Properties of Solutions

Chem 104
The Solution Process
 A solution is formed when one substance (the
solute) disperses uniformly throughout
another (the solvent).

 The mixing of gases is a spontaneous

process, that is, it occurs of its own accord
without input of energy from outside the
system.
Spontaneous mixing of gases
Entropy – A Driving Force
 Entropy (S) is a thermodynamic property of all
substances that is proportional to their degree of
disorder.
 A perfect crystal at 0 K, whose atoms are regularly
arranged in a perfect lattice and are motionless,
has an entropy of zero.
 Gases have large positive entropies because their
molecules are highly disordered and in constant
motion at high speeds.
Entropy
 The formation of a solution disperses
molecules, atoms, or ions of one kind
throughout a second substance, which
generally increases the disorder and results
in an increase in the entropy of the system.

 Thus entropic factors almost always favor

formation of a solution.
Enthalpy – A Driving Force
 Energy is required to overcome the
intermolecular interactions in a solute,
which can be supplied only by the new
interactions that occur in the solution,
when each solute particle is
surrounded by particles of the solvent
in a process called solvation.
Intermolecular Interactions
 Allof the solute–solute
interactions and many of the
solvent–solvent interactions must
be disrupted for a solution to
form.
Solution Formation -- Enthalpy
 Because enthalpy is a state function, we can
use a thermochemical cycle to analyze the
energetics of solution formation. The process
occurs in three discrete steps, indicated by
ΔH1, ΔH2, and ΔH3.
Enthalpy of Solution
 The overall enthalpy change in the formation
of the solution (ΔHsoln) is the sum of the
enthalpy changes in the three steps:

 ΔHsoln = ΔH1+ΔH2+ΔH3
Solution Formation
Solvation Process - Exothermic
 When a solvent is added to a solution, steps 1 and
2 are both endothermic because energy is required
to overcome the intermolecular interactions in the
solvent (ΔH1) and the solute (ΔH2).

 Because ΔH is positive for both steps 1 and 2, the

solute–solvent interactions (ΔH3) must be stronger
than the solute–solute and solvent–solvent
interactions they replace in order for the
dissolution process to be exothermic (ΔHsoln<0).
Solvation Process - Exothermic
 ΔH1 and ΔH2 are both positive because it
requires energy to pull molecules away from
each other.

 That energy cost is due to the intermolecular

forces present within any solute or solvent.

 The forces acting between molecules such as

CH3Cl are largely van der Waals and dipole-
dipole interactions.
Solvation Process - Exothermic
 Some molecules that contain O-H, N-H, or
F-H bonds can form hydrogen bonds that are
relatively strong intermolecular forces.
 Ions of opposite charge, such as in a crystal

of NaCl, are attracted to each other because

of electrostatic forces.
 Each of those forces increase with decreasing

distance. It costs energy to pull molecules

and ions away from each other.
Solvation Process - Exothermic
 When the expanded form of the
solvent and the solute are combined
to form a solution, energy is released,
causing ΔH3 to be negative.

 The solute and solvent can interact

through the various types of
intermolecular forces.
Solvation Process - Endothermic
 A positive value for ΔHsoln does not mean that a solution will
not form. Whether a given process, including formation of a
solution, occurs spontaneously depends on whether the total
energy of the system is lowered as a result. Enthalpy is only
one of the contributing factors.

 A high ΔHsoln is usually an indication that the substance is not

very soluble. Instant cold packs used to treat athletic injuries,
for example, take advantage of the large positive ΔHsoln of
ammonium nitrate during dissolution (+25.7 kJ/mol), which
produces temperatures less than 0°C
Solution Process -Summary
 Solutionswill form only when the
energy of interaction (considering
both enthalpy and entropy)
between the solvent and solute is
greater than the sum of the
solvent-solvent and solute-solute
interactions.
 Factors Affecting Solubility –Like
Dissolves Like
 If a non-polar molecule, such as oil, is mixed with a
polar molecule like water, no solution forms.
 Water's solvent-solvent intermolecular interactions
are mostly hydrogen bonds and dipole-dipole while
oil has only van der Waals.
 Water can satisfy its hydrogen bonds and become
stabilized by dipole-dipole interactions only when
near other water molecules. That is why such a
solution will never form between oil and water.
 Only when the solute and solvent molecules have
several common structural features such as their
polarities will a solution form.
Like Dissolves Like
 Solubility is a continuum.

