CHEM 102 INSTRUCTIONAL OBJECTIVES
Prof. Upali Siriwardane,
Chemistry Program, Louisiana Tech University, Ruston, LA 71272
CHAPTER 15. The Chemistry of Solutes and Solutions
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
15.9
15.10
15.11
Solubility and Intermolecular Forces
Enthalpy, Entropy, and Dissolving Solids
Solubility and Equilibrium
Temperature and Solubility
Pressure and Dissolving Gases in Liquids: Henry's Law
Solution Concentration: Keeping Track of Units
Vapor Pressures, Boiling Points, and Freezing Points of Solutions
Osmotic Pressure of Solutions
Colloids
Surfactants
Water: Natural, Clean, and Otherwise
Objectives are as follows:
Basic Skills
Students should be able to:
1. Describe how liquids, solids, and gases dissolve in a solvent (Section 15. D. Predict
solubility based on properties of solute and solvent (Section 15. 1). Interpret the
dissolving of solutes in terms of enthalpy and entropy changes (Section 15.2).
2. Differentiate among unsaturated, saturated, and supersaturated solutions (Section
15.3).
3. Describe how ionic compounds dissolve in water (Section 15-3).
4. Predict how temperature affects the solubility of ionic compounds (Section 15.4).
5. Predict the effects of temperature (Section 15.4) and pressure on the solubility of
gases in liquids (Section 15.5).
6. Describe the compositions of solutions in terms of weight percent, mass fraction,
parts per million, parts per billion, parts per trillion and molarity (Section 15.6).
7. Interpret vapor pressure lowering in terms of Raoult's Law (15-7).
8. Use molarity to calculate the colligative properties: freezing point lowering, boiling
point elevation, and osmotic pressure (Section 15.7).
9. Differentiate the colligative properties of nonelectrolytes and electrolytes (15.7).
10.Explain the phenomena of osmosis and reverse osmosis and calculate osmotic
pressure (Section 15.8).
11.Describe the various kinds of colloids and their properties (Section 15-9).
12.Explain how surfactants work (Section 15. 10).
13.Discuss the earth's water supply and the sources of fresh water (Section 15. 11).
Discuss how municipal drinking water is purified (Section 15.11).
14.Describe what causes hard water and how it can be softened (Section 15. 11).
15.Explain how household wastes can contaminate groundwater (Section 15.11).
1
CHEM 102 CLASS NOTES
Prof. Upali Siriwardane,
Chemistry Program, Louisiana Tech University, Ruston, LA
71272
CHAPTER 15. The Chemistry of Solutes and Solutions
KEY CONCEPTS
Solutions
Solubility
Solute
Solvent
Saturated solutions
Supersaturated
solutions
Molality
Solubility gases:
Henry's Law
Mass percentage
Molarity
Parts per billion (ppb), Mole fraction
Energy changes of
solution process
Boiling point
Parts per million
(ppm)
Solubility of
substances
Freezing point
Colloids
True solutions
Suspensions
Normality
Proof
15.1
Volume & Mass %
Colligative properties
Vapor pressure
Osmotic pressure
Colligative
properties of
electrolytic
solutions
Solubility and Intermolecular Forces
The dissolving process of solutes in a solvent depends on the relative
strength of three intermolecular attractive forces. The three types of forces
between the solute and the solvent are solute-solute and solvent-solvent and
solvent-solvent must be considered in the solution process. These interactions
could be ion-ion, ion-ion dipole, dipole-dipole and London dispersion forces
between the ions, dipoles or non-polar molecules of the solvents or the solute.
These intermolecular attractions must be broken before new solute-solvent
attractive forces can become effective. Perhaps the bond breaking and bond
forming processes take place simultaneously. A solute will dissolve in a solvent if
the solute-solvent forces of attraction are great enough to overcome the solutesolute and solvent-solvent forces of attraction. A solute will not dissolve if the
solute-solvent forces of attraction are weaker than individual solute and solvent
intermolecular attractions. Generally, if all three of the intermolecular forces of
attraction are roughly equal, the substances will be soluble in each other.
2
Non-polar solute - Non-polar solvent:
In all types of non-polar compounds, about the only intermolecular
attractions are the very weak induced dipole forces. The weak attractive forces
formed by the solute-solvent molecules compensate for breaking those weak
bonds in the two pure non-polar substances. An example is solid iodine (I2)
dissolved in liquid bromine (Br2).
Non-polar solute - Polar solvent:
Non-polar Iodine is not very soluble in water. An intermolecular bond
between an induced dipole (I2) and a polar bond in water is not very strong
compared to the hydrogen bonds in water. The water molecules would rather
remain hydrogen bonded to each other, then to allow iodine molecule come
between them. The water molecules effectively "squeeze" out the non-polar
iodine. The intermolecular forces are not roughly equal, therefore, the "unlike"
substances are not soluble in each other.
Various gases such as O2, N2, H2, CO2 are not very soluble because the gases are
essentially non-polar. Of course you may say that oxygen must be dissolved in
water to sustain fish life -- true, but the solubility is very low. Carbon dioxide is
soluble in water such as carbonated beverages -- again this is true but why does it
fizz when opened or lose the bubbles on standing? Carbon dioxide is not very
soluble in water.
Polar Solute - Polar Solvent:
Polar ammonia molecules dissolve in polar water molecules. These molecules
mix readily because both types of molecules engage in hydrogen bonding. Since
the intermolecular attractions are roughly equal, the molecules can break away
from each other and form new solute (NH3), solvent (H2O) hydrogen bonds. A
wide variety of solutions are in this category such as sugar in water, alcohol in
water, acetic and hydrochloric acids.
Ionic Solute - Polar Solvent:
When an ionic crystal such as NaCl is placed in water, a dissolving reaction
will occur. Initially, the positive and negative ions are only attracted to each other.
