Solubility & Intermolecular Forces
John W. Moore
Conrad L. Stanitski
Peter C. Jurs
http://academic.cengage.com/chemistry/moore
Chapter 14
The Chemistry of Solutes and
Solutions
Solution = homogeneous mixture of substances.
It consists of:
• solvent - component in the greatest amount.
• solute - all other components (may be >1).
Solvent-solute interactions determine if a substance
will dissolve in a particular solvent.
Stephen C. Foster • Mississippi State University
Solubility & Intermolecular Forces
Solutions:
• Exist in all 3 physical states.
• Can be mixtures of solids, liquids and gases.
Type of Solution
Gas in gas
Gas in liquid
Gas in solid
Liquid in liquid
Solid in liquid
Solid in solid
Examples
Air
Carbonated drinks.
Hydrogen in Pd metal.
Motor oil, vinegar.
Ocean water, sugar-water.
Bronze, pewter, 14K gold.
Solute-Solvent Interactions
Like dissolves like
If solute and solvent intermolecular forces are:
• Similar… the pair will be soluble.
 polar dissolves polar;
 non-polar dissolves non-polar.
• Dissimilar… insoluble.
Solute-Solvent Interactions
Solute-Solvent Interactions
Substances dissolve when:
solvent-solute attraction > solvent-solvent attraction,
and
> solute-solute attraction.
Miscible liquids dissolve in all proportions.
e.g. ethanol and water (both H-bonded polar liquids).
Immiscible liquids form distinct separate phases.
e.g. gasoline (non-polar) and water (polar).
NiCl2(aq)
CCl4
C7H16
and
CCl4
NiCl2(aq)
C7H16
Carefully layered
Stir…
…settle
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Solute-Solvent Interactions
Name
methanol
ethanol
1-propanol
1-butanol
1-pentanol
1-hexanol
Formula
CH3OH
C2H5OH
C3H7OH
C4H9OH
C5H11OH
C6H13OH
Solubility (g/100g H2O @20°C)
miscible
miscible
London
miscible
forces
7.9
increasing
2.7
0.6
Solute-Solvent Interactions
Alternate view
Alcohols: polar head (–OH) on a non-polar tail.
The head is hydrophilic (“water loving”)
The tail is hydrophobic (“water hating”)
Water solubility decreases as alcohols grow larger:
• Solute-solute attraction grows.
• Solute-solvent attraction stays ≈ constant.
As the tail gets bigger, it is harder and
harder to dissolve.
Enthalpy, Entropy and Dissolving Solutes
Enthalpy, Entropy and Dissolving Solutes
ΔHsoln= ΔHstep a + ΔHstep b + ΔHstep c
On mixing, solute and solvent molecules spread out
over a larger V.
• This increases entropy (S).
• Processes with ΔS > 0 tend to be product-favored.
For NH4NO3 + H2O:
• ΔHsoln > 0 (unfavorable).
• But ΔSsoln > 0 (drives solution formation).
Solubility and Equilibrium
Solutions are either:
Unsaturated
• [Solute] < solubility.
• All added solid is dissolved.
• Can dissolve more solute.
Solubility and Equilibrium
At saturation, [solute] remains constant, but the
solute is in dynamic equilibrium:
solute + solvent
solution
Solute constantly moves in and out of solution.
Saturated
• No more solute will dissolve.
• Undissolved solid is often present.
Supersaturated
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Solubility and Equilibrium
Supersaturated solutions have more than the
equilibrium amount dissolved.
They can be produced by:
• Dissolving solute at high T.
 solubility usually increases with T.
• Slowly lowering T.
 it can take a long time to occur by chance.
• Too much remains temporarily dissolved.
Dissolving Ionic Solids in Liquids
When an ionic compound dissolves in water, the ions
• overcome the forces holding them in the lattice.
• become hydrated (surrounded by water).
Lattice energy = E required to overcome the forces
holding the ions together in a crystal.
Enthalpy of hydration (ΔHhydration) =
E released as an ion becomes
hydrated.
Supersaturation
“seed” crystal
of sodium
acetate
Supersaturated
sodium acetate
solution
The solution warms…
…heat is released as crystals form
Excess sodium
acetate begins to
crystallize around
the seed
More and more
crystallizes until
a saturated
solution remains.
Dissolving Ionic Solids in Liquids
ΔHsoln= -Lattice E + ΔHhyd(cations) + ΔHhyd(anions)
ΔHsoln may be:
• positive (endothermic)
• negative (exothermic)
▪ cold packs (NH4NO3)
▪ hot packs (CaCl2)
Ionic solids are insoluble in nonpolar solvents
• Ionic compound lattice energies are large.
 need energy to break them apart.
• Nonpolar solvents cannot hydrate ions.
 do not release E to overcome the large lattice E.
Entropy and Dissolving Ionic Solids in Water
ΔSlattice> 0 when a crystal lattice breaks down.
ΔShyd< 0 when H2O molecules order around the ions.
If ΔS = ΔSlattice + ΔShyd > 0; dissolving is favored.
