Transcript No Slide Title

CHM 1046 - General Chemistry 2
Summer A 2011
Dr. Jeff Joens
Classroom: CP 145
Time: M, T, W, R 11:00am to 12:15pm
Office: CP 331
phone 348-3121
web page: www.joenschem.com
email: [email protected]
Homework Problems
Chapter 12 Homework Problems: 2, 5, 8, 12, 20, 23, 32, 34, 36, 42, 52,
64, 72, 74, 78, 80, 84, 89, 98
Note: Homework is not to be turned in. My solutions to the homework
problems are available at my website. Answers (but not detailed
solutions) to the in chapter practice problems and the odd numbered end
of chapter problems are given at the end of the book. There are also
several copies of the solution manual (Solutions Manual for Chemistry:
A Molecular Approach) available at the Reserve Desk (2nd Floor of the
Green Library) under Dr. Graves name. This solution manual gives
detailed solutions to all of the problems in the book.
CHAPTER 12
Solutions
Solutions
A solution is a homogeneous mixture. The major component of a
solution is called the solvent, while the minor components are called
solutes.
There are several common kinds of solutions
kind of solution
example
gas in gas
air (78% N2, 21 % O2, 0.9 % Ar + others)
gas in liquid
O2 in water
liquid in liquid
ethanol in water
solid in liquid
NaCl in water
solid in solid
alloys (bronze - copper + tin)
We will usually focus on solutions where the solvent is a liquid, but all of
the above combinations represent possible types of solutions.
Solubility
The term solubility means the amount of a particular solute that
will dissolve in a given amount of solvent (usually a liquid). There are
two common ways in which solubility is given.
1) Mass solute per mass solvent. Indicates the maximum mass of
solute that will dissolve in a given mass of solvent.
Example: At 25 C the solubility of sodium chloride in water is
36. g NaCl per 100. g water.
2) Molar solubility. The maximum number of moles of solute per
liter of solution.
Example: The molar solubility of silver chloride (AgCl) in pure
water at 25 C is 1.3 x 10-5 mol/L. (Equivalent to 1.9 x 10-4 g per 100 g
water).
There are other ways of expressing solubility, including mass per
volume of solvent, and solubility product.
Soluble and Insoluble
There are several common terms that refer to the ability of a solid
solute to dissolve in a liquid solvent.
Soluble - Indicates that a large amount of solute will dissolve in a
given amount of solvent.
Insoluble - Indicates that a small amount, or no, solute will
dissolve in a given amount of solvent.
In our above examples sodium chloride is soluble in water and
silver chloride is insoluble in water.
In fact, many substances classified as “insoluble” will dissolve to
a small extent in water or other solvents. We will discuss this in more
detail later in the semester.
Miscible and Immiscible Liquids
We can classify liquids in terms of their ability to form solutions
when mixed.
Two liquids are miscible if any amounts of the two liquids will
combine to form a solution (water + ethyl alcohol).
Two liquids are immiscible if they do not mix (water + gasoline).
Thermodynamics of Solution Formation
Consider the following general process for the formation of a
solution by dissolving a liquid or solid solute in a liquid solvent.
M(s)  M(soln)
There are two factors that favor formation of a solution:
1) An increase in the randomness of the system.
2) A decrease in the energy of the system.
For now, we will take energy (E) and enthalpy (H) as meaning
the same thing, as they are usually close in value.
Entropy of Solution (Ssoln)
1) An increase in the randomness or disorder of the system. In
thermodynamics randomness is measured in terms of entropy. For
solutions the change in entropy is called Ssoln.
Consider the dissolution of a solid in a liquid. We would expect
that the solution is less ordered than the starting pure solid and pure
liquid. Therefore, we expect Ssoln > 0 (surprisingly, there are a few
cases where Ssoln < 0, but they are extremely rare).
