4. Solution Composition
Solutions are homogeneous mixtures that can be
gases, liquids or solids.
Solutions can be described as dilute or concentrated.
molarity
mass percent
mole fraction
molality
5. Concentrations
moles
Molarity = M =
liters
æ mass of solute ö
Mass % =ç ÷ x 100%
è mass of solution ø
nA
Mole fraction of A= X A =
nA + nB
moles of solute
Molality =
kg of solvent
6. Practice Problem 1
A Solution is prepared by mixing 1.00g of ethanol
(C2H3OH) with 100.0g of water to give a final volume of
101 ml. Calculate the molarity, mass percent, mole
fraction and molality of ethanol in this solution.
M=.215, %=.990%, X=.00389, m=.217
7. Normality
Normality is defined as the number of equivalents per
liter of solution.
equivalents depends on the reaction taking place
in the solution
example: for an acid base reaction, the equivalent
is the mass of acid or base that can furnish or
accept exactly 1 mole of protons (H+ ions).
H2SO4 or Ca(OH)2 equivalents = molar mass/2
one equivalent of acid reacts with one equivalent
of base.
8. Normality
• For oxidation - reduction reactions, the equivalent
is defined as the quantity of oxidizing or reducing
agent that can accept or furnish 1 mole of
electrons.
one equivalent of reducing agent will react with
exactly one equivalent of oxidizing agent.
9. Example
The equivalent mass of an oxidizing or reducing
agent can be calculated from the number of
electrons in a half reaction.
MnO4- + 5e- + 8H+ Mn2+ + 4H2O
Since the MnO4- ion present in 1 mole of
KMnO4 in acidic solution consumes 5 moles of
electrons, the equivalent mass is the molar
mass divided by 5:
molar mass 158g
= = 31.6g
Eq mass of KMnO4 = 5 5
10. Practice Problem 2
The electrolyte in automobile lead storage batteries is a
3.75 M sulfuric acid solution that has a density of 1.230
g/ml. Calculate the mass percent, molality, and
normality of the sulfuric acid.
7.50 N
12. Solubility
What factors affect solubility?
The cardinal rule of solubility is that like dissolves
like.
Polar solvents are used to dissolve a polar or ionic
solute and nonpolar solvents are used to dissolve
a nonpolar solute.
13. Solubility
Solubility and formation of a solution takes place in
three distinct steps:
1. Separating the solute into its individual components
(expanding the solute).
2. Overcoming intermolecular forces in the solvent to
make room for the solute (expanding the solvent).
3. Allowing the solute and solvent to interact to form
the solution.
14. Solubility
1. Expanding the solute:
endothermic
2. Expanding the solvent:
endothermic
3. Interaction: exothermic
Enthalpy of solution = ΔHsoln =
ΔH1 + ΔH2 + ΔH3
15. Solubility
ΔHsoln could be overall negative sign (exothermic).
ΔHsoln could be overall positive sign (endothermic).
16. Example
Oil slicks to not dissolve in water:
1. ΔH1 is endothermic but small. C chains held
together with LDF need separated.
2. ΔH2 is endothermic and large. Hydrogen bonds in
the water are difficult to separate.
3. ΔH3 will be small since polar and nonpolar
interactions are minimal.
ΔHsoln is an overall endothermic process and unlikely
to produce a solution.
17. Example 2
Most ionic substances dissolve in water:
1. ΔH1 is endothermic and large. Ionic forces must
be overcome.
2. ΔH2 is endothermic and large. Hydrogen bonds in
the water are difficult to separate.
3. ΔH3 will be very exothermic because most ionic
substances interact very well with water.
ΔHsoln is usually small but can be overall exothermic
or endothermic.
18. Enthalpy of Hydration
Enthalpy of hydration (ΔHhyd ) combines the terms ΔH2
(expanding the solvent) and ΔH3 (solvent-solute
interactions).
The heat of hydration represents the enthalpy
change assoicated with the dispersal of a gaseous
solute in water.
H2O(l) + Na+(g) + Cl-(g) Na+(aq) + Cl-(aq)
ΔHsoln = ΔH1 + ΔHhyd
19. Example 3
Heat of solution of NaCl and water:
1. NaCl(s) Na+(g) + Cl-(g) ΔH1=786 kJ/mol
2. & 3. H2O(l) + Na+(g) + Cl-(g) Na+(aq) + Cl-(aq) ΔHhyd=783
kJ/mol
ΔHsoln=3kJ, endothermic but small.
