Applications of Aqueous Equilibria III:The Basis of solubility
Reading: Moore chapter 15, sections 15.1-15.6
Questions for Review and Thought: 26, 28, 30, 32, 38, 42, 43, 50.
Key Concepts and Skills:
•
definition of enthalpy of solution, entropy of solution, hydration, miscible,
immiscible, lattice energy, saturated solution, unsaturated solution, supersaturated
solution, solubility, weight percent
•
Given the structure of a compound, predict solubility in a given solvent; use
Henry’s Law to assess the solubility of a gas in a liquid.
Lecture Topics:
I.
Review of solute-solvent interactions:
solute + solvent  solution Ksp
A solute is the material dissolved in a solution; the solvent is the medium in which a
solute is dissolved to form a solution. A saturated solution (Q=Ksp) is a solution whose
solute concentration equals its solubility; a dynamic equilibrium is established. An
unsaturated solution (Q<Ksp) is one in which the solute concentration is less than its
solubility. Solutions containing more than the equilibrium concentration of a solute are
supersaturated (Q>Ksp) precipitation is spontaneous (but not necessarily rapid)
•Principles of “Like dissolves Like”:
1. Substance of similar non-covalent forces (intermolecular forces, IM) are likely
soluble in each other
2. Solutes do not readily dissolve in solvents whose non-covalent (IM) forces are
quite different from their own.
3. Strong solute-solvent attractions favor solubility; stronger solute-solute or
solvent-solvent attractions reduce solubility.
•Strength of intermolecular forces:
Ion-dipole>dipole-dipole (hydrogen bonding)>dipole-induced dipole> London forces
(dispersion forces, induced dipole-induced dipole)
Case study: “Oil (hydrocarbons) and water don’t mix”
The Intermolecular (attractive) forces holding H2O in the liquid phase: dipole-dipole
(hydrogen bonding)
For hydrocarbons (CnHn+2), London forces hold molecules in liquid phase. London
attractive forces exist when a momentary dipole in one molecule induces a dipole in
another molecule. The strength of the attraction depends on how readily electrons in a
molecule can be polarized; larger molecules have more electrons and are more
“polarizable”
•Compare boiling points as a gauge of the strength of intermolecular forces:
Substance
boiling point
H2O
100°C
hydrogen bonds
C5H12
36°C
London forces
C6H14
69°C
London forces
C7H16
98°C
London forces
C8H18
126°C
London forces - (one can see the additive effect of
multiple weak attractions (induced dipole-induced dipole)
Liquids that dissolve in each other are said to be miscible.
Examples: CCl4 and C6H14 Forces present in both: London forces
CH3OH and H2O; CH3CH2OH and H2O
Forces present in both: hydrogen-bonding
(strong); forces also present in CH3OH and CH3CH2OH: London forces (weak, since only
C1 or C2; see bp trend above)
However: CH3CH2CH2CH2CH2CH2CH2CH2OH and H2O are immiscible, that is, they
form two separate phases which do not mix. The boundary between the two phases is the
interface. Octanol and H2O: Forces present in both: hydrogen bonding; forces present in
octanol: London forces, stronger than hydrogen bonding because C8 (see bp trend above).
• In general, as the hydrocarbon “tail” of the alcohol gets longer, that alcohol becomes
less soluble in water; the additive London forces between the hydrocarbon portions of the
alcohol molecules become sufficiently large that they are stronger than any hydrogenbonding interaction that could be accomplished with water. Solubility in water would
mean that the London forces would be broken up, and this is not favored.
•Immiscible: C6H14 and H2O: Forces present in H2O: hydrogen bonding; forces present
in C6H14: London forces; since different IM forces present in each, they are not soluble in
each other. Physical basis: in mixing hydrocarbon & water, you disrupt the strong dipoledipole interactions in water and replace them with weaker dipole-induced dipole
interactions in the solution. Thus, with such different IM forces for each, water and
hydrocarbon molecules tend to avoid each other and stick to themselves, resulting in
phase separation.
•Immiscible: fats (triacylglycerols) and fatty acids in H2O
Fats and fatty acids contain long (C12-C22) hydrocarbon tails that render them insoluble in
water. Are niacin, vitamin C and vitamin A soluble in fat, or soluble in water?
O
O
O
OH
O
O
O
O
O
HO
OH
OH
O
N
OH
O
OH
niacin
H 3C
vitamin C
CH3
CH3
CH3
OH
CH3
vitamin A
fatty acid
fat (triacyl glycerol)
II.
Thermodynamics of mixing processes
Mixing of solid solutes in solvents: Enthalpy of mixing (ΔHsoln) consists of a three-step
process:
1.) breaking solvent-solvent interactions (for H2O, this involves breaking hydrogen
bonds) ΔH1>0
2.) break solute-solute interactions (for ionic solids such as NaCl, this involves breaking
ionic bonds and supplying the lattice energy) ΔH2>0
3.) establishing solvent-solute attractive forces (for NaCl in H2O this means establishing
ion-dipole forces, hydration energy) ΔH3<0
•ΔHsoln=ΔH1+ΔH2+ΔH3>0 endothermic dissolution;
example:
NH4Cl (s)  NH4+(aq) + Cl- (aq)
ΔHsoln > 0
•ΔHsoln=ΔH1+ΔH2+ΔH3<0 exothermic dissolution
example:
HCl (g) + H2O(l)  H3O+ (aq) + Cl-(aq) ΔHsoln<0
CaCl2(s)  Ca2+(aq) + 2Cl-(aq)
ΔHsoln<0
For ionic compounds, the competition between lattice energy and hydration energy
determines the overall enthalpy of mixing!