Temperature and Solubility
Solubility of Gases
Le Chatelier’s principle:
Gas + solvent
sat. solution
usually ΔHsoln< 0
Gas solubility almost always decreases as T increases.
+1 & -1 ion compounds are often soluble (ΔS > 0).
H2O can be highly organized around small +2 and +3
ions (ΔShyd << 0). So ΔS < 0 (favors insolubility).
e.g. CaO = slightly soluble (0.13 g/100 mL at 10 °C)
Al2O3 = insoluble.
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Temperature and Solubility
Pressure and Dissolving Gases
More gas dissolves when P of the gas above a liquid
increases.
Gas + solvent
saturated solution
Solubility of Solids
Usually
sually increases as T increases.
Le Chatelier: More gas on the reactant side…
…shift toward products (more gas in solution).
Henry’s Law
Gas solubility is directly proportional to P of the gas.
Sg = kH Pg
Solubility
Henry’s Law
Henry’s law constant
Solution Concentration: Units
Soft drinks “fizz” when opened. Calculate [CO2] when
a drink is bottled (P = 4.5 atm), and after it’s opened.
For CO2 in water kH = 3.4 x 10-2 M/atm. The partial P
of CO2 in the atmosphere = 3.1 x 10-4 atm.
Sg = kH Pg
When Pg = 4.5 atm
Sg = 3.4 x 10-2 M (4.5 atm) = 0.15 M
atm
When opened
Sg = 3.4 x 10-2 M (3.1 x 10-4 atm) = 1.1 x10-5 M
atm
Parts per Million, Billion and Trillion
% = Parts per hundred =
mass solute
x 102
mass solution
Very dilute = low mass fraction.
Units for high dilution (trace solute):
Parts per million (ppm) =
mass solute
x 106
mass solution
Parts per billion (ppb) =
mass solute
9
mass solution x 10
Parts per trillion (ppt)
mass solute
x 1012
mass solution
=
Mass fraction =
Mass Solute
Total Mass of Solution
Weight percent = Mass fraction x 100%
Example
Saline solutions, NaCl(aq), are often used in medicine.
What is the weight percent of NaCl in a solution of 4.6 g
of NaCl in 500. g of water?
4.6 g
= 0.0091
mass fraction =
500. g + 4.6 g
weight percent = 0.0091 x 100% = 0.91 %
Parts per Million, Billion and Trillion
1 ppm of solute in water = 1 mg / 1000 g of solution
Since 1L of water has a mass ≈ 1000 g.
1 ppm ≈ 1 mg/L
Similarly
1 ppb ≈ 1 μg/L
1 ppt ≈ 1 ng/L
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Molarity and Molality
Converting Units
Molarity = M = moles of solute
liters of solution
Molality is another concentration scale:
Molality = m =
moles of solute
kilograms of solvent
A mass-based unit.
• Uses solvent mass (not solution).
• It is T independent (unlike molarity).
Commercial 30.0% hydrogen peroxide has density =
1.11 g/mL at 25°C. What is its molarity?
30.0 % H2O2 = 30.0 g H2O2 in 100.0 g of solution
Moles of H2O2 =
30.0 g
= 0.8821 mol
34.01 g mol-1
Volume of solution =
100.0 g = 90.09 mL
1.11 g/mL
[H2O2] = 0.8821 mol = 9.79 M
0.09009 L
Converting Units
Converting Units
Sea water is 10,600 ppm Na+. Calculate the mass
fraction and molarity of sodium ions in sea water. The
density of sea water is 1.03 g/mL.
…10,600 ppm Na+. Calculate the molarity of Na+. Density = 1.03 g/mL.
Mass fraction
10,600 ppm = 10,600 g Na+ in 106 g of solution
Mass fraction =
Volume of 106 g of solution
= (106 g/1.03 g mL-1)
= 9.709 x 105 mL = 970.9 L
10,600 g Na+
106 g of solution
= 0.0106
Mass percent = 1.06 %
[Na+] = (461.1 mol Na+)/(970.9 L) = 0.475 M
Vapor Pressure Lowering
Solvent vapor P drops if non-volatile solute is added.
Raoult’s law:
law
P1 = X1 P°1
vapor pressure of solvent
over the solution
Molarity
Moles of Na+ in 106 g of solution
= (1.06 x 104 g)/(22.99 g mol-1)
= 461.1 mol
vapor pressure of
pure solvent
mole fraction
of the solvent
X1 = moles of 1 =
n1
total moles
(n1+n2+…)
Lower purity solvent = lower vapor P.
Vapor Pressure Lowering
A solution of urea in water has a vapor P of 291.2 torr.
The vapor P of pure water is 355.1 torr. Calculate the
mole fraction of urea in the solution.
Urea is non-volatile. Water obeys Raoult’s law:
Pwater = XwaterP°water
291.2 torr = Xwater(355.1 torr)
Xwater= 291.2/355.1 = 0.820
Xurea = 1 – 0.820 = 0.180
(Remember: X1 + X2 + … = 1)
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Boiling Point Elevation
Freezing Point Lowering
Non-volatile solutes increase the b.p. of a solvent.