3
The water molecules are hydrogen bonded to each other. If the crystal is to
dissolve, these bonds must be broken.
Negative chloride ions on the surface are attracted by neighboring positive
sodium ions and by the partially positive hydrogen atom in the polar water
molecule (See the graphic on the right).
Similarly, the positive sodium ions are attracted by both chloride ions and
the partially negative oxygen atom in the polar water molecule. (See the graphic
on the right).
A "tug-of-war" occurs for the positive and negative ions between the other
ions in the crystal and the water molecules. Whether the crystal dissolves is
determined by which attractive force is stronger. If the internal ionic forces in the
crystal are the strongest, the crystal does not dissolve. This is the situation in
reactions where precipitates form. If the attractions for the ions by the polar water
molecules are the strongest, the crystal will dissolve. This is the situation in
sodium chloride.
Once the ions are released from the crystals, the ions are completely surrounded
by water molecules. Note that the proper atom in the water molecule must "point"
toward the correct ion. The charge principle and the partial charges in the polar
molecule determine the correct orientation.
15.2
Enthalpy, Entropy, and Dissolving Solids
Typically the dissolving of a solid in water will involve measurable heat either
exothermic or endothermic. There is not absolute agreement on whether dissolving itself
should be categorized as wholly physical, partly chemical, etc., but intermolecular forces
are certainly involved (at the very least).
Regardless of the appropriate "label" for dissolving as a process there is an overall energy
term. It is known as the heat of solution, Hsoln.
Hsoln
could be broken into few
steps.
1. solute particles are separated from the solid mass (energy is absorbed, H1)
4
2. solvent particles move apart to make space for dissolved solute (energy is absorbed,
H2)
3. solute and solvent particles are attracted to one another (energy is released, DH3)
For most solids dissolving in water, the sum of the first two terms is greater than
the third and thus dissolving is frequently endothermic (Hsoln = +) and solubility
generally increases with increasing temperature. When heats of solution become very
highly positive it is often because the solute and solvent are dissimilar and, in the extreme
case, immiscible. The old rule of "like dissolves like" is an approximation, but a good one.
Energy Changes and dissolution
Dissolution is either exothermic or endothermic
Energy is gained or lost during dissolution
Substance
LiCl
NaCl
KCl
KOH
H°soln (kJ/mol)
-37.0
3.9
17.2
-57.6
Entropy of Solution
When solids dissolve in liquids the entropy which is measure of disorder of the
system nearly always increases. This owes mainly to the increased freedom of movement
of the solute particles as the forces, which hold the solid together are overcome.
Dispersed in the solvent, the molecules and complex ions can translate and rotate as well
as vibrate. The energy states for the many possible translational and rotational motions
are more closely spaced than those for the vibrational states in the solid. Dispersal of
energy is what entropy is all about. So the entropy change for the first step in our
dissolving "process" is positive (S1 = +). Also occurring in the system, the solvent
molecules must move about to make space for the solute. In general this results in
disruptions of the solvent-solvent forces and more freedom of movement for the solvent
molecules, hence additional pathways for energy dispersal. So S2 is typically positive.
In the final step solvent and solute interactions decrease the free movement of both
species to some extent and the entropy may decrease. This leaves two unknowns: what
are the relative magnitudes of these three steps which occur in the system AND what
about Ssurr?
5
The question about Ssurr takes us back to the question about the overall Hsoln
(actually Hsys) since the flow of heat between the system and the surroundings will
determine Ssurr. If heat moves into the surroundings then the entropy there will rise. If
the heat flow is in the other direction then DSsurr will be negative. Taking urea as an
example, urea certainly dissolves and therefore: .Ssys + .Ssurr >0. But if the solvent-solute
interactions are strong enough, Hsys (i.e., Hsoln) might be negative. This would make
DSsurr positive but would also tend to make S3 more negative. If, on the other hand,
Hsys were positive, then Ssurr would be negative and so S1 + S2 > |S3 + Ssurr|. As
already mentioned, it is possible to find the heat of solution in the lab but determining the
entropy of solution (Ssoln) is a more complicated matter. This is true for most entropy
determinations, which are typically indirect. In this experiment the thermodynamic
requirements for the equilibrium condition (Kc ) can be exploited to determine DSsoln
from the readily found heat of solution and the free energy change, G.
Gsoln = Hsoln -TSsoln
For a spontaneous process G should be negative.
Gosoln = -RT ln Kc
When gases dissolve in water, the hydrated molecules represent a higher degree of order
than the random distribution in the gas. S for the dissolving of a gas is, therefore,
negative.
Soluble gases, then, must have a negative H. Experiment shows that the dissolving of
gases is, in fact, an exothermic process.
15.3
Solubility and Equilibrium
When crystals are first placed in a solvent, many particles may leave the surface and go
into solution. As the number of solute particles in solution increases, some of the
dissolved particles return to the surface of the crystal.
Eventually the number of particles leaving the crystal surface equals the number
returning to the surface. This point is called solution equilibrium.
At a specific temperature, there is a limit to the amount of solute that will dissolve in a
given quantity of solvent. For instance, at 20oC, a maximum of 64.2 grams of nickel
6
chloride will dissolve in 100 cm3 of water. This quantity is called the solubility of the
substance at 20oC.
Solubility is an equilibrium process leading either to unsaturated, saturated, or supersaturated solutions.
Unsaturated solutions: An unsaturated solution has a lower amount of solute compared
to a saturated solution.
Saturated solution. A saturated solution contains the maximum amount of solute that
can be dissolved in a solvent at a particular temperature.
Supersaturated solution: A supersaturated solution is a special case of saturated
solutions that holds more solute than it would normally hold. Supersaturation results
when a solution is cooled and the excess solute is not precipitated out. If you add a tiny
crystal of a solute to a super saturated solution precipitation occurs immediately.
