 4 (aq.)  Mn 2 (aq.)
reductant: C2 O 4 (aq.)  CO 2 (g)
 balance first with respect to atom undergoing redox
 for rest: in aq. acid, plenty of H+ and H2O available (in aq. base, plenty of OH- and
H2O avail.)
 balance O with H2O:
MnO4 (aq.)  Mn 2 (aq.)  4H2O(l)
 then balance H with H+ (from acid):
8H  (aq.)  MnO4 (aq.)  Mn 2 (aq.)  4H2O(l)
 balance charge with e- :
5e   8H   MnO 4 (aq.)  Mn 2 ( aq.)  4H 2 O (l)
 similarly for oxalate:
Chem 59-110 (’02), ch. 6, Solutions
9
C2 O 24 (aq. )  2 CO 2 (g)
 no O, H balancing needed
 charge:
C2 O 24 (aq. )  2 CO 2 (g)  2 e 
 multiply half-reactions as appropriate to give same number of electrons in each (gain =
loss), then add:
10e   16H  (aq.)  2 MnO 4 (aq.)  2 Mn 2 (aq.)  8H 2 O(l)
5C 2 O 24 (aq.)  10CO 2 (g)  10e 
 overall (may need to simplify in some cases):
16H  (aq.)  2MnO 4 (aq.)  5C2 O 24 (aq.)  2Mn 2 (aq. )  8H 2 O(l)  10CO 2 (g)
 likewise, Examples 6.8 (in acid) and 6.9 (in base)
 if in aqueous base, treat initially as if in acid, then add equal amounts of OH- to each side to
neutralize acid (your text suggests a slightly different method; you choose one method)
 Example in base:
Al(s) + S(s)  Al(OH)3(s) + HS-(aq)

half reactions:
oxidation: Al(s)  Al(OH)3(s)
reduction: S(s)  HS-(aq)

for mass, redox-active atoms already balanced
 balance O with H2O:
Al(s) + 3 H2O(l)  Al(OH)3(s)
S(s)  HS-(aq)

balance H, first with H+:
Al(s) + 3 H2O(l)  Al(OH)3(s) + 3 H+(aq)
S(s) + H+(aq)  HS-(aq)

then, since in base, add hydroxides to both sides to neutralize acid on one side:
Al(s) + 3 H2O(l) + 3 OH-(aq)  Al(OH)3(s) + 3 H2O(aq)
simplified: Al(s) + 3 OH-(aq)  Al(OH)3(s)
S(s) + H2O(aq)  HS-(aq) + OH-(aq)
Chem 59-110 (’02), ch. 6, Solutions
10


balance charge with e-:
Al(s) + 3 OH-(aq)  Al(OH)3(s) + 3 eS(s) + H2O(aq) + 2 e-  HS-(aq) + OH-(aq)
multiply half-reactions to balance numbers of electrons on opposite sides:
by 2x: 2 Al(s) + 6 OH-(aq)  2 Al(OH)3(s) + 6 eby 3x: 3 S(s) + 3 H2O(aq) + 6 e-  3 HS-(aq) + 3 OH-(aq)
 add half-reactions and simplify, if necessary:
2 Al(s) + 3 S(s) + 3 H2O(aq) + 6 OH-(aq)  2 Al(OH)3(s) + 3 HS-(aq) + 3 OH-(aq)
simplify: 2 Al(s) + 3 S(s) + 3 H2O(aq) + 3 OH-(aq)  2 Al(OH)3(s) + 3 HS-(aq)
Redox Titration

