Solutions Chapter 12 Guide

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Solution Formation
1. Types of Solutions
2. Solubility and the Solution Process
3. Effects of Temperature and Pressure on
Solubility
Contents and Concepts

Colligative Properties
4. Ways of Expressing Concentration
5. Vapor Pressure of a Solution
6. Boiling-Point Elevation and Freezing-Point
Depression
7. Osmosis
8. Colligative Properties of Ionic Solutions
Colloid Formation
9. Colloids

Solution Formation
1. Types of Solutions
a. Define solute and solvent.
b. Define miscible fluid.
c. Provide examples of gaseous solutions,
liquid solutions, and solid solutions.
Learning Objectives

2. Solubility and the Solution Process
a. List the conditions that must be present to
have a saturated solution, to have an
unsaturated solution, and to have a
supersaturated solution.
b. Describe the factors that make one
substance soluble in another.
c. Determine when a molecular solution will
form when substances are mixed.
d. Learn which conditions must be met to
create an ionic solution.

3. Effects of Temperature and Pressure on
Solubility
a. State the general trends for the solubility of
gases and solids with temperature.
b. Explain how the solubility of a gas changes
with temperature.
c. Apply Henry’s law.

Colligative Properties
4. Ways of Expressing Concentration
a. Define colligative property.
b. Define molarity.
c. Define mass percentage of solute.
d. Perform calculations with the mass
percentage of a solute.
e. Define molality.
f. Calculate the molality of a solute.
g. Define mole fraction.
h. Calculate the mole fractions of components.

i. Convert molality to mole fraction.
j. Convert mole fraction to molality.
k. Convert molality to molarity.
l. Convert molarity to molality.
5. Vapor Pressure of a Solution
a. Explain vapor-pressure lowering of a
solvent.
b. State Raoult’s law.
c. Calculate vapor-pressure lowering.
d. Describe an ideal solution.

6. Boiling-Point Elevation and Freezing-Point
Depression
a. Define boiling-point elevation and freezing-
point depression.
b. Calculate boiling-point elevation and
freezing-point depression.
c. Calculate the molecular mass of a solute
from molality.
d. Calculate the molecular mass from
freezing-point depression.

7. Osmosis
a. Describe a system where osmosis will take
place.
b. Calculate osmotic pressure.
8. Colligative Properties of Ionic Solutions
a. Determine the colligative properties of ionic
solutions.

Colloid Formation
9. Colloids
a. Define colloid.
b. Explain the Tyndall effect.
c. Give examples of hydrophilic colloids and
hydrophobic colloids.
e. Describe coagulation.
f. Explain how micelles can form an
association colloid.

A solution is composed of two parts: the solute and
the solvent.
Solute
The gas (or solid) in a solution of gases (or solids),
or the component present in the smaller amount.
Solvent
The liquid in the case of a solution of gases or
solids, or the component present in the larger
amount.

Fluids that mix with or
dissolve in each other in
all proportions are said
to be miscible (left).
Fluids that do not
dissolve in each other
are said to be
immiscible (right).

a.The 5 g of Mo is the solute; the 80 g of Cr is the
solvent.
b.MgCl2 is the solute; H2O is the solvent.
c. O2 and N2 are the solutes; Ar is the solvent.
Concept Check 12.1
Identify the solute(s) and solvent(s) in the following
solutions.
a. 80 g of Cr and 5 g of Mo
b. 5 g of MgCl2 dissolved in 1000 g of H2O
c. 39% N2, 41% Ar, and the rest O2

A saturated solution is
in equilibrium with
respect to the amount
of dissolved solute.
The rate at which the
solute leaves the solid
state equals the rate at
which the solute
returns to the solid
state.

The solubility of a solute is the amount that
dissolves in a given quantity of solvent at a given
temperature.
An unsaturated solution is a solution not in
equilibrium with respect to a given dissolved
substance and in which more of the substance can
be dissolved.

A supersaturated solution is a solution that
contains more dissolved substance than a
saturated solution does. This occurs when a
solution is prepared at a higher temperature and is
then slowly cooled. This is a very unstable
situation, so any disturbance causes precipitation.

Solubility can be understood in terms of two
factors:
1. The natural tendency toward disorder
favors dissolving.
2. The relative forces between and within
species must be considered.
Stronger forces within solute species oppose
dissolving.
Stronger forces between species favor dissolving.