20. Solubility
The dissolving process often requires a small amount of
energy (stirring, heat), but ionic substances, even
though their expansion is very endothermic, dissolve
due to their tendency toward increased probability.
Processes that require large amounts of energy tend
not to occur.
21. Practice Problem 3
Decide whether liquid hexane (C6H14) or liquid
methanol (CH3OH) is the more appropriate solvent for
the substances grease (C2oH42) and potassium iodide
(KI).
Hexane works best for grease and methanol serves
as a better solvent for potassium iodide
23. Structure and Solubility
Since polarity is a big factor in solubility and molecular
structure determines polarity, structure has a lot to do
with solubility.
Vitamins are divided into fat soluble and water
soluble due to their structures:
24. Pressure Effects
Pressure has little effect on the solubilities of solids or
liquids, but it does significantly increase the solubility of
a gas.
25. Henry’s Law
Henry’s law states that the amount of a gas dissolved in
a solution is directly proportional to the pressure of the
gas above the solution. The relationship between gas
pressure and the concentration of dissolved gas is
given by:
C=kP
C = the concentration of the dissolved gas, k is a
constant characteristic of a particular solution and P
represents the partial pressure of the gaseous solute
above the solution.
Henry’s law is obeyed most accurately for dilute
solutions that do not dissociate or react with the
solvent.
26. Practice Problem 4
A certain soft drink is bottled so that a bottle at 25°C
contains CO2 gas at a pressure of 5.0 atm over the
liquid. Assuming that the partial pressure of CO2 in the
atmosphere is 4.0 x 10-4 atm, calculate the equilibrium
concentrations of CO2 in the soda both before and after
the bottle is opened. The Henry’s law constant for CO2
in aqueous solution is 3.0 x 10-2 mol/Latm at 25°C.
.16 mol/L and 1.2 x 10-5 mol/L
27. Temperature Effects
Solubility doesn’t always increase with temperature.
The dissolving of a solid occurs more rapidly at higher
temperatures, but the amount of solid that can be
dissolved may increase or decrease with increasing
temperature.
Predicting the temperature dependence of solubility is
very difficult. The only sure way to determine the
temperature dependence of a solid’s solubility is by
experiment.
29. Temperature Effects
The behavior of gases dissolving in water typically
decrease with increasing temperature.
Thermal pollution in lakes and rivers
Boiler scale –
2Ca2+ + HCO3(aq) H2O + CO2(aq) + CaCO3(aq)
32. Vapor Pressure and
Solutions
Liquid solutions have physical properties significantly
different from those of the pure liquid solvent.
Antifreeze with water delays freezing and boiling.
Salt and water lowers freezing point.
33. Vapor Pressure and
Solutions
A nonvolatile solute lowers the
vapor pressure of a solvent.
1. Vapor pressure of the pure
solvent is greater than that of
the solution.
2. The equilibrium vapor
pressure of the pure solvent
(water) is greater than that of
the solution equilibrium vapor
pressure.
3. Water vaporizes and adds to
solution.
34. Vapor Pressure and
Raoult’s Law: P
Solubility
=X P
soln solvent
o
solvent
• Psoln is the observed vapor pressure of the solution,
Xsolvent is the mole fraction of solvent, and Posolvent is
the vapor pressure of the pure solvent.
• In a solution consisting of half nonvolatile solute
molecules and half solvent molecules (typically
water), the observed vapor pressure is half of that of
the pure solvent, since only half as many molecules
can escape.
• If vapor pressures are known, moles can be
determined.
35. Practice Problem 5
Calculate the expected vapor pressure at 25°C for a
solution prepared by dissolving 158.0 g of
sucrose, molar mass= 342.3 g/mol, in 643.5cm3 of
water. At 25°C, the density of water is 0.9971 g/cm3
and the vapor pressure is 23.6 torr.
23.46 torr
36. Vapor Pressure and
Solubility
The lowering of vapor pressure depends on the number
of ionic solute particles present in the solution.
Example: 1 mole of sodium chloride dissolved in
water lowers the vapor pressure approximately twice
as much as expected because the solid has two ions
per formula unit, which separates when it dissolves.