Explain the following terms used to describe solutions.
solvent b) solute c) solubility d) saturated solution
e) supersaturated solution f) dilute solutions e) concentrated solutions
a) Solvent: Solvent is the liquid which make up the bulk of a solution. Aqueous solutions
are solid, liquid and gaseous solutes dissolved in water. It is important to realize that the
definition of solution includes any homogenous phase, gas, liquid and solids. A gas
dispersed in a liquid is called an aerosol. There could be solid solutions (solid-solid) of
metals that are called alloys such as bronze. Bronze is a homogenous solid solution of
copper and tin.
b) Solute: Solute is the other substance in a solution once the solvent is identified.
Solvent has the highest concentration in the solution.
c) Solubility: This is a parameter to indicate the quantity of solute that will dissolve in a
given quantity of a solvent. Solubility is expressed conveniently as grams of solute per
100 mL of solution.
7
d) Saturated solution. A saturated solution contains the maximum amount of solute
that can be dissolved in a solvent at a particular temperature.
e) Supersaturated solution: A supersaturated solution is a special case of saturated
solutions that holds more solute than it would normally hold. Supersaturation results
when a solution is cooled and the excess solute is not precipitated out. I f you add a tiny
crystal of a solute to a super saturated solution precipitation occurs immediately.
f) Dilute solutions: A dilute solution has a lower amount of solute compared to another
solution.
e)Concentrated solutions: A concentrated solution has a higher amount of solute
compared to another solution.
Solubility Equilibrium
Consider an ionic solid, MX(s), consisting of M+ cations and X- anions, in equilibrium with
its ions in aqueous solution:
MX(s) <===> M+(aq) + X-(aq)
AgCl(s) <===> Ag+(aq) + Cl-(aq)
The equilibrium constant for this process, called the solubility product constant, Ksp, is in
Ksp = [M+][X-] or Ksp = [Ag+][Cl-]
Le Chatelier's Principle is used in predicting the effect of temperature on solubility. It has
been used to estimate the solubility of a salt in water as a function of temperature. As
you learn later the temperature affects solubility product equilibrium and it.
15.4
Temperature and Solubility
As you learn later solubility product equilibrium and it is related to temperature.
The effect of temperature on the solubility of a solid in water may be predicted using Le
Principe du Chatelier if it is known whether the process is endo- or exothermic. If the
solution process is endothermic, solubility increases with increasing temperature; if
exothermic, solubility decreases. Le Chatelier's Principle permits us to make only
qualitative predictions of solubility for a particular solute-solvent system.
8
15.5
Pressure and Dissolving Gases in Liquids: Henry's Law
Pressure has little effect on solution process of solids and liquids. When the solute is
a gas pressure increase the solubility. The amount of gas, which dissolves in a given
amount of solvent, is greater at high pressure than it is at low pressure.
The mass of a gas, which will dissolve in a liquid at a given temperature, varies directly
with the partial pressure of that gas. This statement is Henry's Law, named in honor of
William Henry, the English chemist who first discovered this relationship.
Sg = kHPg
where Sg is the solubility
kH is the Henry’s the Law constant
Pg is partial pressure of gas
Increasing the pressure of a gas above a liquid increases its solubility
Table: Molar Henry's Law Constants for Aqueous Solutions at 25oC
Gas
Constant
(Pa/(mol/litre))
He
O2
N2
CO2
NH3
282.7 x 10+6
74.68 x 10+6
155. x 10+6
2.937 x 10+6
5.69 x 10+6
Constant
(atm/(mol/litre))
2865.
756.7
1600.
29.76
56.9
Describe the effect of the following on solubility of a solute in a solvent:
a) Temperature b) Pressure.
Consider following cases: KNO3(solid solute)/water; mixture of N2 and
O2(gaseous solutes)/H2O (Henry's law).
a) Temperature
i) Solid solutes: Solubility of solid solutes in water normally increases with increasing
temperature. For example, solubility of KNO3 increases with increasing temperature.
9
ii) Gaseous solutes: Solubility of gaseous solute normally decreases with increasing
temperature. For example, the solubility of N2 and O2 in water decreases with
temperature.
b) Pressure
i) Solid solutes: The pressure above a solution has little or no effect on the solubility of
the solid solute. For example, the pressure has no effect on the solubility of KNO3 in
water.
ii)Gaseous solutes: Solubility of gaseous solute normally increases with increasing
temperature. Soft drinks are made by dissolving CO2 in water under high pressure. The
effect of solubility of gases in liquids are governed by the Henry's Law:
Sg = kH pg
pg= partial pressure of gas above the solution in atm.
Sg= concentration of gas in the solution in mol/L.
kH= Henry's constant which is characteristic of the gas and the solvent.
kH is obtained from the slope of a graph where kH is plotted against pg.
Henry's constant, kN2 for N2/H2O solution is greater than kO2 for O2/H2O solution.
Therefore, at higher pressures more N2 dissolves in water (blood) than. This solubility
difference at high pressures is the cause of "divers bend" found among deep water divers.
15.6
Solution Concentration: Keeping Track of Units
Describe and define the following terms used for solutions:
a) Solvent
b) Solute
c) Molarity (M)
d) Molality (m)
e) Mole fraction (Ca)
f) Mass percent (% weight)
g) Volume percent (% volume)
h) "Proof"
i) ppm and ppb
a) Solvent: Solvent is the liquid which make up the bulk of a solution. Aqueous solutions
are solutions of solids, liquids and gaseous solutes in water. It is important to realize that
the definition of solution includes any homogenous phase including gases, liquids and
10
solids. A gas dispersed in a liquid is called an aerosol. There could be solid solutions
(solid-solid), which are called alloys. There could be solid solutions such as bronze, which
is a homogenous solid solution of copper and tin.
b) Solute: Solute is other substance in the solution once the solvent is identified.
c) Molarity(M)
Molarity (M) =
d) Molality (m)
Molaity (m) =
moles of solute
-----------------Liters of solution
moles of solute
-----------------------kg of solvent
e) Mole fraction (Ca)
Mole fraction (Ca) =
moles of solute
-----------------------moles of solute + solvent
f) Mass (w/w) % or Weight %
Mass of solute
Mass (w/w) % = -----------------------Mass of solution
g) Volume (v/v) % or Volume %
Volume of solute
Volume (v/v) % = -----------------------Volume of solution
x 100
x 100
h) Proof = Volume % x 2
i) ppm or ppb (w/w or v/v)
ppm (w/w or v/v) =
ppb (w/w or v/v) =
Mass (volume) of solute
---------------------------Mass (volume) of solution
Mass (volume) of solute
-----------------------------Mass (volume) of solution
x 106
x 109
A solution containing 58.5 g of NaCl in 2206 g of water has a density of 1.108
g/cm3. Calculate the Molarity of the solution.