For molecular solutions,
this can be summarized
as “Like dissolves like.”
In other words, solutes
dissolve in solvents that
have the same type of
intermolecular forces.
An immiscible solute
and solvent are
illustrated at right.

Concept Check 12.2
You have two molecular compounds, X and Y.
Compound X has stronger intermolecular forces
than compound Y. X has no dipole-dipole bonding of
any kind, and Y exhibits hydrogen bonding. Predict
the relative solubility of compounds X and Y in water
and in a nonpolar solvent. Explain your reasoning.

When considering ionic solutes in water, we need
to examine the hydration energy and the lattice
energy.

The stronger ion-dipole
force between the ion
and the solvent—that is,
hydration energy—
favors dissolving.
A stronger force
between ions—that is,
lattice energy—opposes
dissolving.

The force of attraction
between water and
both a cation and an
anion is illustrated to
the left with lithium
fluoride, LiF.

The process of
dissolving occurs at
the surfaces of the
solid. Here we see
water hydrating
(dissolving) ions.

The hydration energy for AB2 must be greater than
the hydration energy for CB2.
Concept Check 12.3
The hypothetical ionic compound AB2 is very
soluble in water. Another hypothetical ionic
compound, CB2, is only slightly soluble in water. The
lattice energies for these compounds are about the
same. Provide an explanation for the solubility
difference between these compounds.

In general, solubility depends on temperature.

In most cases, solubility increases with increasing
temperature. However, for a number of
compounds, solubility decreases with increasing
temperature.
The difference is explained by differences in the
heat of solution.

When dissolving absorbs heat (is endothermic),
the temperature of the solution decreases as the
solute dissolves. The solubility will increase as
temperature increases.
When dissolving releases heat (is exothermic), the
temperature of the solution increases as the solute
dissolves. The solubility will decrease as
temperature increases.

Hot packs use an
exothermic solution
process.
Cold packs use an
endothermic solution
process.

Henry’s law describes the effect of pressure on
gas solubility: The solubility of a gas in a liquid is
directly proportional to the partial pressure of the
gas above the solution.
This is expressed mathematically in the equation
S = kHP
where
S = gas solubility
kH = Henry’s law constant for the gas
P = partial pressure of the gas over the solution

In general, pressure has
little or no effect on the
solubility of solids or
liquids in water.
The solubility of a gas
increases as pressure
increases, as illustrated
at right.

?
Helium–oxygen mixtures are sometimes
used as the breathing gas in deep-sea
diving. At sea level (where the pressure is
1.0 atm), the solubility of pure helium in
blood is 0.94 g/mL. What is the solubility of
pure helium at a depth of 1500 feet?
Pressure increases by 1.0 atm for every 33
feet of depth, so at 1500 feet the pressure is
46 atm. (For a helium–oxygen mixture, the
solubility of helium will depend on its initial
partial pressure, which will be less than 1.0
atm.)

P1 = 1.0 atm P2 = 46 atm
S1 = 0.94 g/mL S2 = ?
1
2
12
P
P
SS ×=
atm1.0
atm46
mL
g0.94
2
×=S
mL
g43
2
=S

At high elevations, the partial pressure of oxygen
decreases, decreasing the solubility of oxygen in
water. Fish require a certain minimum level of
dissolved oxygen to survive.
Concept Check 12.4
Most fish have a very difficult time surviving at
elevations much above 3500 m. Howcould Henry’s
law be used to account for this fact?

The concentration of a solute can be quantitatively
expressed in several ways:
1. Molarity
2. Mass percentage of solute
3. Molality
4. Mole fraction

Molarity is the moles of solute per liter of solution.
It is abbreviated as M.
solutionofliters
soluteofmoles
=M

Mass percentage of solute is the percentage by
mass of solute in a solution.
100%
solutionofgrams
soluteofgrams
soluteofpercentageMass
×
=

?
An experiment calls for 36.0 g of a
5.00% aqueous solution of potassium
bromide. Describe how you would
make up such a solution.

