Section Eight pp. 1
Section Eight: Solutions
Reading: Tro Chapter 12
Recommended Problems: Mastering Chemistry Online Homework
I.
Solution – any substance that is evenly dispersed or distributed throughout
another substance.
A.
B.
C.
D.
II.
homogeneous mixture – mixture in which the components are uniformly
mixed and cannot be visually distinguished.
heterogeneous mixture – mixture characterized by observable segregation
of component substances.
A solution contains both a solvent (dissolving medium) and solute
(substance dissolved in the solvent).
Solid solutions with a metal as the solvent are called alloys. For example,
solid brass is a mixture of Cu and Zn.
Solubility
Solute + Solvent
A.
dissolve
crystallize
Solution
A substance is said to be miscible when there is no apparent limit to the
solubility of one substance in another. That is, if the intermolecular forces
of attraction are of the same type and of equal strength, we predict that an
ideal solution forms, and its enthalpy of solution (ΔHsolution) is very close
to zero. That is, nearly ideal behavior is often observed when the solutesolute, solvent-solvent, and solute-solvent interactions are very much
alike.
Example: pentane and hexane
B.
Two liquids that do not mix are said to be immiscible or a nonideal
solution. In this case, the two liquids mix endothermically, indicating that
the solute-solvent interactions are weaker than the interactions among the
molecules in the pure liquids. More energy is required to expand the
liquids than is released when the liquids are mixed. Therefore, ΔHsolution is
positive.
Example: pentane and H2O
Section Eight pp. 2
C.
When a solute and solvent release large quantities of energy in the
formation of a solution, that is when ΔHsolution is large and negative, one
can assume that strong interactions exist between the solute and the
solvent.
Example: acetone and H2O
D.
E.
F.
G.
“Like dissolves like” – if both solutions are nonpolar or if both solutions
are significantly polar, solvation will occur.
1.
Once again, substances with similar intermolecular forces tend to
be soluble in one another.
2.
Nonpolar solvents tend to be insoluble in polar liquids.
A solution that cannot dissolve any more solute at a given temperature is
said to be saturated. Referring to the equation above, when dissolution
and crystallization occur at the same rate, the solution is in a state of
dynamic equilibrium. The quantity of dissolved solute remains constant
with time.
A solution that is able to dissolve more solute is called unsaturated.
Under suitable conditions, it is sometimes possible to form solutions that
contain a greater amount of solute than that needed to form a saturated
solution. Such solutions are said to be supersaturated. Suppose we
prepare a saturated solution at one temperature and then change the
temperature to a value at which the solubility is lower (at a lower
temperature). Usually, the excess solute crystallizes from solution, but
occasionally all the solute may remain in solution.
Problem #1: Predict whether the substances listed below are more likely to dissolve in
CCl4 or H2O. Explain your predictions.
A.
C7H16
B.
Na2SO4
C.
HCl
D.
I2
Section Eight pp. 3
Problem #2: Biochemical Application! Consider the structures of Vitamins A and C
shown below. Which is more likely to be soluble in the fatty tissue of the body? Briefly
explain.
OH
Vitamin A
III.
O O
OH
OH
HO OH
Vitamin C
Factors Affecting Solubility
A.
B.
Temperature
1.
The solubility of most solid solutes in water increases with
elevated temperatures. That is, solubility increases as temperature
increases
2.
In contrast to solid solutes, the solubility of gases decreases in
water with increasing temperature (recall: increasing temperature =
increase in Kinetic Energy = greater ease for gas molecules to
escape from solution). For solutions of gases in organic solvents,
the situation is often reversed; that is, gases become more soluble
at higher temperatures.
Pressure
1.
The solubility of a gas in any solvent is increased as the pressure of
the gas over the solvent increases. Think about a freshly opened
can of soda pop as it fizzes! Why does the soda go flat after time?
2.
As more pressure is applied to gas molecules, the rate at which
they strike the surface to enter the solution phase increases.
3.
By contrast, the solubilities of solids and liquids are not
appreciably affected by pressure.
The solubility of a gas is directly proportional to its partial pressure. This
observation was first observed by William Henry and is now known as
Henry’s law:
C (solubility) = k Pgas
where k is called Henry’s constant (depends on the gas, solvent, and
temperature), and P is the partial pressure of the gas. The solubility of a
gas is proportional to its partial pressure because an increase in pressure
corresponds to an increase in the rate at which gas molecules strike the
surface of the solvent.
Section Eight pp. 4
Henry’s law fails for gases at extremely high pressures, and it also fails if
the gas ionizes in water or reacts with water. We expect Henry’s law to
apply only to equilibrium between molecules of a gas and the same
molecules in solution.
Problem #3: Calculate the mol of CO2 that will dissolve in enough water to form 900
mL of solution at 20 °C if the partial pressure of CO2 is 1.00 atm. NOTE: k for CO2 in
water at 20 °C = 2.3 x 10-2 mol L-1atm-1
IV.