37. Practice Problem 6
Predict the vapor pressure of a solution prepared by
mixing 35.0 g solid Na2SO4 (molar mass= 142.05
g/mol) with 175 g water at 25°C. The vapor pressure of
pure water at 25°C is 23.76 torr. (how many moles of
solute particles are present?)
22.1 torr
38. Nonideal Solutions
When both solvent and solute are volatile, both
contribute to the vapor pressure over the solution. A
modified form of Raoult’s law is used:
Ptotal = PA + PB = XAPoA + XBPoB
• Ptotal is the total vapor pressure of solution AB. XA
and XB are the mole fractions of A and B. PA and PB
are the partial pressures of A and B.
39. Ideal vs. Nonideal
A liquid-liquid solution that obeys Raoult’s law is called
an ideal soution. Raoult’s law is to solutions what the
ideal gas law is to gases. Nearly ideal behavior is often
observed when solutes and solvents are similar.
• When strong interactions occur (ΔHsoln is very
exothermic), a negative deviation of Raoult’s law
results
• When weak interactions occur (ΔHsoln is
endothermic), a positive deviation of Raoult’s law
results.
41. Practice Problem 7
A solution is prepared by mixing 5.81 g acetone
(C3H6O, molar mass = 58.1 g/mol) and 11.9 g
chloroform (HCCl3, molar mass = 119.4 g/mol). At
35°C, this solution has a total vapor pressure of 260
torr. Is this an ideal solution? The vapor pressures of
pure acetone and pure chloroform at 35°C are 345 and
293 torr, respectively.
Acetone (CH3)2CO and chloroform CHCl3
319 is the expected total vapor pressure; this is not
an ideal solution. This is a negative deviation,
therefore interactions must be strong.
43. Vapor Pressure and
Freezing/Boiling Points
Since changes of state depend on vapor pressure, the
presence of a solute also affects the freezing point and
boiling point of a solvent.
• Freezing point depression, boiling point elevation
and osmotic pressure are called colligative
properties.
• These properties are dependent on the number of
solute particles and not the identity of the particles.
44. Boiling Point Elevation
The normal boiling point of a liquid occurs at the
temperature at which the vapor pressure is equal to 1
atmosphere. A nonvolatile solute lowers the vapor
pressure of the solvent; therefore such a solution must
be heated to a higher temperature than the ‘pure’
boiling point for the vapor pressure to reach 1
atmosphere.
• A nonvolatile solute elevates the boiling point of the
solvent.
45. Boiling Point Elevation
Water and a nonvolatile water
solution. The boiling point
increases and the freezing
decreases with the solution.
The effect of a nonvolatile
solute is to extend the liquid
range of a solvent.
46. Boiling Point Elevation
The magnitude of the boiling point elevation depends
on the concentration of the solute. The change in
boiling point can be calculated by:
ΔT = Kbmsolute
• ΔT is the boiling point elevation, Kb is a constant that
is characteristic of the solvent and is called the molal
boiling-point elevation constant. msolute is the
molality of the solute in the solution.
• An observed boiling point elevation can determine
molar mass.
48. Practice Problem 8
A solution was prepared by sissolving 18.00 g glucose
in 150.0g water. The resulting solution was found to
have a boiling point of 100.34°C. Calculate the molar
mass of glucose. Glucose is a molecular solid that is
present as individual molecules in solution.
180 g/mol
49. Freezing Point
Depression
Vapor pressures of ice and liquid water are the same at
0°C (freezing point). When a nonvolatile solute is
dissolved in water, the vapor pressure of the solution
lowers and therefore the freezing point of the solution is
lower than that of the pure water.
ΔT = Kfmsolute
• ΔT is the freezing point depression, Kf is a constant
that is characteristic of the solvent and is called the
molal freezing-point depressionconstant. msolute is
the molality of the solute in the solution.
50. Practice Problem 9
What mass of ethylene glycol (C2H6O2, molar mass =
62.1 g/mol), the main componenet of antifreeze, must
be added to 10.0 L water to produce a solution for use
in a car’s radiator that freezes at -10.0°F (-23.3°C)?
Assume the denisty of water is exactly 1 g/mL.
