Transcript Unit Powerpoint - Horton High School

 Solutions
(homogeneous mixtures) are a part
of our everyday life. Do you use mouthwash,
toothpaste, cough syrup?
 My swimming pool is one very large solution.
The presence of algae makes my pool green
and slimy. I then need to add an algicide
which is another solution. Bromine is in
solution as well as other chemicals (total
alkalinity).
 Solubility
is of fundamental importance in a
large number of practical applications
ranging from ore processing to the use of
medicines and the transport of pollutants.
 Solubility is often said to be one of the
“characteristic properties of a substance”.
 used to describe a substance
 indicates its polarity
 helps distinguish it from another substance
 becomes a guide for the applications of that
substance
 You will often see in a compounds
description: “is insoluble in water or alcohol
but soluble in concentrated sulfuric acid”
 Solubility
of a substance is useful when
separating mixtures. The synthesis of
chemical compounds, by the milligram in a
lab or by the ton in industry makes use of the
relative solubility of the end product.
 Cooks,
chemists, farmers, pharmacists and
gardeners need to know which compounds
are soluble and which are insoluble.
 You
will separate a mixture or five
substances for first lab using the property of
solubility as an important property.
 What
properties do solutions have?
 How do we describe the concentrations?
 Why do some substances dissolve while
others do not?
 Solution
– a homogeneous mixture that is
uniform throughout. The same substances
will be in the same relative amounts.
Solutions are not pure substances because
they can have variable composition (example
salt water).
 Solvent
– any substance that has other
substances dissolved in it. The substance that
is present in the largest amount (by volume,
mass or number of moles) is usually the
solvent.
 Solute
– a substance dissolved into the
solvent.
 Concentrated – a solution with a high level of
solute ( > 1.0 M)
 Dilute – a solution with a low level of solute
(< 1.0 M)
 A solution can be a gas, solid (alloy) or a
liquid.
 Aqueous – a liquid solution in which water is
the liquid (aq)
 Miscible – liquids which will dissolve in each
other
 Immiscible – liquids which do not dissolve in
each other.
 The
ability of a solvent to dissolve a solute
depends on the forces of attraction between
the particles. There is always some
attraction between the solvent and the
solute.
 Solubility – the mass of the solute that
dissolves in a given quantity of solvent, at a
certain temperature. May be g/mL or molar
solubility (moles of solute in 1 L of solution).
 Saturated – no more solute will dissolve in a
solution. Excess solute can be seen.
 Unsaturated
– a solution which is not yet
saturated. More solute can dissolve.
 Sparingly soluble or slightly soluble – the
solubility is between insoluble (less than 0.1g
per 100 mL) and soluable (1g per 100 mL).
 The solubility of gasses do not use the same
terms as that for solids and liquids. An
example is the amount of dissolved oxygen in
water. 0.0009 g/100 mL is the solubility of
oxygen at 20oC in fresh water. This is very
soluble.


Substances of “like” polarity tend to be more soluble in each
other than substances that differ in polarity.
Remember – polarity is a measure of how electrons are shared
between bonding elements. It also is dependent on shapes of
molecules determined by VESPR.
“Like dissolves like.”
Polar solvents dissolve other polar molecules
and ionic compounds.
Nonpolar molecules dissolve other nonpolar
molecules.
Alcohols which have characteristics of both
tend to dissolve both polar and nonpolar but
will not dissolve ionic solids.
 Step
1 – the forces between the particles in
the solid must be broken. Either between the
ions or between the molecules. This requires
energy.
 Step 2 – Some of the intermolecular forces
between the particles in the liquid must be
broken. This also requires energy.
 Step 3 – There is an attraction between the
particles of the solid and the particles of the
liquid. This step gives off energy.
 Sugar
(polar) in water (polar).
 Iodine (nonpolar) in benzene (nonpolar)
 Electronegativity:
Electronegativity
Difference
Type of
Bond
0.0 – 0.4
nonpolar
H-H
0.0
0.4 – 2.0
polar
H-Cl
0.9
> 2.0
ionic
NaCl
2.1
 Polar,
a)
b)
c)
d)
nonpolar or ionic
N and H
F and F
Ca and O
Al and Cl
3.0 – 2.1
4.0 - 4.0
3.5 -1.0
3.0 – 1.5
0.9
0
2.5
1.5
polar
nonpolar
ionic
polar
 Not
all substances dissolve in the same
manner or amount.
 1.
The solubility of most substances increases
as the temperature of the solvent increases.
For some substances, the opposite is true.
At 25oC NaCl 36.2 g/100g H2O
At 100oC NaCl 39.2 g/100g H2O
Sodium sulfate
At 40oC 50g/100g H2O
At 100oC 41g/100g H2O
 Gases
have a higher solubility in cold water
than in hot water.
 …….So
temperature affects the rate at which
something will dissolve.
 2.
Pressure
The solubility of a gas increases as the
pressure of the gas above the solution
increases. Soft drinks are bottled under high
pressure.
Henry’s Law
At a given temperature the solubility of a
gas in a liquid (S) is directly proportional
to the pressure of the gas above the
liquid (P). S1 = S2
P1 = P 2
P then
S

