Solutions (Chapter 12)
The Solution Process
Why do things dissolve?
-- driving force toward more random state (entropy)
-- attractive forces between solute and solvent (enthalpy)
“like dissolves like”
(ionic and polar substances tend to be water soluble)
Intermolecular Forces!!!
Intermolecular Forces in Solutions
Solubility
• the amount of a substance that will
dissolve in a given amount of solvent
• How much will dissolve?
– Units are often: g solute per 100 g of
solution
– “saturated” solution: maximum amount of
solute is dissolved
• Supersaturated: more dissolved than saturated
– miscible: soluble in all proportions
• Recall dissociation equations
– e.g. NaCl(s)  NaCl(aq) = Na+(aq) + Cl–(aq)
molecular form
ionic form
solute(undissolved) ⇆ solute(dissolved)
Solubility
• the amount of a substance that will
dissolve in a given amount of solvent
• How much will dissolve?
– Units are often: g solute per 100 g of
solution
– “saturated” solution: maximum amount of
solute is dissolved
• Supersaturated: more dissolved than saturated
– miscible: soluble in all proportions
• Recall dissociation equations
– e.g. NaCl(s) ⇆ NaCl(aq) = Na+(aq) + Cl–(aq)
molecular form
ionic form
solute(undissolved) ⇆ solute(dissolved)
Heats of Solution
•
DHsoln = DHsolute + DHsolvent + DHmix
>0
>0
<0
DHhydration
•
DHsoln is a combination of two opposing effects:
Lattice Energy (DHsolute) -- endothermic
– energy required to separate solid particles
Hydration (solvation) Energy (DHhydration) -- exothermic
– energy released as gaseous solute particles are surrounded
by solvent molecules
– e.g.
K+(g) + Cl–(g)  K+(aq) + Cl– (aq)
DHhydration = -819 kJ/mol
∴ DHsoln ≈ DHsolute + DHhydration (DHsoln can be positive or negative)
Factors Affecting Solubility
• Effect of Temperature on Solubility
– Most solids are more soluble at higher temp, most gases are
less soluble at higher temp
• Effect of Pressure on Solubility
– No significant effect for solid or liquid solutes, but major effect
with gaseous solutes dissolved in liquid solvents
Henry’s Law:
Sg ∝ Pg
gases are more soluble at higher
pressure (e.g. carbonated beverage)
or
Sg = kHPg
or
S1/S2 = P1/P2
where S = solubility, P = pressure, kH = Henry’s law constant (depends on gas)
Concentrations of Solutions I
(See Table 12.5, p 529)
• Molarity (M)
– M = moles of solute/liters of solution
• Mole Fraction (and mole percent)
– XA = moles A/[moles A + moles B + …]
– mole % = XA x 100%
– Mixtures of gases: XA ∝ nA ∝ PA (at constant temp)
where X = mol fraction, n = mol, P = pressure
so, XA = PA/Ptotal
– (i.e. the mol fraction will equal the pressure fraction!)
• Mass %, parts per million, etc
– ppm, ppb can be by mass or by volume, e.g.
mass solute
x multiplication factor
mass solution
• Multiplication factor = 100
• Multiplication factor = 106
• Multiplication factor = 109
mass %
ppm
ppb
Concentrations of Solutions II
• Weight Fraction (and weight percent)
– WFA = mass A/mass of solution
– Wt % = WFA x 100%
e.g. a 5.00% (by weight) solution of NaCl contains:
5 g NaCl in 100 g of solution (5.00 g NaCl and 95.00 g H2O)
• molality (m) -- don’t confuse it with Molarity (M)!!!
– m = moles solute/kg of solvent
– Independent of temperature
e.g. molality of above 5.00% NaCl solution?
m = [(5.00 g NaCl x (1 mole NaCl/58.4 g NaCl)]/(0.09500 kg H2O)
= 0.90 mol NaCl/kg = 0.901 m NaCl
Conversions Between Concentration Methods
Example: Commercial hydrobromic acid, HBr, is 40.0% by
weight. The density of this solution is 1.38 g/mL. Calculate
the molality, molarity, and mole fraction of this HBr solution.
•
40.0% HBr means that 100 g of solution contains:
40.0 g HBr and 60.0 g H2O
Moles HBr = 40.0 g x (1 mole/80.9 g) = 0.494 mol
Moles H2O = 60.0 g x (1 mole/18.0 g) = 3.333 mol
XHBr = 0.494/(0.494 + 3.333) = 0.129 (or 12.9 mole %)
m = moles HBr/kg H2O = 0.494 mole/0.0600 kg = 8.23 m
•
To find molarity, need volume of solution (from density):
Volume of 100 g of solution = 100 g x (1 mL/1.38 g) = 72.5 mL
M = mole HBr/L soln = 0.494 mol/0.0725 L = 6.82 M
Sample Problem
Commercial sulfuric acid is 96.0% H2SO4 (formula mass =
98.07 g/mole) by weight and has a density of 1.85 g/mL.