Enthalpy of Solution (Hsoln)
2) A decrease in the energy of the system. If mixing lowers
energy of a system then it is more likely to occur. Since solutions
usually prepared under conditions of constant pressure, we look at
enthalpy of solution, Hsoln, defined as the change in enthalpy for
process
M(s)  M(soln)
the
are
the
the
Hsoln
To decide whether Hsoln will be positive, negative, or
approximately zero we need to consider three type of attractive forces.
solvent-solvent - Attractive forces acting between solvent
molecules.
solute-solute - Attractive forces acting between solute molecules.
solvent-solute - Attractive forces acting between a solvent
molecule and a solute molecule.
Steps In Solution Formation
We can think solution formation as a three step process:
1) Separate the solute particles
Hsolute > 0
2) Separate the solvent particles
Hsolvent > 0
3) Mix solute and solvent particles Hmixing < 0
____________________________________________
Formation of solution
Hsolution
From Hess’ law Hsolution = Hsolute + Hsolvent + Hmixing
Examples of Solvent-Solute Attractive Forces
Hsoln For Various Solvent + Solute Combinations
solvent
polar
polar
nonpolar nonpolar
solute
polar
nonpolar
polar
nonpolar
solvent-solvent attraction
strong
strong
weak
weak
solute-solute attraction
strong
weak
strong
weak
solvent-solute attraction
strong
weak
weak
weak
enthalpy change (Hsoln)
~0
positive
positive
~0
Solution Formation
Hsoln
Ssoln
Will a solution form?
negative
positive
Yes.
~ zero
positive
Yes.
positive
positive
Maybe.
A general consequence of the above is the general rule in mixing
liquids that “like dissolves like”. Two polar liquids or two nonpolar
liquids will usually mix (because Hsoln  0), but a polar and a nonpolar
liquid will not usually mix (because Hsoln >> 0).
Solutions of Ionic Compounds in Water
For solutions formed by dissolving ionic compounds in water,
the energy change for solution formation depends on two factors:
|Hlattice| - the lattice energy (energy required to convert the ionic
solid into gas phase ions
|Hhydration| - the hydration energy (energy associated with
converting gas phase ions to hydrated ions in solution
The change in energy for solution formation, Hsoln, is
Hsoln  |Hlattice| - |Hhydration|
Hsoln < 0 (exothermic) Favors solution formation.
Hsoln > 0 (endothermic) Favors not forming a solution.
Example - KF(s) in water
Since |Hlattice|  |Hhydration|, then Hsoln  0. (Actually
endothermic by ~ 2 kJ/mol).
Classification of Solutions
Solutions can be classified according to the relative amount of
dissolved solute.
Unsaturated solution - Contains less
than the maximum equilibrium amount of
dissolved solute.
Saturated solution - Contains the maximum equilibrium amount of dissolved
solute.
Supersaturated solution - Contains
more than the maximum equilibrium amount
of dissolved solute.
Note that a supersaturated solution is
unstable. It will spontaneously form a precipitate until it becomes a saturated solution.
Effect of Temperature On Solubility
Note that the solubility of most solids in water or other liquids
increases with increasing temperature.
Recrystallization
The increase in the solubility of a solid in a liquid solvent is the basis of
a purification technique called recrystallization. The solid is dissolved in
a minimum amount of hot solvent. When the solvent cools, the solute
precipitates to form pure crystals.
Gas Solubility
Gases can also dissolve in liquids. Gas solubility is described by
Henry’s law
[A] = kH pA
[A] = concentration of dissolved A (in mol/L)
pA = partial pressure of A above the solution
kH = Henry’s law constant
The value for the Henry’s law
constant depends on the solvent, the gas,
and temperature. For most gases solubility is low (exceptions are gases that can
chemically interact with the solvent) and
solubility decreases as temperature increases.
Example:
The Henry’s law constant for water is kH = 2.02 x 10-3 mol/L.atm
at T = 20 C. What is the concentration of dissolved oxygen in water in
equilibrium with the Earth’s atmosphere (pO2 = 0.21 atm)?
Example:
The Henry’s law constant for water is kH = 2.02 x 10-3 mol/L.atm
at T = 20 C. What is the concentration of dissolved oxygen in water in
equilibrium with the Earth’s atmosphere (pO2 = 0.21 atm)?