If the solubility of a gas in water is 0.77 g/L at
3.5 atm of pressure, what is the solubility in g/L
at 1.0 atm of pressure? The temperature is
constant at 25oC.
 If
S1 = S2 then
___ ____
P1
P2
S2 = S1 x P2
_______
P1
= 0.77 g/l x 1.0 atm
_________________
3.5 atm
= 0.22 g/L
 3.
Surface Area
In order to dissolve, the solvent must come
in contact with the solute. Smaller pieces
increase the amount of surface of the solute
interacting with the solvent.
Term Used for Solubility
Solubility (g/L)
Very Soluble
Greater than 100 g
Soluble
Between 10 g and 100 g
Slightly Soluble
Between 1 g and 10 g
Insoluble
Less than 1 g (.1g /100 ml)
 When
a solution of NaOH is added to a
solution of CuSO4 a blue solid precipitate
forms. The mixture of solutions contains Na+,
OH-, Cu 2+, and SO42- ions.
 When we consult a solubility table we find
that the Cu(OH)2 is insoluble, the precipitate
formed must be Cu(OH)2. An equation for this
reaction can be written as:
2NaOH (aq) + CuSO4 (aq)
Na2SO4 (aq) + Cu(OH)2 (s)
 In
solution these compounds exist as
dissociated ions as shown below:
2Na+ (aq) + 2OH- (aq) + Cu2+ (aq) + SO42- (aq)
Na+ (aq) + SO42- (aq) + Cu(OH)2 (s)

Notice that the Na+ (aq) and SO42- (aq) remain unchanged
by the reaction. These ions are called spectator ions
because they do not react. So if we show only the ions
changed by the reaction, the equation becomes:
Cu2+ (aq) + 2OH- (aq)
Cu(OH)2 (s)
This is called the net ionic equation.

Precipitation Reactions and Solubility Rules
A solid dissolves in water because the water
molecules around the solid’s surface collide with
the particles of the solid. The attraction
between polar water and a partially charged
particle or ion enables the water molecules to
pull these particles away from the crystal, a
process called dissociation. This is followed by
solvation.
Predicting Precipitates
We can predict the formation of a precipitate if
we know the solubility of the compounds.
Patterns in solubility have lead to the creation of
solubility rules. Solubility rules are based on
what happens when salts are placed into water,
they either dissolve or they do not dissolve.
Precipitation occurs because two SOLUBLE
substances containing solvated ions are mixed
and a pair of ions form a substance that is
INSOLUBLE. The insoluble substance is the
precipitate (solid) that is formed.
 When you mix two solutions containing ionic
compounds you can predict the products based
on DOUBLE REPLACEMENT reaction.
 If one of the predicted products is not soluble, it
will precipitate out and a reaction occurs.
 If both predicted products are soluble, no
reaction occurs (all ions remain solvated).

Simple Rules for the Solubility of Ionic
Compounds in Water
1. Most nitrate (NO3-) salts are soluble.
2. Most salts containing the alkali metals cations
(Li+, Na+, K+, Rb+, Cs+) and the ammonium (NH4+)
cation are soluble.
3. Most chloride, bromide, and iodide salts are
soluble.
Except those containing Ag+, Pb2+, and Hg+
4. Most sulfate salts are soluble.
Except those containing Ba2+, Pb2+, Hg2+, Sr2+ and
Ca2+.
5. Most hydroxide salts are insoluble except those
of the alkali metal cations and a few alkaline
earth metals: Ba2+, Sr2+, and Ca2+.
6. Most sulfides (S2-), carbonate (CO32-), chromate
(CrO42-), and phosphate (PO43-) salts are
insoluble.

Example:
Will a precipitation reaction occur when Ba(NO3)2 (aq) and Na2SO4
(aq) are mixed?
Possible products from a double displacement reaction:
BaSO4 (s) and NaNO3 (s)

From the solubility table BaSO4 is not soluble so it would form a
precipitate.
Ba(NO3)2 (aq) + Na2SO4 (aq)
BaSO4 (s) + 2 NaNO3 (aq)
We can also write a complete (total) ionic equation:
Ba2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + SO42- (aq)
2Na+ (aq) + 2NO3- (aq) + BaSO4 (s)
or finally we can write a net ionic equation which shows only the
ions which precipitate from solution.
Ba2+ (aq) + SO42- (aq)
BaSO4 (s)