A 5.00% aqueous solution of KBr has 5.00 g KBr
per 100. g solution. The remainder of the 100. g is
water: 95 g.
We can use this ratio to determine the mass of KBr
in 36.0 g solution:
Since 1.80 g KBr is required for 36.0 g of solution,
the remainder consists of 34.2 g water.
We make the solution by mixing
1.8 g KBr in 34.2 g water.
KBrg1.80
solution100.g
KBrg5.00
solutiong36.0 =×

Molality is the moles of solute per kilogram of
solvent. It is abbreviated as m.
solventofkilograms
soluteofmoles
=m

?
Iodine dissolves in a variety of organic
solvents. For example, in methylene
chloride, it forms an orange solution.
What is the molality of a solution of
5.00 g iodine, I2, in 30.0 g of methylene
chloride, CH2Cl2?

Mass of solute = 5.00 g I2
Mass of solvent = 30.0 g CH2Cl2
kg1
g10
Ig253.8
Imol1
solventg30.0
Ig5.00 3
2
22
××=m
kg
mol
0.657=m

Mole fraction is the moles of component over the
total moles of solution. It is abbreviated Χ.
solutionofmolestotal
soluteofmoles
=Χ

?
A solution of iodine, I2, in methylene
chloride, CH2Cl2, contains 5.00 g I2 and
56.0 g CH2Cl2. What is the mole fraction
of each component in this solution?

Mass of solute = 5.00 g I2
Mass of solvent = 56.0 g CH2Cl2
mol0.01970
Ig253.8
Imol1
Ig5.00
soluteMoles
2
2
2
=×
=
mol0.6594
ClCHg84.93
ClCHmol1
Ig56.0
solventMoles
22
22
2 =×
=
mol0.6791
mol0.6594mol0.01970molesTotal
=
+=

0.02902I =Χ
totalmol0.6791
Imol0.01970 2
I2
=Χ
0.97122ClCH =Χ
2 2
2 2
CH Cl
0.6594 mol CH Cl
0.6791mol total
Χ =

?
A bottle of bourbon is labeled 94 proof,
meaning that it is 47% by volume of
alcohol in water. What is the mole
fraction of ethyl alcohol, C2H5OH, in the
bourbon? The density of ethyl alcohol
is 0.80 g/mL.
One liter of bourbon contains 470 mL of alcohol
and 530 mL of water. To solve this problem, we
will convert the volume of ethyl alcohol to mass
using density, and then convert to moles using
molar mass.

mol8.16
g46.08
mol1
mL1
g0.80
OHHCmL470 52
=××
0.22ethanol =Χ
mol29.4
g18.02
mol1
mL1
g1.00
OHmL530 2 =××
mol37.6mol29.4mol8.16molesTotal =+=
totalmol37.6
OHHCmol8.16 52
ethanol =Χ

?
A 3.6 m solution of calcium chloride,
CaCl2, is used in tractor tires to give
them weight. The addition of CaCl2 also
prevents water in the tires from
freezing at temperatures above –20°C.
What are the mole fractions of CaCl2
and water in such a solution?
The 3.6 m solution contains 3.6 mol CaCl2 in 1.0 kg
of water. To solve this problem, we will convert the
mass of water to moles, and then compute the
mole fractions.

Moles solute = 3.6 mol CaCl2
0.0612CaCl =Χ
OHmol55.5
g18.02
mol1
kg1
g10
OHkg1.0
solventMoles
2
3
2 =××
=
mol59.1mol55.5mol3.6molesTotal =+=
totalmol59.1
mol3.6
2CaCl =Χ
0.94OH2
=Χ
totalmol59.1
mol55.5
2H =OΧ

Converting between molality and mole fraction is
relatively simple, because you know the masses or
moles of both the solute and the solvent.

?
A solution contains 8.89 × 10-3
mole
fraction of I2 dissolved in 0.9911 mole
fraction of CH2Cl2 (methylene chloride).
What is the molality of I2 in this
solution?
We will assume we have 1 mole of the solution, so
we begin with 8.89 × 10-3
mol I2 and 0.9911 mol
CH2Cl2. Next, we will convert the moles of solvent
into grams, and then into kilograms. Finally, we will
compute the molality of the solution.

Moles solute = 0.00889 mol I2
kg
mol
0.106=m
22
3
ClCHkg0.08417
g10
kg1
mol1
g84.93
mol0.9911solventKilograms
=
××=
kg0.08417
mol0.00889
=m

Converting between molality and molarity requires
knowing the density of the solution. This enables
you to calculate the mass or volume of the
solution. You can then distinguish the amount of
solute from the amount of solvent, or combine
them to find the volume of solution.
These types of conversions are illustrated in the
following problems.