[O2] = kH pO2 = (2.02 x 10-3 mol/L.atm) (0.21 atm)
= 4.2 x 10-4 mol/L (or 0.0135 g/L)
While small, this is enough dissolved oxygen to support fish life.
Concentration Units - Molarity
There are several common units for solutions.
Molarity (M) = moles solute
liters solution
Molarity is the most common concentration unit used for
solutions because it is easy to prepare solutions with know values of
molarity.
Note that molarity depends on temperature, because the volume
occupied by a solution changes when temperature changes.
Concentration Units - Molality and Mole Fraction
Solution concentrations can also be given in terms of molality
and mole fraction.
Molality (m) = moles solute
kg solvent
Mole fraction (Xi) =
ni
ntotal
= moles of ith substance
total number of moles
Molality and mole fraction are used in some specialized
applications, such as in equations for some of the colligative properties
of solutions and in Dalton’s law of partial pressure. Note that molality
and mole fraction are temperature independent.
Concentration Units - Percent By Mass, Parts By Mass
or Volume, and Related Units
We can also give solution concentrations in terms of the masses
of substances present in the solution.
Percent by mass (mass %) = mass of solute x 100 %
mass of solution
Parts per million by mass (ppm) =
mass of solute x 106
mass of solution
Parts per million by volume (ppm) =
volume of solute x 106
volume of solution
By analogy with ppm (parts per million) we can also define ppb
(parts per billion; multiply by 109) or parts per trillion (ppt; multiply by
1012). We always should indicate if parts per… is by mass or by volume.
We can also define percent by volume, or mole percent (percent by
moles).
Solution Calculations
There are two general types of problems you should be able to do
in connection with solutions.
1) Given information on how a solution is prepared, you should
be able to find the solute concentration (molarity, molality, mole fraction,
mass percent, and so forth).
2) Conversion from one concentration unit to a different
concentration unit.
Example: A solution is prepared by mixing 40.00 g ethylene
glycol (C2H6O2, M = 62.07 g/mol) and 60.00 g water (H2O, M = 18.02
g/mol). The density of the solution is D = 1.0514 g/mL. What are the
molarity, molality, mole fraction, and percent by mass ethylene glycol in
the solution?
Example: A solution is prepared by mixing 40.00 g ethylene
glycol (C2H6O2, MW = 62.07 g/mol) and 60.00 g water (H2O. MW =
18.02 g/mol). The density of the solution is D = 1.0514 g/mL. What are
the molarity, molality, mole fraction, and percent by mass ethylene
glycol in the solution?
nE = 40.00 g
1 mol = 0.6444 mol E
62.07 g
nW = 60.00 g 1 mol = 3.3296 mol W
18.02 g
XE =
nE
nE + nW
=
(0.6444 mol)
= 0.1622
(0.6444 + 3.3296)mol
mE = mol E = 0.6444 mol = 10.74 mol/kg
kg H2O
0.06000 kg
The total mass of solution is m = 40.00 g + 60.00 g = 100.00 g.
Since the density of the solution is D = 1.0514 g/mL, the volume of the
solution is
V = 100.00 g 1 mL
= 95.11 mL
1.0514 g
So ME = 0.6444 mol = 6.775 mol/L
0.09511 L
Finally, the mass percent ethylene glycol is
. 100 % =
. 100 % = 40.00 %
mass % =
mE
40.00 g
mE + mW
(40.00 + 60.00)g
Solutions of Volatile Liquids - Raoult’s Law
Raoult’s law is a relationship that applies to some solutions of
volatile liquids. A volatile liquid is a liquid that has a significant vapor
pressure. Liquids that do not have a significant vapor pressure are called
non-volatile.
Consider a solution consisting of two volatile liquids A and B.
Raoult’s law predicts that the partial pressure of vapor from each liquid
above the solution is given by the relationships
pA = XA pA
pB = XB pB
where pA, pB are the partial pressures
of A and B above the solution
XA, XB are the mole fraction
of A and B in the solution
pA, pB are the vapor
pressures of pure A and pure B
Example: At 20C the vapor pressure of pure benzene is 24. torr
and the vapor pressure of pure toluene is 76. torr. Consider a solution
with XB = 0.40. Assuming that benzene and toluene form an ideal
solution, find the partial pressure of benzene and toluene and the total
pressure above the solution.
