SOLUTIONS
We all want them – but few of us have them.
Solutions
A solution is a completely homogeneous
mixture.
 Solutions have one or more solutes (dissolvee)
and a solvent (dissolver).
 When a substance dissolves it is soluble
 Liquids that mix are miscible, those that don't
are immiscible

Characteristics of solutions
Fluid solutions are clear (transparent).
 Dissolving is a physical change. Dissolved
substances retain their chemical identities.
 The physical attributes of the solvent and solute
will be altered (physical state, mp, etc).

Types of solutions
Solutions can involve any state of matter.
 Liquid/liquid
 Solid/solid
 Gas/gas
 Solid in liquid
 Gas in liquid

Aqueous solutions - ionic

Solvation of ionic substances
Aqueous solutions - ionic

Solvation involves dissociation of the solid ionic
solute.
Aqueous solutions - ionic
The energy needed to break strong ionic
bonds comes from the energy liberated when
solvent/solute interactions are formed.
 Water, being polar, has partially positive and
negative regions which are attracted to the
positive and negative ions of the ionic solute.

Aqueous solutions - ionic

Ionic salts are insoluble in non-polar solvents
Aqueous solutions - molecular
Solvation of molecular substances.
 Molecular substances do not have strong
interactions to overcome.

 Dissociation
in solution does not occur.
 Neutral molecules are separated from each other.

Molecules with polar regions will be soluble in
water.
 Examples
– ethanol (CH3CH2OH), sucrose
Aqueous solutions - molecular

Sucrose
Aqueous solutions - molecular
Polar substances are soluble in water because
the polar water molecule can attach to the
polar portions of the solute molecule.
 Nonpolar molecules are not soluble in water
because water is more attracted to itself than
the potential solute.
 Nonpolar molecules usually do not contain
oxygen – example methane (CH4).

Aqueous solutions - molecular



Molecule polarity depends on bond polarity and
geometry.
Bond polarity depends on electronegativity
differences.
Miscibility – Liquid solutes
Liquids which mix in all proportions are said to be miscible.
 The liquid present in larger amounts is considered the
solvent.
 When mixing unlike liquids, volumes do not necessarily add.

Solubility

Rules of Thumb
 Like
dissolves like
 Polar dissolves polar, nonpolar dissolves nonpolar

Measuring solubilty
 Accepted
unit is g/100 g solvent.
 Temperature must be specified – may be 0º C, may
be 20º C, may be 100ºC
Solubility graph
Concentration

Per cent concentration
 Weight
% (w/w)
 w/w%
= (mass solute)/(mass solution)x100%
 mass solution = mass solute + mass solvent
 Volume
 v/v%
% (v/v)
= (volume solute)/(volume solution)x100%
 volume solution ≠ volume solute + volume solvent!!
% concentration problem

A solution is prepared by mixing 1.00 g
ethanol (C2H5OH) with 100.0 g water to give
a final volume of 101 mL. Calculate the w/w
and v/v % concentrations. (The density of
ethanol is 0.789g/mL)
Solution to concentration problem
w/w:
[1.00g eth/101g soln]x100% = 0.990%w/w
 v/v:
volume of ethanol is 1.00g/0.789g/mL =
1.27mL
[1.27mL/101mL]x100% = 1.26%v/v

Molarity
Unit: moles/(liter solution)
 Symbol: M
 Making a solution: Solute is measured,
dissolved in a small amount of water and
diluted to desired volume in a volumetric flask.

Molarity examples
What mass of KCl is necessary to make 300
mL of a 0.15 M solution?
 Solution: 0.15 M x 0.300 L = 0.045 mol KCl
molar mass of KCl = 74.6 g/mol
0.0450 mol x 74.6 g/mol = 3.4 g KCl

Molarity example
You have 458 mL of a 0.29 M solution of
sodium hydroxide. What mass of sodium
hydroxide is contained therein?
 Solution: 0.458 L x 0.29 M = 0.13 mol NaOH
0.13 mol x 40. g/mol = 5.2 g NaOH

Molality
Unit: (moles solute)/(kg solvent)
 Symbol: m
 Examples: How would you prepare a 0.17 m
solution of sodium phosphate using 800. mL
water?
 Solution: 0.17 mol/kg x .800 kg = 0.14 mol
Molar mass of Na3PO4 =163.9 g/m
0.14 mol x 163.9 g/mol = 23g Na3PO4