11
moles of solute
Molarity(M) = -----------------------Liters of solution
solute = NaCl; m.w = 58.44 g/mole
moles of NaCl = ?
58.5 g NaCl
1 mole NaCl
= 1.00 mole NaCl
58.44 g NaCl
To get liters of solution:
Total mass of solution = 58.5 g + 2206 g = 2264.5 g
use g/cm3 (density) to convert g -- cm3 and then 1L/1000cm3 to cm3 ---- L
2264.5 g solution 1000 cm3 solution 1 L solution
1.108 g solution
1000 cm3 solution
= 2.044 L solution
1.00 mole NaCl
Molarity of NaCl solution = ------------------------- = 0.489 M
2.044 L solution
Calculate the molality of C2H5OH in water solution which is prepared by mixing
75.0 mL of C2H5OH and 125 g of H2O at 20oC. The density of C2H5OH is 0.789
g/mL.
moles of solute
Moles ofC2H5OH
Molality(m) = ----------------------- = -----------------------------kg of solvent
kg of solvent
m.w. (C2H5OH) = 46.08 g/mole d(C2H5OH) = 0.789 g/mL
To get moles of C2H5OH
75.0 mL C2H5OH
0.789 g C2H5OH
1 kg H2O
1 mL C2H5OH
46.08 g C2H5OH
12
= 1.284 mole C2H5OH
To get kg of H2O
125 g H2O
1 kg H2O
= 0.125 kg
H2O
1000 g H2O
1.284 mole C2H5OH
Molality(m) = ------------------------ = 10.27 m
0.125 kg H2O
Calculate the molarity of a solution of water/alcohol containing 35% C2H5OH by
weight. The density of this solution is 1.10 g/mL.
moles of solute
Moles of C2H5OH
Molarity(M) = --------------------- = --------------------L of solution
volume(L) of the solution
m.w (C2H5OH) = 46.08 g/mole d(C2H5OH) = 1.10 g/mL
To get moles of C2H5OH: ( 35% C2H5OH solution has 35g C2H5OH + 65g H2O)
35 g C2H5OH
1 mole C2H5OH
= 0.7595 mol C2H5OH
46.08 g C2H5OH
To get liters of solution:
100 g solution
1 mL solution
1 L solution
1.10 g solution
1000 mL solution
= 0.0901 L
0.7595 mole C2H5OH
Molarity (M) = ------------------------------- = 8.36 M
0.0901 L
Calculate the mole fraction of benzene in a benzene(C6H6)-chloroform(CHCl3)
solution which contains 60 g of benzene and 30 g of chloroform.
For a mixture containing
moles(n) of a
na
Mola Fraction(X a) = --------------------------=
-------------moles(n) of a + moles(n) of b
na + nb
a = C6H6
b = CHCl3
nC6H6
Mola Fraction(X a) = -----------------nC6H6 + nCHCl3
m.w (C6H6) = 78.12 g/mole m.w (CHCl3) = 119.37 g/mole
13
To get moles of C6H6:
60 g C6H6
1 mole C6H6
= 0.768 mole C6H6
78.12 g C6H6
To get moles of CHCl3:
30 g CHCl3
1 mole CHCl3
= 0.251 mole CHCl3
119.37 g CHCl3
X6H6
0.768
=
XCHCl3 =
= 0.754
0.768 + 0.251
1.019
0.251
0.251
0.251 +
0.768
0.754 + 0.246 = 1
15.7
0.768
= 0.246
1.019
X C6H6= 0.754
X cCHCl3 = 0.246
Note: X C6H6+ X CHCl3 = 1
Vapor Pressures, Boiling Points, and Freezing Points of Solutions
What is the effect of concentration of solute (non-volatile) on the following
properties of solvents:
a) Vapor pressure (Raoult's Law); ( Psolution = solvent Posolvent)
b) Boiling point (Tb = Kb msolute)
c) Freezing point (Tf = Kf msolute)
Vapor pressure, boiling point, freezing point and osmotic pressure are called colligative
properties since they are related to number of solute particles in a solution. Number of
non-dissociating solute particles in a solution is expressed as c( mole fraction),
m(molality), or M(molarity).
14
a) Vapor pressure (Raoult's Law); Psolution = solvent Posolvent)
The vapor pressure of a liquid is decreased or lowered by dissolving a solid solute in
water. The vapor pressure a solution is given by the equation
Vapor pressure of a solution is governed by Raoult's law.
Psolution = solvent Posolvent
Note: csolvent= (1-solute) making an equivalent form to the equation:
Psolution = (1- solute)Posolvent
Psolution = vapor pressure over the solution due to solvent molecules
solvent = mole fraction solvent in the solution
Posolvent =vapor pressure of the pure solvent
vapor pressure over a solution increases with the increasing solvent.
b) Freezing point
The freezing (melting or fusion) point of a liquid is decreased by dissolving a solid solute
in water. The boiling point decrease or the depression of a solution is given by the
equation:
Tf = Kf msolute
Tf = boiling point elevation.