 Like forces (solvent) dissolve like forces

(solute)

 Polar liquids tend to dissolve in polar

solvents.

 Non-polar liquids tend to be insoluble in

polar liquids.
Solubility Terms
 Solubility – maximum amt of solute that can
dissolve at a given temperature.
 Saturated – is at equilibrium with the

undissolved solute.
 Unsaturated – dissolved less that the amt

needed at that temperature.

 Miscible – liquids that mix is all proportions.
 Immiscible – liquids that do not dissolve in

one another.
 Factors Affecting Solubility --
Pressure
 Solid and liquid solubility are not appreciably
affected by pressure.

 The solubility of a gas in any solvent is

increased as the partial pressure of the gas
above the solvent increases; a direct
relationship.

S = kPg Henry’s Law

g
Effect of Pressure on Gas Solubility
Solubility Vs Partial Pressure
Reducing Partial Pressure
Factors Affecting Solubility – Temperature

 The solubility of most solid solutes in water

increases as the solution temperature
increases.

 The solubility of gases in water decreases

with increasing temperature.
Solubility of Gases Vs Temperature
Solubility of some Ionic Solids
Solution Concentration
 Solution concentration can be expressed
numerous ways:
◦ Mass percent (pph)
◦ ppm
◦ ppb
◦ molarity
◦ Molality

◦ The terms dilute and concentrated are relative

terms regarding concentration.
Mass Percent

mass of the solute

 Mass % = ----------------- x 100
total mass of sol’n
ppm (106) or ppb (109)
 ppm and ppb are used to express
concentration of very dilute solutions.

mass of solute
 ppm = ------------------ x 106

total mass of sol’n

1 ppm = 1mg/litre
Molarity (M)

moles of solute
 Molarity = ----------------

litres of solution
Molality (m)
 A concentration unit that is not temperature
dependant.
 It is determined as the moles of solute per

kilogram of solvent.

moles of solute
 Molality = -------------------

kilograms of solvent
 Molarity Vs Molality
 Molaritydepends on the volume of
solution.

 Molality depends on the mass of

solvent.
Mass Percent
 Calculate the mass percent of sodium
hypochlorite in commercial bleach, if 1.00
grams of NaOCl (the active ingredient in
bleach) is dissolved in 19.05 grams of
solution.
Mass Percent
 Calculate the grams of NaOCl (5.25% by
mass) in 245 grams of a commercial bleach
solution.
ppm and ppb
 The solubility of AgCl is 0.008 grams/100
grams of solution. What is this concentration
in ppm?

If 35 grams of ethanol is dissolved in 1150

kilograms of solution. What is the
concentration of ethanol in parts per billion
(ppb)?
Molality
 Calculate the molality of each of the following
solutions:a. 2.89 g of NaCl dissolved in 0.159
L of water (density of water is 1.00 g/mL)
0.311 molal NaCl

b. 1.80 mol KCl in 16.0 mol of H2O (6.25 molal KCl)

c. 13.0 g benzene, C6H6 in 17.0 g CCl4

(9.80 molal C6H6)
Colligative Properties
 Lowering the freezing point or raising the
boiling point are physical properties of
solutions that depend on the quantity
(concentration) but not on the kind of solute
particles.
 These are colligative properties

 Colligative – depending on the collection

Boiling Point (BP) Elevation
 IncreaseBP is dependant on the
molality (m) of the solute.

 The BP elevation is proportional to the

total concentration of the solutes
involved whether molecular or ionic.
 BP Elevation
 Represented by ‘i’ the van’t Hoff
factor.