7.76kg
51. Practice Problem 10
Mr. Wieland is trying to identify a human hormone that
controls metabolism by determining its molar mass. A
sample weighing 0.546 g was dissolved in 15.0 g
benzene, and the freezing-point depression was
determine to be 0.240°C. Calculate the molar mass of
the hormone.
776 g/mol
53. Osmotic Pressure
A solution and pure solvent are separated by a
semipermeable membrane, which allows solvent but
not solute molecules to pass through. As time
passes, the volume of the solution increases and
volume of the solvent decreases.
The flow of solvent into the solution through the
membrane is called osmosis.
Eventually the liquid levels stop changing and reach
equilibrium, but there is a greater hydrostatic pressure
on the solution than on the pure solvent. The excess
pressure is called osmotic pressure.
54. Osmotic Pressure
Osmosis can be prevented by applying pressure to the
solution. The minimum pressure that stops the osmosis
is equal to the osmotic pressure of the solution.
55. Osmotic Pressure
Osmotic pressure is primarily dependent on solution
concentrations.
Π = MRT
• Π is the osmotic pressure in atmospheres, M is the
molarity of the solution, R is the gas law constant,
and T is the Kelvin temperature.
56. Practice Problem 11
To determine the molar mass of a certain protein, 1.00
x 10-3 g of it was dissolved in enough water to make
1.00 mL of solution. The osmotic pressure of this
solution was found to be 1.12 torr at 25.0°C. Calculate
the molar mass of the protein.
1.66 x 104 g/mol (proteins often have large molar
masses)
57. Osmosis
Osmosis prevents transfer of all solute particles. Dialysis
(a similar phenomenon) occurs at the walls of most plant
and animal cells. In this case, the membrane allows
transfer of both solvent and small solute molecules and
ions.
Solutions that have identical osmotic pressures are said to
be isotonic solutions.
• Solutions having a higher osmotic pressure are
hypertonic and can cause crenation.
• Solutions having a lower osmotic pressure are
hypotonic and can cause hemolysis.
59. Practice Problem 12
What concentration of sodium chloride and water is
needed to produce an aqueous solution isotonic with
blood (Π = 7.70 atm at 25°C)?
.158 M
60. Reverse Osmosis
When a solution in contact with pure solvent across a
semipermeable membrane is subjected to an external
pressure larger than its osmotic pressure, reverse
osmosis occurs. The pressure will cause a net flow of
solvent from the solution. This process can act as a
molecular filter for many solutions.
• This process is being used to produce fresh water
from sea water by desalination.
62. Colligative Properties
Colligative properties of solutions depend on the total
concentration of solute particles. The relationship
between the moles of solute dissolved and the moles of
particles in solution is usually expressed using the van’t
Hoff factor:
moles of particles in solution
i=
moles of solute dissolved
• The expected value for i can be calculated for a salt
by noting the number of ions per formula unit (NaCl
=2)
• These calculated values assume that when a salt
dissolves, it completely dissociates into its
component ions.
63. Colligative Properties
Not all ions dissociate completely and act independently
in solution; ion pairing occurs in solution and some pairs
will act as a single particle.
• Ion pairing is most important in concentrated solutions.
• Less ion pairing occurs in more dilute solutions.
• The deviation of i is greatest where the ions have
multiple charges.
65. Colligative Properties
To adjust freezing point depression, boiling point
elevation and osmotic pressure calculations, i can be
used to better account for ion pairing.
ΔT = imK
• K is the freezing or boiling point constant
Π = iMRT
66. Practice Problem 13
The observed osmotic pressure for a 0.10 M solution of
Fe(NH4)2(SO4)2 at 25°C is 10.8 atm. Compare the
expected and experimental values for i.
i=4.4; The experimental value for i is less than the
expected value.
69. Colliods
The Tyndall effect is when suspended particles scatter
light. This is used to distinguish between a suspension
and true solution.
A suspension of tiny particles in some medium is called
a colloidial dispersion or colloid.
• Suspended particles are large molecules or ions
from 1-1000nm.
• Suspended particles remain suspended due to
electrostatic repulsion
71. Colloids
Colloids can be destroyed by coagulation; usually by
heating or by adding an electrolyte.
• Heating increases velocities of particles and allows
them to collide with enough energy that particles
aggregate.
• Adding an electrolyte neutralizes the ion layers and
particles precipitate out.