Kf = molal freezing point depression constant-molal means concentration is in given
as molality(m).
msolute= concentration of solute expressed as molality(m).
c) Boiling point
The boiling point of a liquid is increased by dissolving a solid solute in water. The boiling
point increase or the elevation of a solution is given by the equation:
Tb = Kb msolute
Tb = boiling point elevation.
Kb = molal boiling point elevation constant-molal means concentration is in
given as molality(m).
msolute= concentration of solute expressed as molality(m).
Osmotic Pressure of Solutions
d) Osmotic pressure ( = MRT )
Osmotic pressure is a colligative property which depends on the number and size of
solute particles. For example, two solutions, water and salt water, separated by a
semipermeable membrane, a membrane that allows only water molecules to pass
15
through, the pressure exerted by the pure water to push more water into salt solution is
called the osmatic pressure. Osmotic pressure of a solution can be calculated using the
equation:
 = MRT
 = Osmatic pressure of the solution
M= Molarity of the solute in the solution
R = Ideal gas constant (R= 0.08026 or 62.4)
T= Temperature of the solution in Kelvin
Calculation Colligative Properties of Solutions Containing Ionic Compounds
To calculate the colligative properties of a solutions containing ionic compounds or
salts that can dissocite into ions, for example-NaCl --- Na+ and Cl-, these equations,
Psolution = (1-csolute)Posolvent , T = Kb msolute, T = Kf msolute,  = MRT ,
should be multiplied by a factor called Vant Hoff factor (i) to include the effect of extra
particles or more ions generated through the complete dissociation of NaCl. The Vant Hoff
factor or NaCl is equal to 2 (NaCl ---1 Na+ + 1Cl-). The Vant Hoff factor or CaCl2 is equal
to 3(CaCl2 --- 1 Ca2+ + 2 Cl-). For weak electrolytes which dissociate in completely the
Vant Hoff factor is always less than the what is expected from complete dissociation.
The forms of the colligative property equations for dissociating solutes:
Psolution = (1-i csolute) Posolvent
T
= i Kb msolute,
T
= i Kf msolute,
P
= i MRT
i
= Vant Hoff Factor
moles of ions in the solution
i = -------------------------------moles of solute dissolved
The vapor pressure above a glucose-water solution at 25oC is 23.8 torr. What is
the mole fraction of glucose (non-dissociating solute) in the solution. The vapor
pressure of water at 25oC is 30.5 torr.
This is a problem to use Raoult's law [Psolution = (1- csolute)Posolvent] and calculate the mole
fraction of solute, glucose (msolute= ?) given the vapor pressure of the solution (Psolution =
23.8 torr) and the pure solvent(Posolvent = 30.5 torr).
Psolution = (1- csolute) Posolvent
23.8 torr = (1- csolute) 30.5 torr
(1- csolute) = 23.5/30.5
(1- csolute) = 0.7803
-csolute = 0.7803 -1
-csolute= -0.2196
16
csolute= 0.2196
Mole fraction of glucose = 0.2196
A 2.25g sample of a compound is dissolved in 125 g of benzene. The freezing
point of the solution is 1.02oC. What is the molecular weight of the compound.
Kf for benzene = 5.12 oC/m, freezing point = 5.5oC.
This is a problem using the equation, DT f =Kf msolute to calculate the molality of a solute
(msolute= ?) given the freezing point (1.02oC) of a solution, freezing point of the pure
solvent (5.5oC) and Kf (5.12 oC/m), molal freezing point depression constant.
First we have to calculate T f:
Note that DT is expressed as an absolute value: T =5.5 -1.02 = 4.48
T f = Kf msolute
4.48 = 5.12 x msolute
msolute = 4.48/5.12
msolute = 0.875 mol/kg
Once the mole fraction is calculated, moles of solute can be can be calculated
because
moles of compound
Molality(m) = -------------------------kg of solvent
=
Moles of compound
----------------------kg of H2O
Moles of compound = Molality(m) x 1 kg of H2O
Molality = 0.875 mol/kg of H2O
Moles of compound = 0.875 moles
Grams of the compound in 1 kg water: 125 g of water has 2.25 g of the compound.
Therefore, grams of the compound in 1 kg water = (2.25/125) x 1000 = 18 g
Grams of the compound in 1 kg water = 18 g
Once the number of moles and the grams of the of the compound in 1kg of water is
calculated, molecular weight of the compound can be can be calculated from the because
grams of the compound in 1 kg water
18g
Molecular weight(grams/mole) = --------------------------------------------- = -----------moles of the compound in 1kg water
0.875 moles
Molecular weight of the compound =20.57 g/mole
Calculate the osmotic pressure in torr of a 0.500 M solution of NaCl in water at
25oC. Assume a 100% dissociation of NaCl.
17
To calculate the osmotic pressure of a solution containing salts that can be
dissociated into ions, for example: NaCl --- Na+ and Cl-, these equations, P = MRT ,
should be multiplied by a factor called Vant Hoff factor (i) to include the effect of the more
particles or more ions generated through the complete dissociation of NaCl. The Vant Hoff
factor or NaCl is equal to 2 (NaCl ---1 Na+ + 1Cl-), 1 + 1 = 2.
 = i MRT
 = Osmatic pressure of the solution = ?
M= Molarity of the solute in the solution = 0.500 M
R = Ideal gas constant = 62.4 L-torr/mol K
T= Temperature of the solution in Kelvin = 25oC +273.15 = 298.15 K
i = (NaCl ---1 Na+ + 1Cl-), 1 + 1 = 2
 = 2 x 0.500 M x 62.4 L-torr/mol K x 298.15 K
 = 18605 torr
The osmotic pressure of the solution = 18605 torr
Predict the type of behavior (ideal, negative, positive) based on vapor pressure
of the following pairs of volatile liquids and explain it in terms of intermolecular
attractions:
a) Acetone((CH3)2CO)/water (H2O) ; b) Ethanol(C2H5OH)/hexane (C6H14);
c) Benzene (C6H6)/toluene(CH3C6H5).
a) Acetone((CH3)2CO)/water (H2O).