KMnO4 example (dark purple)
MnO4-(aq) + 5 Fe2+(aq) + 8 H+(aq)  Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
Add KMnO4 solution slowly to Fe2+ solution with stirring, slight persistence of purple colour is
the end point (Fig. 6.9)
Disproportionation
 one type of molecule goes in both redox directions in same reaction
eg. 2 H2O2 (l)  2 H2O (l) + O2 (g)
Ox. # (-1)
(-2)
(0)
Example 6.10
6.5 Colligative Properties of Solutions
 solution properties that depend upon number of particles dissolved (i.e.- concentration) in a
given solvent but not on the kind of solute
 four colligative properties:
1. vapour pressure reduction
2. boiling point elevation
3. freezing point depression
4. osmotic pressure (related to 1.)
Changes in Vapour Pressure: Raoult’s Law
 non-volatile solute decreases vapour pressure of volatile solvent above the solution
 illustrated by experiment: 2 beakers with same volume of liquid, one pure solvent, one
with non-volatile solute in same solvent
Chem 59-110 (’02), ch. 6, Solutions
11
 in closed chamber, volume decreases in pure solvent, increases in solution beaker
 result of dynamic equilibria:
 in pure solvent, evaporation and condensation rates equal
 in solution, evaporation rate decreased due to interaction with solute, but condensation
rate same, decreased vapour pressure
 hence net transfer of liquid to solution beaker
 extent of vapour pressure lowering proportional to concentration, independent of which solute
 quantitatively described by Raoult’s Law
PA = X A PAo
PA = vapour pressure of soln
XA = mol fraction of the solvent
PAo = vapour pressure of pure solvent
 example:
 PH2Oo = 17.5 mm of Hg at 20C. If glucose added so that XH2O = 0.8, then:
 PH2O = 0.8 x 17.5 = 14 mm of Hg
 similarly in Example 6.11 calculating vapour pressure of benzene with and without
naphthalene in solution
 alternatively, if vapour pressures of pure solvent and solution known could calculate mol
fraction solvent and, thence, molar mass of solute
 Example: 0.144 g of a compound of unknown molar mass dissolved in 10.00 g benzene
(C6H6); vapor pressure of solution is 94.35 mm Hg at 25oC, compared to 95.00 mm Hg for
pure benzene at the same temp. Calculate molar mass of the unknown.
 moles of benzene:
 1 mol 
10.00 g C6 H6 
 = 0.1280 mol C6 H6
 78.115 g 
P
0.1280 mol benzene
X benzene = benzene
= 0.9932 =
o
Pbenzene
0.1280 mol benzene + ? mol unknown
 solve:
8.764 x 10-4 mol unknown; 
0.144 g
= 164 g / mol
8.764 x 10-4 mol
 Raoult’s Law is for ideal solutions, like treatment of ideal gases earlier (not concerned with
deviations)
 what if two volatile components?
Chem 59-110 (’02), ch. 6, Solutions
12
 application of Raoult’s Law is basis of distillation (later in chapter)
Boiling-Point Elevation
 liquid boils when its vapour pressure = external pressure on its surface (can escape)
 non-volatile solutes lower vapour pressure of solvent, higher temp required to increase
number of solvent particles in gas phase
 hence, solution boils at higher temp than pure solvent, Fig. 6.11
 note, vapour pressure of solution lower than that of pure solvent at all temps
 increase in boiling point proportional to number of solute particles per mol of solvent, i.e.proportional to molality of particles
t bp = K bp msolute
Kbp = molal boiling point elevation constant for the solvent
 see Table 6.2 for Kbp’s (and Kfp’s; see below) of some solvents (note: not in C, as below, but
in K; here these are the same)
 for water, Kbp = 0.51 C/m , therefore a 1 m sucrose solution boils at 0.51 C higher than
water
 but, a 1 m solution of NaCl boils 1.02 C higher. Why?
 Example: calculate the boiling point for the system in Raoult’s Law Example above (later, do
Exercise 6.12)
 unknown is eugenol, molar mass 164.2; calculate its molality, thence boiling point
elevation
 1 mol eugenol 
-4
0.144 g eugenol 
 = 8.77 x 10 mol eugenol
 164.2 g eugenol 
8.77 x 10-4 mol eugenol
= 8.77 x 10-2 mol eugenol
0.01000 kg benzene
t bp = (2.53  C / m)(0.0877 m) = 0.222  C
boiling point = 80.10  C + 0.222  C = 80.32  C
 Example: determine molar mass of oil of wintergreen, given that a solution of 1.25 g of it in
100 g of benzene has a boiling point of 80.31 oC. (later, do Exercise 6.13)
Chem 59-110 (’02), ch. 6, Solutions
13
boiling point elevation = 80.31  C - 80.10 C = 0.21  C
t bp
0.21  C
molality of solution =
=
= 0.083 m
K bp
2.53  C / m
 0.083 mol 
moles of solute = 
.
kg solvent) = 0.0083 mol solute
 ( 0100
. kg solvent 
 100
calc. molar mass:
1.25 g
= 150 g / mol
0.0083 mol
 note beneficial aspects of boiling point elevation in “antifreeze” mixture in auto radiator
Freezing-Point Depression
 upon freezing a solution, crystals of pure solvent usually separate out, eg. pure ice in sea
water
 therefore, if Fig. 6.12 extended down to freezing point, vapour pressure of solid from solution
is same as that from pure solvent
 since vapour pressure line of liquid solution is below that of pure liquid solvent at all
temps, therefore it meets vapour pressure line of solid (i.e.- triple point) at a lower temp
 this depression of freezing point also proportional to molality (of particles)
Tfp = K fp msolute
Kfp = molal freezing point depression constant (also in Table 6.2)
 work through Example 6.14 and 6.15 on own (latter, with Fig. 6.13, illustrates dissociation in
solution)
 analogous use of freezing point depression to calculate molar mass (in lab could use any of
colligative properties, see below, but this is most common)
Osmotic Pressure
 semi-permeable membrane allowing passage of solvent, not solute
 use to separate two solutions of different concentrations - observe net movement of solvent to
solution of higher solute conc’n, Fig. 6.14
 quantitated in terms of pressure necessary to counteract it (recall ideal-gas law);
Chem 59-110 (’02), ch. 6, Solutions
14
 V = nRT
n
 =   RT = MRT
V
 - osmotic pressure in atm
n - number of moles of solute
V - volume of solution
R - gas constant (0.0821 L atm/mol K)
T - temp, K
note: proportional to molarity
 small pressure differences (to 10-3 atm) easily measured, hence used to measure
concentrations down to 10-4 M; much used for biological macromolecules
 Example 6.16
 terms: isotonic, hypotonic, hypertonic
 applications to biological cells, eg. Red blood cell crenation & lysis, pickles & carrots, edema,
food preservation
 note: this water movement from high to low water concentration is spontaneous, termed
“passive”; contrast “active” transport in living cells (requires energy)
6.6 Mixtures & Distillation