?
Citric acid, HC6H7O7, is often used in
fruit beverages to add tartness. An
aqueous solution of citric acid is 2.331
m HC6H7O7. What is the molarity of the
solution? The density of the solution is
1.1346 g/mL.
A 2.331 m solution contains 2.331 mol solute in
1.000 kg solvent. We will use this relationship first
to convert the moles of citric acid to grams and
then to find the mass of solution. Using density,
we can then find the volume of solution. Finally,
we will compute the molarity of the solution.

Moles solute = 2.331 mol HC6H7O7
L
mol
1.827=
g447.88
mol1
g192.14
mol2.331soluteMass =×=
L1.2761
mL
L10
g1.1346
mL1
g1447.88solutionLiters
3
=
××=
−
g1447.88g1000.00g447.88solutionofMass =+=
L1.2761
mol2.331
=M

?
An aqueous solution of ethanol is 14.1
M C2H5OH. The density of the solution
is 0.853 g/cm3
. What is the molality of
ethanol in the solution?
We will work with 1.00 L of solution. First, we will
convert volume to mass using the density. Then,
we will find the masses of the solute and the
solvent. Finally, we will compute the molality.

g649.7
mol1
g46.08
mol14.1soluteofMass =×=
g853
mL
g0.853
L10
mL1
L1.000solutionofMass 3-
=
××=
kg
mol
69.4=
kg0.2033
g10
kg1
g203.3
g649.7g853solventofMass
3
=×=
−=
kg0.2033
mol14.1
=m

Colligative properties of solutions are properties
that depend on the concentration of the solute
molecules or ions in solution but not on the
chemical identity of the solute.
1. Vapor-pressure lowering
2. Boiling-point elevation
3. Freezing-point lowering
4. Osmotic pressure

The vapor pressure of a solution, P, is less than
the vapor pressure of the pure solvent, P°.
When the solute is nonvolatile, the vapor pressure
of a solution is the mole fraction of the solvent
times the vapor pressure of pure solvent.
o
solventsolventsolution PXP =
o
PP solvent<solution

We can explain this relationship by recalling that
evaporation occurs at the surface of the solution.
When a solute is present, less of the surface is
occupied by solvent molecules.

To establish an equilibrium, the gaseous
solvent will condense in the more concentrated
solution until the vapor pressures and
concentrations are equal.

To find the vapor-pressure lowering (a colligative
property), we rearrange the following equation:
solution
o
solvent PPP −=∆
( )
o
solventsolute
solventsolute 1-Since
PXP =∆
= XX
o
solventsolvent
o
solvent PPP X−=∆
( )solvent
o
solvent 1- XPP =∆

Note that vapor-pressure lowering is directly
proportional to the solute concentration, the
definition of a colligative property.
o
solventsolutesolvent PXP =∆

?
Eugenol, C10H12O2, is the chief
constituent of oil of clove. This pale
yellow liquid dissolves in ethanol,
C2H5OH; it has a boiling point of 255°C.
As a result, we know eugenol’s vapor
pressure is very low; it can be
considered nonvolatile. What is the
vapor-pressure lowering at 20.0°C of a
solution containing 8.56 g of eugenol in
50.0 g of ethanol? The vapor pressure
of ethanol at 20.0°C is 44.6 mmHg.

mol0.0521
g164.22
mol1
g8.56soluteMoles =×=
mmHg2.04Δ =P
mol1.085
g46.08
mol1
g50.0solventMoles =×=
mmHg44.60.04581Δ
Δ 0
solventsolvent
×=
=
P
PΧP
mol1.137
mol1.085mol0.0521molesTotal
=
+=
0.04581
mol1.137
mol0.0521
solute
==Χ

When the solute is nonvolatile, it has no
appreciable vapor pressure itself and forms an
ideal solution.
When the solute is volatile, a nonideal solution
results.

The phase diagram on the next slide shows the
changes in freezing and boiling points when a
nonvolatile solute is added to a solvent. The blue
line shows the pure solvent; the purple line shows
the solution.