Example: At 20C the vapor pressure of pure benzene (B) is 24.
torr and the vapor pressure of pure toluene (T) is 76. torr. Consider a
solution with XB = 0.40. Assuming that benzene and toluene form an
ideal solution, find the partial pressure of benzene and toluene and the
total pressure above the solution.
From Raoult’s law, pA = XApA
XB + XT = 1, so XT = 1 – XB = 1 – 0.40 = 0.60
So
pB = (0.40)(24. torr) = 9.6 torr
pT = (0.60)(76. torr) = 45.6 torr
ptotal = psoln = pB + pT = 9.6 torr + 45.6 torr = 55.2 torr
Ideal Solution
An ideal solution is a solution whose components obey Raoult’s
law.
The solvent in a solution will always obey Raoult’s law in the
limit XA  1.
Solutions will generally be close to ideal if they are composed of
similar molecules with similar intermolecular attractive forces.
A = C6H6; B = C6H5CH3
A = CH3COCH3; B = CS2
Solutions With a Nonvolative Ionic Solute
For a solution containing a nonvolatile ionic solute (or any
nonvolatile solute that can dissociate into smaller particles) there is one
additional complicating factor. As before, there will only be solvent
molecules above the solution.
If the solvent obeys Raoult’s law, then
pA = psoln = XA pA
However, we must take into account the
ionization or dissociation of the solute in
calculating the mole fraction of solvent.
XA = moles of solvent particles
total moles of particles
XB = moles of solute particles
total moles of particles
Finding the Moles of Particles (Examples)
Consider the following three substances:
C6H12O6 (sugar)
NaBr (sodium bromide)
K2SO4 (potassium sulfate)
When the above substances are dissolved in water how many
moles of particles will form per mole of dissolved substance?
Finding the Moles of Particles (Examples)
C6H12O6 (sugar)
C6H12O6(s)  C6H12O6(aq)
mol particles = 1.00 mole sugar 1 mole particle = 1.00 mol particles
1 mol sugar
NaBr (sodium bromide)
NaBr(s)  Na+(aq) + Br-(aq)
mol particles = 1.00 mol NaBr 2 mol particles = 2.00 mol particles
1 mol NaBr
K2SO4 (potassium sulfate)
K2SO4(s)  2 K+(aq) + SO42-(aq)
mol particles = 1.00 mol K2SO4 3 mol particles = 3.00 mol particles
1 mol K2SO4
Van’t Hoff Factor (i)
The van’t Hoff factor, i, is defined as
i = effective moles of particles
moles of solute
So effective moles of particles = i (moles of solute)
For dilute solutions the van’t Hoff factor can be found from the
formula for dissociation. However, for more concentrated solutions the
van’t Hoff factor is lower than that expected, and must be found
experimentally. Consider NaCl (NaCl(s)  Na+(aq) + Cl-(aq)).
Molality NaCl
iexpected
iexperimental
0.0001
2.0
1.99
0.001
2.0
1.97
0.01
2.0
1.94
0.1
2.0
1.87
Unless otherwise stated we will use the “dilute” value for i.
Example
The vapor pressure of pure water is pW = 149.4 torr at T = 60.
C. Find the partial pressure of water above each of the following
solutions at T = 60. C. You may assume the solvent (water) behaves
ideally.
a) 1.00 mol sugar (C6H12O6) and 1000. g water.
b) 1.00 mol calcium nitrate (Ca(NO3)2) and 1000. g water.
The vapor pressure of pure water is pW = 149.4 torr at T = 60.
C. Find the partial pressure of water above each of the following
solutions at T = 60. C.
a) 1.00 mol sugar (C6H12O6) and 1000. g water.
moles water = 1000. g water 1 mol water = 55.56 mol
18.0 g water
moles solute particles (sugar) = 1.00 mol
XW =
55.56 mol
= 0.982
(55.56 + 1.00)mol
pW = (0.982) (149.4 torr) = 146.7 torr
The vapor pressure of pure water is pW = 149.4 torr at T = 60.