 For non-electrolytes ‘i’ is 1, for an

electrolyte ‘i’ depends on how the
substance ionizes.

 ‘i’ for NaCl is 2, where CaCl2 is 3

 BP Elevation
 The change in BP for a solution:

 ΔT = iKbm
b

 ‘i’ is van’t Hoff factor

 K is the molal boiling point-elevation
b
constant
 m is molality
Freezing-Point (FP) Depression
 The change in the FP is directly proportional to
solute molality.

 ΔT = -iKfm
f

 Kf is the molal freezing-point depression

constant.
 Because the solution freezes at a lower

temperature than does the pure solvent, the

ΔTf is negative.
BP Elevation Questions
 What is the boiling point elevation when 11.4 g of
ammonia (NH3) is dissolved in 200. g of water? Kb
for water is 0.52 °C/m.

 If 15.0 grams of NaCl dissolve in 100 g of water

what is the BP elevation?

 0.64 g of adrenaline in 36.0 g of CCl4 produces a

bp elevation of 0.49 °C. What is adrenaline's
molecular weight? Kb is CCl4 is 5.02 oC/m
FP Depression Questions
 What is the freezing point depression when
62.2 g of toluene (C7H8) is dissolved in 481 g
of naphthalene? The Kf for naphthalene is
7.00 °C/m.

 What is the freezing point of a solution

prepared by adding 140.0 g trichothecin
(C19H24O5) to 0.746 kg of benzene? The
freezing point of pure benzene is 5.5 °C. The
Kf for benzene is 5.12 °C/m.
FP Depression Questions
 What is the freezing point of a solution
prepared by adding 239.0 g of copper(II)
sulfate pentahydrate to 4.00 liters of water?
The Kf of water is 1.86 °C/m.
Vapour Pressure
 When a liquid evaporates, the gaseous
molecules created escape into the air. If the
liquid is in a closed container, the gaseous
molecules created will not escape but remain
above the liquid.
 These evaporated particles created a pressure

above the liquid, this is known as VAPOUR

PRESSURE.
Raoult’s Law
 If a solid is dissolved into the liquid, solution is
created. The vapour pressure of the solution is
lowered by the addition of the solute.
 Raoult’s Law explains how the vapour pressure of
liquid is altered by the addition of a solute.
P = ( Xsolvent) (Psolvent)
solution
 Psolution – the vapour pressure of the solution
 Xsolvent – the mole fraction of the solvent in solution
 Psolvent – the vapour pressure of the pure solvent at
standard or stated conditions.
Vapour Pressure (VP) Formula
Question
 What is the vapour pressure of a solution at
25oC containing 3.5 moles of glucose in 10.0
moles of water? ( the vapour pressure of pure
water at 25oC is 23.8 torr
VP Question
 What is the vapour pressure of a solution at
25oC when 25.5 grams of glucose (C6H12O6) is
dissolved in 212 grams of water? (the VP of
pure water at this temperature is 23.8 torr)
VP Question
 The vapour pressure of an aqueous solution
is found to be 24.90 mm Hg at 25oC. What is
the mole fraction of solute in this solution?
The vapour pressure of water is 25.756 mm
Hg at 25oC.
VP Question
 At 29.6oC, pure water has a vapour pressure
of 31.1 torr. A solution is prepared by adding
86.8 grams of ‘Y’, a nonvolatile non-
electrolyte to 350.0 grams of water. The
vapour pressure of the resulting solution is
28.6 torr. Calculate the molar mass of ‘Y’.
VP Question
 The vapour pressure of pure water is 23.8
mm Hg at 25oC. What is the vapour pressure
of 2.50 molal C6H12O6?
VP Question
 How many grams of nonvolatile compound B
(molar mass – 97.80 g/mol) would need to be
added to 250.0 grams of water to produce a
solution with a vapour pressure of 23.756
torr? The vapour pressure of water at this
temperature is 42.362 torr.