Expressing Solution Concentration
Recall in section 2 that we learned about one method for expressing solution
concentration—molarity. We defined molarity as moles of solute/liters of
solution. Now we will describe several other methods of expressing concentation,
each of which serves a different purpose.
A.
mass percentage
Mass % of component =
B.
parts per million (for dilute solutions)
ppm of component =
C.
Mass of component in solution
x10 6
total mass of solution
parts per billion (for even MORE dilute solutions!)
ppb of component =
D.
Mass of component in solution
x100
total mass of solution
Mass of component in solution
x10 9
total mass of solution
mole fraction (denoted with the symbol χ)
Mole fraction of component =
E.
Moles of component
total moles of components
molality (denoted with the lowercase letter m)
molality =
moles solute
kilograms of solvent
Section Eight pp. 5
The molality of a given solution does NOT vary with temperature because masses
do not vary with temperature. Molarity, however, changes with temperature
because the expansion or contraction of the solution changes its volume. Thus,
molality is often the concentration unit of choice when a solution is to be used
over a range of temperatures.
Problem #4:
A.
B.
A solution of hydrochloric acid contains 36% HCl by mass.
Calculate the mole fraction of HCl in the solution.
Calculate the molality of HCl in the solution.
Problem #5: A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene and has
a density of 0.876 g/mL. Calculate the molarity of the solution.
V.
Colligative Properties– Those properties of a solvent (i.e. vapor pressure
lowering, freezing point depression, boiling point elevation, osmotic pressure)
that depend on the total concentration of solute particles present.
NOTE: i = van’t Hoff factor, which is determined experimentally and represents
the degree of dissociation of the solute in the solvent.
Molal Boiling Point Elevation/Freezing Point Depression Constants for Various Solvents
Solvent
H2O
Benzene, C6H6
Ethanol, C2H5OH
CCl4
Chloroform, CHCl3
A.
Normal BP (°C)
100.0
80.1
78.4
76.8
61.2
Kb (°C/m)
0.52
2.53
1.22
5.02
3.63
Normal FP (°C)
0.0
5.5
-114.6
-22.3
-63.5
Kf (°C/m)
1.86
5.12
1.99
29.8
4.68
Vapor Pressure Lowering (similar to mole fraction calculation from gas
laws unit)
Psolvent = χsolventPpure ⇐ known as Raoult’s Law
Section Eight pp. 6
where Ppure = vapor pressure of the pure solvent, χsolvent = mole fraction of
the solvent, and P is the vapor pressure of the solvent in the solution.
B.
Boiling Point Elevation – The increase in boiling point of a solution
relative to that of the pure solvent is directly proportional to the number of
solute particles per mole of solvent molecules. That is,
ΔTb = iKb m
where ΔTb = increase in boiling point relative to that of the pure solvent,
Kb is called the molal boiling point elevation constant, and m = molality of
the solution.
C.
Freezing Point Depression – Like the boiling point elevation, the decrease
in freezing point of a solution relative to that of the pure solvent is directly
proportional to the molality of the solute. That is,
ΔTf = iKf m
where ΔTf = decrease in freezing point relative to that of the pure solvent,
Kf is called the molal freezing point depression constant, and m = molality
of the solution.
D.
Osmotic Pressure – The net movement of solvent molecules from a less
concentrated solution into a more concentrated one is known as osmosis.
The pressure required to prevent osmosis is known as the osmotic pressure
(defined by the variable π) and can be calculated as:
π = i MRT
where M = Molarity, R = 0.08206 L atm K-1 mol-1, and T = Temperature
(K).
Problem #6: Determine the vapor pressure of a solution of 92.1 g of glycerin,
C3H5(OH)3, in 184.4 g of ethanol at 40 °C. The vapor pressure of pure ethanol is 0.178
atm at 40 °C, and glycerin is essentially nonvolatile.
Problem #7:
chloroform.
Find the boiling point of a solution of 92.1 g of iodine in 800.0 g of
Section Eight pp. 7
Problem #8: Calculate the freezing point of a solution of 0.724 g of calcium chloride in
175 g of water, assuming complete dissociation by the solute.
Problem #9: Determine the osmotic pressure of a solution with a volume of 0.750 L
that contains 5.0 g of methanol in water at 37 °C.
Problem #10: List the following aqueous solutions in order of their expected freezing
points: 0.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.050 m HC2H3O2, and 0.10 m
C12H22O11.
Problem #11: A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is
found to freeze at 2.32 °C. What is the molar mass of this compound?
Section Eight pp. 8
Problem #12: 0.500 L of an aqueous solution that contains 10.0 g of hemoglobin has an
osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?
Problem #13: A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a
boiling point of 64.5 °C. What is the molar mass of this compound?
Problem #14: An organic compound has a composition of 93.46% C and 6.54% H by
mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C.
The melting point of pure camphor is 178.4 °C, and its freezing point depression constant
is 37.7 °C m-1. What is the molecular formula of the solute?