Why?
A non-volatile solute lowers the f.p. of a solvent:
• Solvent vapor P is lowered.
• Higher T is needed to get the vapor P = external P.
Quantitatively
ΔTb = Kb msolute
Note: the Kb value only depends on the solvent,
the molality, m depends on the solute.
Quantitatively
ΔTf = Kf msolute
Kf is a constant for the solvent.
Ethylene glycol (antifreeze) is added to car radiators
to lower the freezing point of water (stops freezing).
Freezing Point Lowering
Boiling Point Elevation
Calculate the f.p. of an aqueous 30.0% ethylene glycol
mixture. For water Kf = 1.86°C kg mol-1.
At what T will 0.100 m aqueous solutions of urea,
NaCl and sucrose boil? For water Kb= 0.51°C kg mol-1
100.0 g of 30% mix: 30.0 g C2H2(OH)2 + 70.0 g H2O
b.p. of aq. urea = b.p. of aq. sucrose
nglycol = 30.0 g / 62.07 g mol-1 = 0.4833 mol
mglycol = (0.4833 mol / 0.070 kg) = 6.904 molal
ΔTf = 1.86°C kg mol-1(6.904 mol/kg) = 12.8 °C
ΔTb = Kb msolute = 0.51°C kg mol-1(0.100 mol/kg)
= 0.051 °C
b.p. = 100.00°C + 0.051°C = 100.05 °C
Freezing point = 0.00°C – 12.8°C = -12.8°C
Freezing points
are lowered
Colligative Properties of Electrolytes
Vapor pressure lowering, b.p. elevation, and f.p.
depression are colligative properties:
properties
• Depend upon the number of “particles” in solution.
• The type of particle is unimportant.
• 1 M sugar and 1 M urea aqueous solutions have
the same effect.
• 1 M NaCl is different.
 NaCl yields 2 particles in solution (Na+ and Cl-).
 Sugar and urea do not dissociate in solution.
0.100 m aqueous NaCl has a higher b.p…
Colligative Properties of Electrolytes
In aqueous solution:
1 mol sucrose → 1 mol particles
1 mol NaCl → 2 mol particles (Na+ and Cl-)
1 mol CaCl2 → 3 mol (Ca2+ and 2 Cl-)
Modify the formulas by replacing
msolute with
isolutemsolute
where i = the number of particles per formula unit.
= van’t Hoff factor
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Boiling Point Elevation
Colligative Properties of Electrolytes
At what T will 0.100 molal NaCl(aq) boil? Kb = 0.51°C kg mol-1
A simple correction, but only works at low [ion].
• Ions attract each other in solution.
The b.p. of 0.100 molal NaCl:
• Only act as separate units at very low [ion].
ΔTb = Kb isolutemsolute
= 0.51°C kg mol-1(2)(0.100 mol/kg) = 0.10 °C
b.p. = 100.00°C + 0.10°C = 100.10 °C
• i is smaller than expected in many solutions:
[MgSO4]
iexpected
iobserved
0.0005 M
2
2.0
0.005 M
2
1.72
0.50 M
2
1.07
Colligative Properties of Electrolytes
Osmotic Pressure of Solutions
Arrange these solutions in increasing b.p order:
0.10 m BaCl2(aq)
0.12 m K2SO4(aq)
0.12 m KBr(aq)
Semipermeable membrane
• Allows passage of small “particles”.
• Stops large “particles”.
Similar molalities, so similar deviations from iexpected
m iexpected:
BaCl2 = 0.10(3) = 0.30
K2SO4 = 0.12(3) = 0.36
KBr = 0.12(2) = 0.24
b.p order:
KBr < BaCl2 < K2SO4
Osmotic Pressure
absolute T
P=cRTi
net solvent flow
Large molecule cannot pass.
e.g. animal bladders and cell membranes.
Osmosis
Movement of solvent through a semipermeable
membrane from dilute to more concentrated solution.
Osmotic Pressure
Osmotic pressure = P that must be applied to stop
osmosis.
gas constant
molarity of solution
Ion + H2O coordination sphere
– too large to pass through
P=cRTi
particles / formula unit
Easy to remember. Very similar to the ideal gas law:
Height of the
column of
solution is a
measure of P.
Pure
water
semipermeable
bag of 5% sugar
water
P =n R T = cR T
V
Water enters
the bag,
increasing
the P…
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Osmotic Pressure
A cell can be exposed to 3 kinds of solution:
Hypertonic
Isotonic
Hypotonic
[solute]out= [solute]in [solute]out > [solute]in [solute]out< [solute]in
Net flow out.
No net flow.
Net flow in.
Reverse Osmosis
Applied
Pressure
Used to purify water
Water
Brine
Normal Osmosis
Osmotic
pressure
P
Reverse Osmosis
Water molecules cross the
Flow is reversed if P > P is
membrane diluting the brine. applied.
Flow stops when P = P is
applied.
Pure water can be separated
from brine.
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