Two (or more) volatile components in solution: if ideal (Raoult’s Law for “solvent”, Henry’s
Law for “solute”), vapour pressure of components proportional to their mole fractions
 Vapour above this mixture enriched in the the more volatile one (higher Po, lower boiling
point), relative to the liquid, Fig. 6.16
 If that condensed, then re-evaporated, further enrichment
 Actually not constant T as in Fig. 6.16, but transformed to constant P, Fig. 6.17
 Many stages of this leads to fractional distillation; see column Fig. 6.18
 Egs separate O2 from N2; hydrocarbons; etc.

Non-ideal behaviour
 Constant-boiling mixtures: azeotropes
 Eg. ethanol – water, minimum-boiling at 96% ethanol
6.7 Colloidal Suspensions
 particles larger than molecules, but do not settle under gravity
= colloidal dispersions = colloids
 size range 1.0 - 200 nm
Chem 59-110 (’02), ch. 6, Solutions
15
 between “true” solutions and heterogeneous mixtures, examples aerosols, paints, milk,
mayonnaise, etc.
 aqueous colloids:
 hydrophilic: enzymes, antibodies, hemoglobin - surface studded with polar/ionic groups,
interior largely non-aqueous, eg. “globular” protein
 hydrophobic: need to be stabilized by detergents - washing, digestion of foods, metal
salts (PbCrO4, like Fig. 6.22)
 process of flocculation to bring colloidal particles together into aggregates that settle
Suggested Problems
1 – 55, odd; 61, 63
Chem 59-110 (’02), ch. 6, Solutions