The boiling point of a solution is higher than the
boiling point of pure solvent. The boiling-point
elevation, ∆Tb, is given by the following equation:
∆Tb = mKb

The freezing point of a solution is lower than the
freezing point of pure solvent. The freezing-point
depression, ∆Tf, is given by the following equation:
∆Tf = mKf

?
A solution is made up of eugenol,
C10H12O2, in diethyl ether (“ether”). If the
solution is 0.575 m eugenol in ether,
what are the freezing point and the
boiling point of the solution? The
freezing point and the boiling point of
pure ether are –116.3°C and 34.6°C,
respectively. The freezing-point
depression and boiling-point elevation
constants are 1.79°C/m and 2.02°C/m,
respectively.

m = 0.575 m
Kf = 1.79°C/m Kb = 2.02°C/m
Tf° = –116.3°C Tb° = 34.6°C
∆Tf = m × Kf ∆Tb = m × Kb
∆Tf = 0.575 m × 1.79°C/m
∆Tf = 1.03°C This is the freezing-point depression.
Tf = –116.3 – 1.03 = –117.3°C
∆Tb = 0.575 m × 2.02°C/m
∆Tb = 1.16°C This is the boiling-point elevation.
Tb = 34.6 + 1.16 = 35.8°C

?
In a freezing-point depression
experiment, the molality of a solution
of 58.1 mg anethole in 5.00 g benzene
was determined to be 0.0784 m. What
is the molar mass of anethole?
First, we will use the freezing-point depression data
to find the molality of the solution. Next, we will use
the molality and mass of solvent to find the moles of
solute. Finally, we will use the moles of solute and
mass of solute to find the molar mass.

Solute mass = 58.1 mg
Solvent mass = 5.00 g
m = 0.0784 mol/kg
mol103.92
kg0.00500m0.0784
solventkgsolutemol
4-
×=
×=
×= m
solventkg
solutemol
=m
g/mol.148ismassmolarThe
g/mol148
mol103.92
g1058.1
massMolar 4-
-3
=
×
×
=

?
An 11.2-g sample of sulfur was
dissolved in 40.0 g of carbon disulfide.
The boiling-point elevation of carbon
disulfide was found to be 2.63°C. What
is the molar mass of the sulfur in the
solution? What is the formula of
molecular sulfur? The boiling-point
elevation constant, Kb, for carbon
disulfide is 2.40°C/m.

Solute mass = 11.2 g
Solvent mass = 40.0 g
∆Tb = 2.63°C
Kb = 2.40°C/m
First, we will use the definition of the colligative
property to calculate the molality. Next, we will use
the definition of molality to calculate the moles of
solute. Finally, using the mass and moles of
solute, we will find the molar mass and the
molecular formula of the sulfur.

m
m
K
T
m 1.096
C
2.40
C2.63Δ
f
b
=
°
°
==
fbΔ KmT ×=
mol0.04383
kg0.04001.096
solventkgsolutemol
=
×=
×=
m
m
solventkg
solutemol
=m

.8SformulamolecularThe
g/mol255.5
mol0.04383
g11.2
massMolar ==
g/mol.32.065ismassformulaThe
S.issulfurforformulaempiricalThe
8
32.065
255.5
≈=n

Osmosis is the phenomenon of solvent flow
through a semipermeable membrane to equalize
the solute concentration on both sides of the
membrane.
A semipermeable membrane allows solvent
molecules to pass through but not solute
molecules.

Water is the
solvent.
Water will flow
from the left to
the right.

Water is the solvent.
It flows from the beaker into
the thistle tube.

Osmotic pressure, π, is equal to the pressure that,
when applied, just stops osmosis. Osmotic
pressure is a colligative property of a solution.
π = MRT

?
Dextran, a polymer of glucose units, is
produced by bacteria growing in
sucrose solutions. Solutions of dextran
in water have been used as a blood
plasma substitute. At 21°C, what is the
osmotic pressure (in mmHg) of a
solution containing 1.50 g of dextran
dissolved in 100.0 mL of aqueous
solution, if the average molecular mass
of the dextran is 4.0 × 108
amu?