Entropy of solution (ΔS soln): when solutes and solvents mix to form solutions, the
increased disorder that occurs in solution is a natural tendency. Except for gases
dissolving in liquids, ΔSsoln >0. Entropy can be a strong driving force for dissolution.
•For NH4Cl (s)  NH4+(aq) + Cl- (aq), although ΔHsoln > 0, ΔSsoln >>0, since disorder is
greatly increased by disrupting the hydrogen-bonds of water and by separating the ions in
solid NH4Cl. Thus overall dissolution is a favorable process. Because of the
endothermicity of dissolving such solid solutes in solvents, solubility increases with
increasing temperature.
For HCl (g) + H2O(l)  H3O+ (aq) + Cl-(aq) ΔSsoln<0 since for dissolving a gas in a
liquid the gas becomes more ordered relative to the gas phase. However, ΔH soln<<0
because of the establishment of strong ion-dipole interactions with water, thus overall
dissolution is favored. Because of the exothermicity of dissolving gases in solvents, one
would expect gas solubility to decrease with increasing temperature. ΔGsoln=ΔHsoln TΔSsoln
•There are two contributions to ΔSsoln 1.) the disorder achieved upon breaking the lattice
positions in the solid ΔSlattice>0, and 2.) the order resulting from orientation of solvent
(H2O) molecules around the solute. ΔS2<0. ΔSsoln=ΔSlattice+ΔS2
•For ΔSsoln=ΔSlattice+ΔS2>0, dissolution is entropy-favored (usually for +1 and –1 ions, eg
NaCl, NH4Cl, etc.)
•For ΔSsoln=ΔSlattice+ΔS2<0, dissolution is entropy-disfavored (usually for +2, -2 and
higher ions, Mg2+, Ca2+ salts)
example: CaCl2(s)  Ca2+(aq) + 2Cl-(aq) Ca2+ ions have a strong ordering effect on
water, orienting several layers of H2O molecules around the individual ions, thus
ΔSsoln<0; however, ΔHsoln<0, so dissolution process is favored overall. Solubility of this
salt therefore decreases with increasing temperature!
•For NaOH (s)  Na+(aq) + HO-(aq)
favored at all temperatures!
ΔHsoln<0, ΔSsoln>0 Thus dissolution is always
III.
Henry’s law
The solubility of any gas in a liquid increases as the (partial) pressure of the gas increases
Sg = kHPg where Sg is the solubility of the gas, moles/liter; Pg is the (partial) pressure of
the gas, and kH is Henry’s constant which depends on solute, solvent and T.
Example: a gas in contact with water contains 1 mol% CO2. The temperature is 20°C and
the total gas pressure is 2 atm. What is the solubility of CO2 in water? (kHCO2 =4.45x10-5
mol/LmmHg)
Pg = (0.01)(2atm)(760 mmHg/atm) = 15.2 mmHg
Sg = kHPg = (4.45x10-5 mol/LmmHg)(15.2 mmHg)= 6.764 x 10-4 moles/Liter
Application: Soda cans and soda “fizz”; once a can is opened, decreased partial pressure
of CO2 leads to decreased solubility of CO2in H2O, eventually resulting in “flat” soda.
Additional Problems:
1. Calculate the mole fraction of each component in a mixture composed of the
following substances:
a. 10g C2H5OH, 10g H2O, and 10g (CH2OH)2
b. 10 g CCl4, 20 g C6H6, 30 g C10H22
2. Another form of Henry’s Law is the following: Pg= KX, where X is the mole
fraction of the gas in solution, Pg is the partial pressure of the gas above the
solution, and K is the Henry’s law constant in mmHg. Hydrogen is soluble in
water at 20°C only to the extent of 0.000164 g/100g at atmospheric pressure. Find
the mole fraction of hydrogen in water and the Henry’s Law constant for
hydrogen in torr. The mole percent of hydrogen in air is 0.00005.
3. Rank the following substances in terms of INCREASING normal boiling point:
a. CH3CH2CH2CH2OH
b. CH3CH2OCH2CH3
c. CH3CH2CH2CH3 d.
CH3CH2(CO)CH3
4. Which pair will form the strongest hydrogen bonds with each other?
a. C2H5OH and CH3OCH3
b. HOCH2CH2OH and H2O
c. HOCH2CH2OH and CH3OH
d. CH3OCH3 and H2O
5. Which statement(s) is (are) correct?
a. The C-Cl bonds in CCl4 are polar, and CCl4 molecules are polar.
b. The C-Cl bonds in CCl4 are polar, but CCl4 molecules are non-polar
c. The C-Cl bonds in CCl4 molecules are polar
d. The C-Cl bonds in CCl4 molecules are non-polar
6. Which of the following explains why bromine is soluble in Carbon tetrachloride?
a. Both substances are liquids
b. Both substances have densities that are nearly equal
c. Both substances are made up of non-polar molecules and have similar
intermolecular forces
d. Both Br and Cl have high electronegativities
7. For each of the following pairs, select the substance that would be predicted to
have the higher normal boiling point:
a. CH2Cl2 or CH2I2
b. SO2 or CO2
8. Would you expect calcium sulfate (CaSO4) to have a positive or negative ΔSsoln?
Explain. Ksp (CaSO4)=2.4 x10-5 at 25°C. If ΔH soln <0 at 25°C, will an increase or
decrease in temperature result in greater solubility?