Molality example
How would you make 1200 g of a 0.235 m
calcium chloride solution?
 Solution: CaCl2: 111 g/mol
 1.200 kg = mass solvent + mass solute
 mass solute = x
 0.235mol CaCl2 (111) =
x/111g/mol (111)
1 kg solvent
(1.200 kg – x)
 Multiply both sides by 111

Molality example solution
0.0261kg CaCl2 =
x_____
1 kg solvent
(1.200 kg – x)
(Note that the mass of the solute must be in kg)
 .0261(1.200 – mass solute) = mass solute
 .03132–.0261(mass solute) = mass solute
 .03132/1.0261 = mass solute
 = 0.0305 kg CaCl2 (30.5g)

Mole Fraction





XA = nA/(nA + nB)
No unit (all units cancel)
Example. What is the mole fraction of HCl in
concentrated hydrochloric acid (37%w/w)?
37% = 37g HCl/(37g HCl + 63g H2O)
Change all masses to moles
(37/36.5)/(37/36.5 + 63/18)
XHCl = 0.22 (22% of the particles in concentrated HCl
are hydrogen chloride, neglecting dissociation)
Factors affecting solubility

Temperature
 Increase
in temperature usually means increased
solubility for a solid or liquid solute.
 Supersaturation
A
solution carefully cooled below temperature where
solute is soluble
 Disturbing solution or adding seed crystal causes rapid
crystallization with attendant evolution or absorption of
heat
 Gases
become less soluble with increasing
temperature.
Factors affecting solubility

Nature of solute and solvent
 Relative

polarities (“like dissolves like”)
Temperature – higher temperature means higher
solubility (solids) or lower solubility (gases)

Factors affecting rate of solution
 Temperature
– higher temperature means faster
dissolving
 Particle size – smaller particles mean faster solution
 Stirring – stirring increases solution rate
Factors affecting solubility

Pressure
 Pressure
does not affect the solubility of solid or
liquid solutes.
 Gases become more soluble with increasing
pressure
 Henry's Law: Gas solubility is directly related to
pressure
P = kC
P = pressure of dissolved gas over solution k = constant characteristic of
system
C = concentration (solubility)
Henry’s Law problem

A soft drink is bottled so that a bottle at 25ºC
contains CO2 at a pressure of 5.0 atm over the
liquid. Assuming that the partial pressure of
CO2 in the atmosphere is 4.0x10-4atm,
calculate the equilibrium concentration of CO2
in the soda both before and after the bottle is
opened. The Henry's Law constant for CO2 in
aqueous solution is 32 Latm/mol at 25ºC.
Solution to Henry’s Law problem
Before opening: P = kC; k = 32 Latm/mol,
and P = 5.0 atm.
 So CCO2 = P/k = 5.0atm/32 Latm/mol
= 0.16M
 After opening: P = kC; k = 32 Latm/mol, and
P = 4.0x10-4 atm.
 So CCO2 = P/k = 4.0x10-4atm/32 Latm/mol
= 1.3x10-5M

Reactions in solution

Precipitates and ionic equations
 Overall
equation
Pb(NO3)2(aq)+2KI(aq)PbI2(s)+2KNO3(aq)
 Ionic
equation
Pb+2(aq)+2NO3-(aq)+2K++2I-(aq)PbI2(s)+2K++2NO3-(aq)
 Net ionic equation – eliminate spectator ions
Pb+2(aq) + 2I-(aq) PbI2(s)
Reactions in solution

Other reactions may be driven by the
formation of a gas.
2HCl(aq) + Na2CO3(aq)  CO2(g) + H2O(l) + 2NaCl(aq)
overall equation
2H++2Cl-(aq)+2Na++CO3-2(aq)CO2(g)+H2O(l)+2Na+ +2Cl-(aq)
ionic equation
2H+ + CO3-2(aq)  CO2(g) + H2O(l)
net ionic equation
Reactions in solution

Write overall, ionic and net ionic equations for
the following reaction.
zinc metal reacts with hydrochloric acid to give
aqueous zinc chloride and hydrogen gas
Zn + 2HCl(aq)  ZnCl2 + H2
Zn + 2H+(aq) + 2Cl-(aq)  Zn+2(aq) + 2Cl-(aq) + H2(g)
Zn + 2H+(aq)  Zn+2(aq) + H2(g)
Reactions in solution

Reactions can also be driven by formation of a
complex, usually brightly colored.
Cu2+(aq ) + 4 NH3(aq )  Cu(NH3)42+(aq )
0.1M Cu(II)
0.1M Cu(II) with ammonia added
Reactions in solution