C. Find the partial pressure of water above each of the following
solutions at T = 60. C.
b) 1.00 mol calcium nitrate (Ca(NO3)2) and 1000. g water.
moles water = 1000. g water 1 mol water = 55.56 mol
18.0 g water
moles solute particles (Ca(NO3)2) =
1.00 mol Ca(NO3)2 3 mol particle
1 mol Ca(NO3)2
XW =
55.56 mol
= 0.949
(55.56 + 3.00)mol
pW = (0.949) (149.4 torr) = 141.8 torr
Ca(NO3)2(s)  Ca2+(aq) + 2 NO3-(aq)
= 3 mol particles
Colligative Property
A colligative property is any property of a solution consisting of a
volatile solvent and a nonvolatile solute that depends at most on two
things:
1) The physical properties of the solvent
2) The number or concentration of solute particles
There are four colligative properties
- vapor pressure lowering
- boiling point elevation
- freezing point depression
- osmotic pressure
The equations given for colligative properties apply to ideal conditions
and are therefore expected to be correct in the limit Xsolvent  1.
Vapor Pressure Lowering
Consider the situation below. The left side contains pure solvent
A, while the right side contains a solution of A and a nonvolatile solute.
pleft = pA
pright = XA pA
We may define the vapor pressure lowering p = pleft - pright, and so
p = pA - XA pA = (1 - XA) pA
But XA + XB = 1, so 1 - XA = XB, and so
p = XB pA , the final expression for vapor pressure lowering.
Measurement of Vapor Pressure Lowering
The apparatus shown below can be used to accurately measure
the vapor pressure of a solution relative to that of a pure liquid. Note
that the working fluid in the manometer is often chosen to be a nonvolatile liquid of lower density than mercury, such as sulfuric acid or
mineral oil.
Boiling Point Elevation and Freezing Point Depression
Consider the situation below. The left side contains pure solvent
A, while the right side contains a solution of A and a nonvolatile solute.
The boiling point of the solution will be higher than the boiling point of
the pure liquid, and the freezing point of the solution will be lower than
the freezing point of the pure liquid.
Tb = Tb - Tb° = Kb mB  boiling point elevation
Tf = Tf° - Tf = Kf mB  freezing point depression
Kb, Kf = boiling point elevation and freezing point depression constants
Tb, Tf = boiling and freezing
point for the solution
Tb°, Tf ° = boiling and freezing point for the pure liquid
mB = molality of solute particles
Example
A solution is prepared by dissolving 1.00 mol sodium chloride
(NaCl) in 1000. g of water. What are the normal freezing point and
normal boiling point for this solution.
Tb = 100.00 C
Tf = 0.00 C
Kb = 0.512 C/m
Kf = 1.86 C/m
A solution is prepared by dissolving 1.00 mol sodium chloride
(NaCl) in 1000. g of water. What are the normal freezing point and
normal boiling point for this solution.
NaCl(s)  Na+(aq) + Cl-(aq)
Tb = 100.00 C
Tf = 0.00 C
Kb = 0.512 C/m
Kf = 1.86 C/m
molality NaCl = 1.00 mol NaCl 2 mol particles = 2.00 mol particles
1.000 kg water 1 mol NaCl
kg water
So
Tb = (0.512 C/m) (2.00 m) = 1.02 C
Tf = (1.86 C/m) (2.00 m) = 3.72 C
Tb = 100.00 C + 1.02 C = 101.02 C
Tf = 0.00 C - 3.72 C = - 3.72 C
Phase Diagram For Solutions
We can picture the differences between the behavior of a pure
liquid and a solution of the liquid and a nonvolative solute in terms of the
corresponding phase diagram.
In a sense all that has
changed for the solution
is the location of the
triple point, which affects
the locations of the phase
boundaries, as indicated
in the phase diagram at
right.
Osmotic Pressure
Consider the apparatus below. The right chamber contains a pure
liquid, while the left chamber contains a solution of the liquid and a nonvolatile solute. The two chambers are separated by a semipermeable
membrane that allows solvent molecules to pass but prevents the passage
of solute molecules.