Acetone and water will form strong intermolecular forces of attraction because of their
capacity to form hydrogen bonds. This solution, therefore, shows negative deviation
from the Raoults law.
Negative deviation: Lower vapor pressure than the sum of the individual vapor
pressures.
18
b) Ethanol (C2H5OH)/hexane (C6H14);
Ethanol and hexane molecules have different bonding. Ethanol have polar O-H
bond and the hexane have non polar covalent C-H bonds. Because of the unlike nature of
molecules, they repel each other and lead to an increase in the vapor pressure of the
solution called positive deviation from the Raoults law.
Positive deviation: Higher vapor pressure than the sum of the individual vapor
pressures.
c) Benzene (C6H6)/toluene CH3C6H5.
Benzene and toluene are two similar molecules both having non polar covalent C-H
bonds. Because of their like nature, molecules do not repel or attract each other and the
vapor pressure of the solution is the sum of the vapor pressures of the two individual
components. The no change in the pressure of the solution is called ideal behavior
according to the Raoults law
19
.
In the diagram below shows the indicated plot of vapor pressure versus mole
fraction of one component of a solution made from two volatile liquids.
Categorize the behavior of this solution as based on Raoult's law as ideal,
positive deviation or negative deviation.
a) shows ideal behavior
behavior
(b) shows positive behavior
(c) shows negative
Define the Van't Hoff factor (i). Which of the following solutions will show the
highest osmotic pressure:
a) 0.2 M Na3PO4 b) 0.2 M C6H12O6 (glucose) c) 0.3 M Al2(SO4)3 d) 0.3 M CaCl2
e) 0.3 M NaCl
Vant Hoff factor ( i ) is defined as:
moles of ions in the solution
20
i = ------------------------moles of solute dissolved
To calculate the osmotic pressure of a solution containing salts that can be
dissociated into ions, for example: NaCl --- Na+ and Cl-, the equation,
 = MRT ,
should be multiplied by a factor called Vant Hoff factor ( i ) to include the effect of the
more particles or more ions generated through the complete dissociation of NaCl. The
Vant Hoff factor or NaCl is equal to 2NaCl ---1 Na+ + 1Cl-). The Vant Hoff factor or CaCl2
is equal to 3(CaCl2 --- 1 Ca2+ + 2 Cl-). For weak electrolytes which dissociate in
completely the Vant Hoff factor is always than the one expected from complete
dissociation.
The forms of the colligative property equations for dissociating solutes:
 = i MRT
i = Vant Hoff Factor
a) 0.2 M Na3PO4
= i MRT
i = Vant Hoff Factor = 4; (Na3PO4 dissociates in water into three Na+ and one PO43- ions
giving
total of 4 particles).
M = 0.2; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
= 4 x 0.2 x 62.4 x 298.15 = 14,884 torr
b) 0.2 M C6H12O6 (glucose)
 = i MRT
i = Vant Hoff Factor = 1; (C6H12O6does not break up in the solution. It's a covalent
compound)
M = 0.2; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
 = 1 x 0.2 x 62.4 x 298.15 = 3,721 torr
c) 0.3 M Al2(SO4)3
 = i MRT
i = Vant Hoff Factor = 5 ; Al2(SO4)3 dissociates in water into two Al3+ and three SO42- ions
giving
total of 5 particles.
M = 0.3; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
 = 5 x 0.3 x 62.4 x 298.15 = 27,907 torr
0.3 M Al2(SO4)3 shows the highest osmotic pressure.
d) 0.3 M CaCl2
21
 = i MRT
i = Vant Hoff Factor = 3 ; CaCl2 dissociates in water into one Ca2+ and two Cl - ions giving
total of 3 particles
M = 0.2; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
 = 3 x 0.3 x 62.4 x 298.15 = 16,744 torr
e) 0.3 M NaCl
= i MRT
i = Vant Hoff Factor = 2 ; NaCl dissociates in water into one Na+ and one Cl - ions giving
total of 2 particles
M = 0.2; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
 = 3 x 0.3 x 62.4 x 298.15 = 16,744 torr
15.9
Colloids
Colloids and Suspensions
Describe the following terms:
a) True solutions b) Colloids (Tyndall effect) c) Suspensions.
a) True solutions
Normally light passes through true solutions or they transparent to visible light.
Because of small solute particle sizes they do not scatter light.. True solutions have
solute particles with diameters less than 1 x103
pm. Colloids, on the other hand have
solute particles with diameters in the range 1 x103 to 1 x105 pm. Suspensions have larger
particle diameters which are greater than 1 x105 pm.
22
b) Colloids (Tyndall effect)
Colloids have solute particle diameters in the rage 1 x103 to 1 x105 pm. Colloids
scatter light and the solution become opaque and relative degree of opacity depends on
the sizes and amount of the particles. The scattering of light by colloidal particles is called
Tyndell effect which has been used to distinguish between true solutions and colloids.
This effect will increase with increasing solute particle diameter of a colloid. For example,
milk is opaue because of the higher diameters of the solute particles such as proteins in
the milk.
c) Suspensions.
Suspensions have particle diameters greater than 1 x105 pm. The particles in a
suspension are large enough to be affected by gravity and settle to the bottom of the
mixture with time. For example, muddy water.
15.10
Surfactants
Surfactants constitute the most important group of detergent components.
Generally, these are water-soluble surface-active agents comprised of a hydrophobic
portion, usually a long alkyl chain, attached to hydrophilic or water solubility enhancing
functional groups. The hydrophilic end, which is either polar or ionic, dissolves readily in
water. The hydrophobic, or non-polar, end, however, does not dissolve in water. In fact,
the hydrophobic, or "water-hating" end will move as far away from water as possible. This
phenomenon is called the hydrophobic effect. Because one end of a surfactant resists
water and the other end embraces it, surfactants have very unique characteristics.