T = 21°C + 273 = 294 K
Mass solute = 1.50 g
Liters solution = 100.0 × 10-3
L
Molecular mass = 4.0 × 108
amu
Molar mass = 4.0 × 108
g/mol
mol103.75
g104.0
mol1
g1.50soluteMoles 8
9−
×=
×
×=
mol/L103.75
L10100.0
mol103.75
Molarity 8-
3-
-9
×=
×
×
=

8 mol L atm
3.75 10 0.08206 294 K
L mol K
π − •
= × × ×
•
MRTπ =
4
6.9 10 mmHgπ −
= ×
atm1
mmHg760
atm109.05 7
××= −

One can show that 760.0 mmHg is equivalent to
the pressure exerted by a column of water 10.334
m high. Thus each 1.00 mmHg of pressure is
equivalent to the pressure of a 1.36-cm column of
water. If the density of this dextran solution is
equal to that of water, what height of solution
would exert a pressure equal to its osmotic
pressure?
4
9.4 10 cmπ −
= ×
4 1.36 cm
6.9 10 mmHg
1.00 mmHg
π −
= × ×

Ionic solutes dissolve to form more than one
particle per formula unit. We alter the colligative
property equations to account for this fact by
including i, the number of ions per formula unit:
∆P = iP°AXB
∆Tb = iKbm
∆Tf = iKfm
π = iMRT

?
What is the osmotic pressure at 25.0°C
of an isotonic saline solution (a solution
having an osmotic pressure equal to
that of blood) that contains 0.900 g
NaCl in 100.0 mL of aqueous solution?
Assume that i has the ideal value
(based on the formula).

T = 25°C + 273 = 298 K
Mass solute = 0.900 g NaCl
Volume solution = 100.0 mL = 100.0 × 10−3
L
Molar mass solute = 58.44 g/mol
i = 2
mol0.01540
g58.44
mol1
g0.900soluteMoles =×=
2
3
1.540 10 mol
Molarity 0.1540 mol/L
100.0 10 L
−
−
×
= =
×

K298
Kmol
atmL
0.08206
L
mol
0.15402 ×
•
•
××=π
iMRTπ =
atm7.53=π

A colloid is a dispersion of particles of one
substance (the dispersed phase) throughout
another substance or solution (the continuous
phase). The dispersed particles range from 1000
pm to 200,000 pm in size—much larger than single
molecules or single ions.
Fog is an example of a colloid. In fog, water
droplets are dispersed through air.

Colloids exhibit the Tyndall effect.
The path of the light is visible through a colloid
because the light is reflected by the relatively
larger-sized particles in the dispersed phase.

Colloid Solution

Colloids are characterized according to the state
(solid, liquid, or gas) of the dispersed phase and
the state of the continuous phase.
• Fog and smoke are aerosols, which are liquid
droplets or solid particles dispersed throughout
a gas.
• An emulsion consists of liquid droplets
dispersed throughout another liquid (for example,
particles of butterfat dispersed through
homogenized milk).
• A sol consists of solid particles dispersed in a
liquid.

Colloids in which the continuous phase is water
are categorized into two major classes: hydrophilic
colloids and hydrophobic colloids.
Hydrophilic colloid
A colloid in which there is a strong attraction
between the dispersed phase and the continuous
phase (water).
Hydrophobic colloid
A colloid in which there is a lack of attraction
between the dispersed phase and the continuous
phase (water).

Coagulation is the process by which the
dispersed phase of a colloid is made to aggregate
and thereby separate from the continuous phase.
It is analogous to precipitation from a solution.
Curdled milk is an example of coagulation.

When molecules or ions that have both a
hydrophobic end and a hydrophilic end are
dispersed in water, they associate, or aggregate,
to form colloidal-sized particles called micelles.
A colloid in which the dispersed phase consists of
micelles is called an association colloid.

Soap consists of compounds such as sodium
stearate. Sodium stearate is an example of a
molecule with hydrophobic and hydrophilic ends.
Hydrophobic
end
Hydrophilic
end

In water solution, the stearate ions associate to
form micelles in which the hydrocarbon ends point
inward toward one another and away from the
water, and ionic carboxyl groups are on the outside
of the micelle facing the water.
The cleansing action of soap occurs because oil
and grease can be absorbed into the hydrophobic
centers of soap micelles and washed away.

Far left:
Vegetable oil floating
on water (dyed
green).
Left:
When the mixture is
shaken with soap, an
emulsion forms as
the oil droplets are
absorbed into soap
micelles.

Synthetic detergents also form association
colloids. Sodium lauryl sulfate is a synthetic
detergent present in toothpastes and shampoos.

The detergent molecules we have discussed so far
are classified in the trade as anionics, because
they have a negative charge at the hydrophilic
end.
Other detergent molecules are classified as
cationics, because they have a positive charge at
the hydrophilic end.

Many cationic
detergents
also have
germicidal
properties and
are used in
hospital
disinfectants
and in
mouthwashes.

Solutions Chapter 12 Guide

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