The osmotic pressure,
, is defined as the additional
pressure that must be applied
to the solution to keep the
level of liquid in both chambers the same.
Semipermeable Membrane
A semipermeable membrane is a barrier that allows some
particles to pass through but which prevents the passage of other
particles. The discrimination is usually based on the size of the particles
(or, for ions, the size of the hydrated or solvated ions).
Calculation of Osmotic Pressure
For dilute solutions the osmotic pressure is given by the
relationship
 = [B]RT , where [B] is the molarity of solute particles (mol/L)
Example: What is the osmotic pressure for a 0.0100 M solution
of sugar (C6H12O6) and a 0.0100 M solution of sodium chloride (NaCl)
in water at T = 25. C? Recall that NaCl(s)  Na+(aq) + Cl-(aq) gives
two solute particles for every formula unit of NaCl.
Example: What is the osmotic pressure for a 0.0100 M solution
of sugar (C6H12O6) and a 0.0100 M solution of sodium chloride (NaCl)
in water at T = 25. C? Recall that NaCl(s)  Na+(aq) + Cl-(aq) gives
two solute particles for every formula unit of NaCl.
 = [B]RT , where [B] molarity of solute particles (mol/L)
For sugar [B] = 0.0100 M.
For NaCl, [B] = 0.0200 M
(sugar) = (0.0100 mol/L)(0.08206 L.atm/mol.K)(298 K)
= 0.245 atm ( = 186. Torr)
(NaCl) = (0.0200 mol/L)(0.08206 L.atm/mol.K)(298 K)
= 0.489 atm ( = 372. Torr)
Applications to Molecular Mass Determination
Any of the four colligative properties may be used to determine
the molecular mass of a nonvolative solute. In practice osmotic pressure
or vapor pressure lowering are usually used, as they are more sensitive
and can be used to find the molecular mass of high molecular mass
molecules.
Example: A solution is prepared by dissolving 0.1438 g of a
nonvolatile solute in water to form a solution of final volume V = 25.00
mL. The osmotic pressure of the solution, measured at T = 37.0 C, is
21.6 torr. What is the molecular mass of the solute?
Example: A solution is prepared by dissolving 0.1438 g of a
nonvolatile solute in water to form a solution of final volume V = 25.00
mL. The osmotic pressure of the solution, measured at T = 37.0 C, is
21.6 torr. What is the molecular mass of the solute?
M = grams = (g/L)
moles
(mol/L)
c = 0.1438 g = 5.752 g/L
0.02500 L
 = [B]RT = 21.6 torr (1 atm/760 torr) = 0.02842 atm
[B] =  =
0.02842 atm
= 1.117 x 10-3 mol/L
RT (0.08206 L.atm/mol.K) (310. K)
M=
5.752 g/L
= 5150. g/mol
1.117 x 10-3 mol/L
Reverse Osmosis
One interesting application of a colligative property is the use of
reverse osmosis to purify seawater. The high pressure forces water
through the semipermeable membrane while trapping ions and larger
molecules. The process is usually more expensive than other sources of
potable water, but can be economically viable in some cases.
Medical Applications of Osmotic Pressure
The membranes of red blood cells are semipermeable
membranes. Therefore, these cells can experience an osmotic pressure
when placed in a liquid with a different molality than blood plasma,
which has a molality m  0.30 mol/kg, about the same molality as a 0.9 g
NaCl per 100. g water solution.
isotonic - same molality as blood plasma
hypertonic - higher molality than blood plasma (crenation)
hypotonic - lower molality than blood plasma (hemolysis)
Colloid
A colloid is a mixture of small particles suspended within a
substance. The particles in a colloid are larger than the solute molecules
or ions in a solution. The size of colloid particles is generally in the
range 1 - 1000 nm.
End of Chapter 12
“If you're not part of the solution, you're part of the precipitate.”
- unknown
“How can you distinguish science from junk? Science posits
hypothesis and tests them. Pseudoscience assumes conclusions and finds
evidence to back them up.” - Wendy Kaminer, Sleeping With ExtraTerrestrials