Emulsifiers
A surfactant is also called an emulsifier. Emulsifiers do help oil and water remain in
stable emulsions. Aggregates of oil and emulsifiers form micelles and stay dispersed in
water solution. Examples of emulsifiers include sodium dodecyl sulfate.
23
All surfactants possess the common property of lowering surface tension when
added to water in small amounts, as illustrated in the plot below. The characteristic
discontinuity in the plots of surface tension against surfactant concentration can be
experimentally
determined.
The
corresponding
surfactant
concentration
at
this
discontinuity corresponds to the critical micelle concentration (CMC). At surfactant
concentrations below the CMC, the surfactant molecules are loosely integrated into the
water structure (monomer see figure below). In the region of
the
CMC,
the
surfactant-water
structure is changed in such a way
that the surfactant molecules begin
to build up their own structures
(micelles
in
the
interior
and
monolayers at the surface).
When a small concentration
of surfactant is added to water, the
hydrophobic end will immediately
rise to the surface. The surfactant is then stable. The polar end is happily immersed in the
polar solvent while the non-polar end rises above the surface. This configuration is called
a monolayer. If more surfactant is added and there is not enough room for all of the
hydrophobic ends to stick out of the water, a bilayer will form. However, if the
concentration of surfactant is large enough, even the bilayer configuration will not be
stable. At this point, a micelle forms, in which a group of hydrophilic ends surround their
hydrophobic "tails" and shield them from the water. The micelle, therefore, consists of a
hydrophilic shell and a hydrophobic core.
A very important effect of surfactants in cleaning products is the wetting effect.
Because of the reduced surface tension, the water can be more evenly distributed over
the surface and this improves the cleaning process. The emulsifying effect of surfactants
is important for both cleansing and washing of textiles. Due to the hydrophobic and
hydrophilic parts, surfactants can sorb to non-polar and polar materials at the same time.
During cleansing and washing, the non-polar materials are kept in emulsions in the
aqueous solution and removed by rinsing. By varying the hydrophobic and hydrophilic
24
part of a surfactant, a number of properties may be adjusted, e.g. wetting effect,
emulsifying effect, dispersive effect, foaming ability and foaming control.
Soap the most common surfactant
One of the organic chemical reactions known to ancient man was the preparation of
soaps through a reaction called saponification. Natural soaps are sodium or potassium
salts of fatty acids, originally made by boiling lard or other animal fat together with lye or
potash (potassium hydroxide). Hydrolysis of the fats and oils occurs, yielding glycerol and
crude soap.
In the industrial manufacture of soap, tallow (fat from animals such as cattle and
sheep) or vegetable fat is heated with sodium hydroxide. Once the saponification reaction
is complete, sodium chloride is added to precipitate the soap. The water layer is drawn
off the top of the mixture and the glycerol is recovered using vacuum distillation.
The crude soap obtained from the saponification reaction contains sodium chloride,
sodium hydroxide, and glycerol. These impurities are removed by boiling the crude soap
curds in water and re-precipitating the soap with salt. After the purification process is
repeated several times, the soap may be used as an inexpensive industrial cleanser.
Sand or pumice may be added to produce a scouring soap. Other treatments may result
in laundry, cosmetic, liquid, and other soaps. Soap is an anionic surfactant.
Detergent
The term "detergent" is used for a surfactant material which cleans (or is used for
cleaning), but in this definition soap is not included. Even so, this is still a wide definition,
because, of course, it can refer to the active ingredient, or the solid, liquid, paste or
powder compounded from this active matter.
Surfactants are grouped according to their ionic properties in water:
25

Anionic surfactants have a negative charge

Nonionic surfactants have no charge

Cationic surfactants have a positive charge

Amphoteric surfactants have positive or negative charge dependent on pH A list of
surfactants which are commonly used in biochemistry is presented in the table
below.
Structures of common surfactants:
Chemical or Trade Name
Sodium dodecylsulfate (SDS)
Sodium cholate
Sodium deoxycholate (DOC)
N-Lauroylsarcosine Sodium salt
Lauryldimethylamine-oxide (LDAO
Cetyltrimethylammoniumbromide (CTAB)
Bis(2-ethylhexyl)sulfosuccinate Sodium salt
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15.11
Water: Natural, Clean, and Otherwise
Fresh Water
Fresh water is an essential ingredient of modern life. Thought it’s often available as
the result of natural processes (natural water), there are times when it must be
extracted from impure water, typically salt water. In some middle eastern countries where
rain water is scarce, desalinated sea water through reverse osmosis is the main source
of drinking water.
Water Purification
Any extraction process that purifies water must separate water molecules from
contaminating liquids, solids, or gases.
Desalination: Removing Salt from Sea water
Distillation
One way to purify water is by distillation. Distillation is a general technique for
separating various chemicals from one another. The chemicals are heated to form
a vapor and that vapor is condensed to form a new mixture of chemicals. Because
the various chemicals have different tendencies to form vapors at a particular
temperature, the newly formed mixture has a different balance of the chemicals
from the original mixture. In some cases, the condensed liquid contains primarily
a single chemical—all of the other chemicals are left behind in the original liquid.
Freezing
Freezing salt water to form pure ice works best in cold climates where low temperatures
are available directly. Active refrigeration can also freeze salt water
to obtain fresh water, but it’s expensive. Because of water’s latent heat of melting,
you must remove a large amount of heat from salt water to freeze it. Although
refrigerated water desalination plants have been built, they have proven to be
less economical than distillation plants.
Reverse Osmosis
Another way to desalinate water is by reverse osmosis, a process resembling filtration
except that it takes place at the molecular scale. In effect, salt water is converted
27
to fresh water by filtering the impurities out of it with an incredibly fine filter. This filter is
a semipermeable membrane. It is a surface that only allows certain molecules to pass
through it.