Calculate the molarity (M) and the molality (m) of the H2SO4
solution.
Sample Problem
Commercial sulfuric acid is 96.0% H2SO4 (formula mass =
98.07 g/mole) by weight and has a density of 1.85 g/mL.
Calculate the molarity (M) and the molality (m) of the H2SO4
solution.
2.4 x 102 m and 18.1 M
Colligative Properties
(depend on number of solute particles)
• Vapor Pressure
– (related to mole fraction of solvent)
– Vapor pressure of solution is always less than the pure solvent
– For solutions of non-volatile solutes, Raoult’s Law applies:
Psolution = Xsolvent • P°solvent
• where Psolution = vapor pressure of soln, Xsolvent = mol fraction of
solvent, P°solvent = vapor pressure of pure solvent
OMIT -- mixtures of two or more volatile components (p 539-541)
Freezing and Boiling Points
• Freezing Point Depression and Boiling Point Elevation
– (related to molality of the solution)
– Change in freezing and boiling points:
DTf = Kfm
DTb = Kbm
where Kf and Kb are properties of the solvent:
Kf = molal freezing point depression constant
Kb = molal boiling point elevation constant
e.g. for water:
Kf = 1.86 °C/m
Kb = 0.51 °C/m
Example Problem
• A solution of 6.400 g of an unknown compound in 100.0 g of
benzene (C6H6) boils at 81.7 °C. Determine the molecular
mass of the unknown.
Data for benzene:
Kf = 5.07 °C/m
Tf = 5.07 °C
Kb = 2.53 °C/m
Tb = 80.2 °C
DTb = Kbm
DTb = 81.7 - 80.2 = 1.5 °C
m = DTb/Kb = (1.5 °C)/(2.53 °C/m) = 0.593 m
0.593 mol cmpd
=
kg benzene
0.1000 kg benzene x
0.593 mol cmpd
= 0.0593 moles cmpd
kg benzene
Molecular mass = g/mole =
6.400 g cmpd
0.0593 mol cmpd
= 1.1 x 102 g/mol
Sample Problem
Automobile antifreeze is a concentrated aqueous solution of
ethylene glycol, C2H6O2 (formula mass = 62.0 g/mol). Calculate
the weight percent of an antifreeze solution that would have a
freezing point of -25 °C (equivalent to -13 °F). The Kf constant
for water is 1.86 °C/m and the freezing point of water is 0.00
°C.
Sample Problem
Automobile antifreeze is a concentrated aqueous solution of
ethylene glycol, C2H6O2 (formula mass = 62.0 g/mol). Calculate
the weight percent of an antifreeze solution that would have a
freezing point of -25 °C (equivalent to -13 °F). The Kf constant
for water is 1.86 °C/m and the freezing point of water is 0.00
°C.
• Answer: 45%
Osmotic Pressure
(related to molarity)
• osmosis -- passage of solvent through a “semipermeable
membrane” into a solution
• osmotic pressure (P) -- back pressure required to stop osmosis
P ∝ M (at constant temp)
• van’t Hoff equation:
PV = nRT
– since n/V = M, then P = MRT
• Used for determining MM of unknowns, especially large
molecules, e.g. polymers, proteins, etc. Important in
medical solutions; cell walls are semipermeable
membranes!
– hyperosmotic (P > body), hypoosmotic (P < body), isosmotic, or
isotonic (P = body)
Osmotic Pressure Measurement
Real Solutions
• Strong electrolytes do not always dissociate 100%. van’t
Hoff factors correct for ion pairing and other effects.
i=
mol of particles in solution
mol of formula units dissolved
• “Corrected equations”
– DTf = imKf
– DTb = imKb
– P = iMRT
e.g.,
NaCl(s) --> Na+(aq) + Cl–(aq)
# moles of ions = 2 x (moles of NaCl)
so, colligative properties are
about twice as large
Tyndall effect:
scattering of light
by a colloidal dispersion
Micelle Formation
how soaps work -- micelle formation
long hydrocarbon "tail"
hydrophobic
O
O- Na+
-O C
2
CO2-
-O C
2
CO2-
-O C
2
-O C
2
-O C
2
anionic "head"
hydrophilic
CO2-
CO2CO2CO2CO2
micelle
How Soaps Work
Sample Problem
A sample of a protein is dissolved in water to give a solution
that contains 5.00 mg of protein per 1.00 mL. At 20.0 °C, this
solution is found to have an osmotic pressure of 0.760 torr.
Calculate the molecular mass of the protein.
Sample Problem
A sample of a protein is dissolved in water to give a solution
that contains 5.00 mg of protein per 1.00 mL. At 20.0 °C, this
solution is found to have an osmotic pressure of 0.760 torr.
Calculate the molecular mass of the protein.
1.20 x 105 g/mole