Iron solutions form a complex with thiocyanate
(SCN-)
Fe+3 + SCN-  FeSCN+2
0.1M Fe+3
0.1M Fe+3 with thiocyanate added
Reactions in solution
Formation of a small covalent molecule will
also drive a reaction.
2KOH(aq) +H2SO4(aq) K2SO4(aq) +H2O
 Use the solubility table (Appendix D) to decide
how to write reaction equations for
precipitation reactions.
 K2S(aq) + Pb(NO3)2(aq)  PbS(s) + 2KNO3(aq)
 NaOH(aq) + CuSO4(aq)  Na2SO4(aq)+Cu(OH)2(s)

Colligative properties
From Latin “colligare”, to bind together
 Colligative properties depend on the number
of solute particles present

 Molecular
substances give one mole of particles
per mole substance
 Salts give more than one mole particles per mole
substance because of dissociation
 Number of particles per formula unit = i (van’t Hoff
factor)
Colligative properties

Vapor pressure lowering
 Solute
particles take the place of some of the
solvent particles at the surface
 Vapor pressure of liquid is lowered by presence of
a solute
 Extent of lowering depends on number of solute
particles present
Colligative properties

Boiling point elevation
 Solutions
have higher boiling points than pure
solvents. This is true with solid solutes and heavier
liquid solutes.
 Other liquid solutes may form azeotropes, which
are mixtures with lower boiling points than either
solute or solvent – example 95% ethanol/water.)
 Solutes raise the boiling point of liquids because
they lower the vapor pressure. When Pv = Pa,
vaporization (boiling) occurs.
Colligative properties

Boiling point elevation depends on the number
of particles present. (van't Hoff factor, i)
 Sodium
chloride elevates the bp of water twice as
much per mole as sucrose, for it makes two particles per
mole. (i = 2 for dilute solutions)
NaCl  Na+ + Cl-
Colligative properties

Freezing point depression
 Solutes
lower the freezing point of liquid solvents.
Solutes interfere with solvents’ ability to associate as a
solid and crystallize.
 Freezing point depression also depends on number of
particles.

Osmosis and osmotic pressure
 Solutions
of different concentrations on either side
of a semi-permeable membrane exert a net
pressure toward the more concentrated solution
Colligative properties

Solvent moves so as to dilute the more
concentrated solution.
Osmotic pressure is the pressure exerted by the
greater height of solution on the concentrated side.
Colligative properties

Reverse osmosis is used to purify water.
Colligative properties

Calculations
 Colligative
property calculations use molality as the
unit of concentration, for it does not depend on
volume.
 Colligative properties vary directly with
concentration.

BP elevation:
 Tb
= ikbm, where kb = the boiling point elevation
constant for that liquid.
Colligative properties
kb for water is 0.51ºC/molal
 FP depression:

 Tf
= ikfm
 kf for water is 1.86ºC/molal

Compare the freezing points of 1.5 molal
aqueous solutions of NaCl and CaCl2.
Tf = (2)1.86ºC/m (1.5m) = 5.6ºC. FP = - 5.6ºC
 CaCl2: Tf = (3)1.86ºC/m (1.5m) = 8.4ºC FP = - 8.4ºC
 NaCl:
Colligative properties

Molecular weight calculations
 Solving
for molality can yield the molecular weight of a
substance if the mass is known.

Example. 1.235 g of a molecular substance
dissolved in 100.0 g benzene lowers the
freezing point of benzene by 0.468ºC. What
is the molecular weight of the substance? kf for
benzene is 5.12 ºC/molal.
Colligative properties

Solution:
Tf = ikfm
m = mol/Kg solvent; mol = mass/mm
Tf = ___ikfmass___
(mm)(Kg solvent)
mm = ___ikfmass___
(Tf)(Kg solvent)
mm = _(1)(5.12ºC/m)(1.235g)____ = 135g/mol
(0.100Kg benzene)(0.468ºC)
Nonhomogeneous mixtures

Suspensions
larger than 103 nm (1 m or 10-6 m)
 Particles will settle out on standing
 Often opaque
 Examples: paint, muddy water, orange juice
 Particles

Colloids
 Particles
between 1 nm and 1m
 Particles will not settle out on standing
Nonhomogeneous mixtures

May appear clear, but exhibits Tyndall effect

Examples: fog, blood, milk, soapy water