However, because it operates at such a tiny size scale, reverse osmosis encounters
some peculiar pressure effects that you don’t see with larger filters where gravity or
suction is the main force. The way to stop osmosis from diluting concentrated salt water
is to increase the pressure of the salt water side. Increasing that pressure tends to drive
water molecules back out of the salt water and through the semipermeable membrane to
pure water side. The two fluids will still reach a phase equilibrium, but the salt water will
retain a higher concentration of immobile molecules than the fresh water. Maintaining a
large difference in concentrations between the two fluids require a huge pressure
difference of many tens or hundreds of atmospheres.
Municipal or Utility Water Treatment
Most municipal water found in a city or community today has been treated extensively.
Specific water treatment methods and steps taken by municipalities to meet local, state,
national, or international standards vary but are categorized below.
Coarse filtration
A coarse screen, usually 50 to 100 mesh (305 to 140 microns), at the intake point of a
surface water supply, removes large particulate matter to protect downstream equipment
from clogging, fouling, or being damaged.
28
Addition of alum or lime
Addition of alum or lime produces aggregate suspended solids (microbes, organic matter,
toxic contaminants) which is then removed by sedimentation in a settling basin. The
sedimentation process is aided by the application of alum (aluminum sulfate) to the raw
lake water this material when added to water containing normal debris such a mud,
particles and color causes these materials to clump and settle to the bottom of
sedimentation tanks where it removed. The clear water at the very top of these units is
skimmed off leaving the alum and debris behind.
Settling or Clarification
Settling or clarification is generally a multi-step process to reduce turbidity and suspended
matter. Steps include the addition of chemical coagulants or pH-adjustment chemicals
that react to form floc. The floc settles by gravity in settling tanks or is removed as the
water percolates through a gravity filter. The clarification process effectively removes
particles larger than 25 microns. The clarification process is not 100% efficient; therefore,
water treated through clarification may still contain some suspended materials.
Sand filtration
Sand filtration is a "physical and sometimes a chemical process for separating suspended
and colloidal impurities from water by passage through a bed of granular sand material.
The purpose of this filtration is to remove any particulate matter left over after
flocculation and settling. Water fills the pores of the filtering sand medium, and the
impurities are adsorbed on the surface of the grains or trapped in the openings." The key
to this process is the relative grain size of the filter medium.
Aeration
Aeration consists of breaking the incoming water into small droplets (spray) into the air,
drawing fresh air through that spray, collecting the water into a storage tank.
The air
drawn though the spray must be vented outside the house -- remember, it can have
odors, toxic substances. Although this system necessitates another pump to repressurize
your supply, you are not adding any chemicals to your water, which makes it attractive.
29
This system is low maintenance and no chemicals to purchase. Initial cost may be higher,
and space requirements may be greater.
Chlorination or Ozonation: Disinfection
Disinfection is one of the most important steps to municipal water treatment. Usually,
chlorine gas is fed into the supply after the water has been clarified and/or softened. The
chlorine kills bacteria. In order to maintain the "kill potential", an excess of chlorine is fed
into the supply to maintain a residual. The chlorine level must be constantly monitored to
assure that no harmful levels of chloramines or chlorinated hydrocarbons develop.
Ozonation
A chemical-free disinfection is known as ozonation. Ozonation relies on oxygen to ensure
that our purified water remains free of any possible microbiological contamination. The
ozonation process involves taking basic molecular oxygen (O2) and passing this oxygen
through a special chamber in which it is exposed to a high voltage electrical charge. (This
type of ozone generation is called cold-plasma discharge.) The electricity causes the
oxygen molecule to split and recombine in a higher-energy form known as ozone (O3).
This ozone is then continuously circulated through the purified water. Ozone is a very
powerful disinfectant and is capable of oxidizing a very broad range of contaminants. In
fact, ozone is highly effective against many types of impurities and organisms, such as
cryptosporidium, that are utterly impervious to chlorination.
Other Chemical Treatment





pH adjustment. Certain chemicals, membranes, ion exchange resins and other
materials are sensitive to specific pH conditions. An example of this is to prevent
acid corrosion in boiler feed water by adjusting the pH so it is between 8.3 to 9.0.
Dispersants. Dispersants are added when scaling may be expected due to
concentration of specific ions in the stream. Dispersants disrupt the scale formation,
preventing growth of precipitate crystals.
Sequestering (chelating) agents. Sequestering agents are used to prevent the
negative effects of hardness, preventing the deposition of Ca, Mg, Fe, Mn and Al.
Oxidizing agents. Oxidizing agents have two distinct functions: as a biocide, or to
neutralize reducing agents.
Ion Exchangers: Strong base anion resin can be used for removing anionic metal
complexes from acidic waters like ZnCl42- in spent pickle solution or Cr(VI) in rinse
water after chromating (Tan). Weakly acidic exchangers (-COO- ) have shown good
30


separation performance for Zn in plating waste (Uy) and for nickel (Halle), however
weakly acidic resins have no widespread applications in the plating industry.
Potassium permanganate. Potassium permanganate (KMnO4) is a strong
oxidizing agent used in many bleaching applications. It will oxidize most organic
compounds and is often used to oxidize ferrous iron to ferric for precipitation and
filtration.
Reducing agents. Reducing agents, like sodium metabisulfite (Na2S2O5), are
added to neutralize oxidizing agents such as chlorine or ozone. In membrane and
ion exchange systems, they prevent the degradation of certain membranes or
resins, which are sensitive to oxidizing agents.
This hypermedia page is prepared by Upali Siriwardane
Chemistry Program, P.O. Box 10348 T.S.,
Carson Taylor Hall, Room 311,Louisiana Tech University,
Ruston, LA 71272-0001
This page was last modified on March 31, 2009
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