Example:
Will a precipitation reaction occur if 150 mL of
1.0M NaI (aq) is added to 200 mL of 0.5M AgNO3
(aq)?
The solubility table shows that AgI is insoluble.
AgNO3 (aq) + NaI (aq)
AgI (s) + NaNO3 (aq)
Net ionic equation
Ag+ (aq) + I- (aq)
AgI (s)
 Molarity
(M) is the most common method for
describing the concentration of a solution,
and is defined as the number of moles of
solute present per litre of solution.
M = n/V (V in litres)
Example: Calculate the molarity of 75.0 g of
KBr (s) dissolved in 500 mL of water.
M= mol solute/ V (in litres)
 Mol
KBr = 75.0 g / 119 g/mol
= 0.630 mol
V solution = 500 mL / 1000 mL
= 0.5 L
M = 0.630mol/0.5 L
= 1.26 M KBr (aq)
 Example:
What are the concentrations of
Ca2+ (aq) and NO3- (aq) in 1.5 M Ca(NO3)2
(aq)?
Ca(NO3)2
Ca2+ (aq) +
2NO31(aq)
1.5 M
1.5 M
+
3.0 M
Solutions prepared by dissolving a precise mass
of solute in a precise volume of solvent are
known as standard solutions. Volumetric flasks
are used to prepare standard solutions.
 Dilutions
It is often necessary to dilute a certain solution.
Dilution is the process by which a solution is
made less concentrated by the addition of more
solvent. When a solution has been diluted its
solute particles are spread out more. When
solutions are diluted, the amount of solute
particles remains the same. Since n=MV
(concentration x volume) we can use the
following equation:
M1V1 = M2V2

 Example:
200.0 mL of 3.0 M NaOH (aq) is
diluted to 400 mL. What is the concentration
of the diluted solution?
Initail # moles = final # moles
M1V1
=
M2V2
(3.0 M x 0.2 L) = ( M2 x 0.4L)
M2
= 1.5 M NaOH (aq)
 Example:
What volume of a stock solution
of 6.0 M HCl (aq) is needed to make 900 mL
of a 1.3 M solution?
M1V1 = M2V2
(6.0 M x V1) = (1.3 M x 0.9 L)
V1 = 0.2 L or 200 mL
 Molality
(m) is the number of moles of solute
per kilogram of solvent.
 M= n/kg
 Example: A solution contains 15.5 g of
NH2CONH2 in 74.3 g H2O. Calculate the
molality.
 Molality = mol solute/kg solvent

Mol NH2CONH2 = 15.5 g / 60.0 g/mol (molar mass)
= 0.258 mol

Kg H2O = 74.3 g x 1 kg
1000 g
= 0.0743 Kg

m = mol solute
Kg solvent
= 0.258 mol
0.0743 Kg
= 3.48 m
Mole fraction (no units) is the number of moles
of one compound in a mixture divided by the
total number of moles of all components in the
mixture.
Xa = na/ (na + nb)
 Example: What are the mole fractions of an
ethanol and water solution prepared by
dissolving 3.45 g C2H5OH in 21.1 g H2O?

Mol ethanol = 3.45 g / 46.0 g/mol = 0.075 mol
Mol of water = 21.0 g / 18.0 g/mol = 1.18 mol
Total # moles = mol of ethanol + mol of water
= 0.075 mol + 1.18 mol
= 1.255 mol
 Mol
fraction of ethanol = 0.075 / 1.255
= 0.05976
Mol fraction of water = 1.18 / 1.255
= 0.9402
 Mol
fraction ethanol + water = 1
% by mass

mass of solute
x 100
(mass of solute + mass of solvent)

Example: A 5 % NaCl solution would consist of
5.0 g of NaCl (s) and 95 g of water.
( I chose 100 g because it was easier to take 5%).
1 mL of H2O = 1 g H2O.
So 5.0 g NaCl (s) in 95 mL of H2O (l).
Example: 24 g of NaCl (s) is dissolved in 152 g
H2O (l).
Calculate the % by mass of this solution.
Mass % NaCl =
24 g
x 100
(24 g + 152g)
= 24/176 x 100 = 14 %

 Properties
of solutions which depend on the
number of solute particles but not on their
nature.
 They are dependent on the number of solute
particles dissolved in a given volume of
solvent. They include: vapour pressure
lowering, boiling point elevation, freezing
point depression and osmotic pressure.
 The
presence of a solute lowers the freezing
point of a solution relative to that of the
pure solvent. For example, pure water
freezes at 0°C (32°F); if one dissolves 10
grams of sodium chloride (table salt) in 100
grams of water, the freezing point goes
down to −5.9°C (21.4°F). If one uses sucrose
(table sugar) instead of sodium chloride, 10
grams in 100 grams of water gives a solution
with a freezing point of −0.56°C (31°F).
 The
reason that the salt solution has a lower
freezing point than the sugar solution is that
there are more particles in 10 grams of
sodium chloride than in 10 grams of sucrose.
Since sucrose, C 12 H 22 O 11 has a molecular
weight of 342.3 grams per mole and sodium
chloride has a molecular weight of 58.44
grams per mole, 1 gram of sodium chloride
has almost six times as many sodium chloride
units as there are sucrose units in a gram of
sucrose. In addition, each sodium chloride
unit comes apart into two ions (a sodium
cation and a chloride anion ). The freezing
point depression of a solution containing a
dissolved substance, such as salt dissolved in
water, is a colligative property.
One can calculate the change in freezing point
(Δ T f ) relative to the pure solvent using the
equation:
Δ T f= K fm
where K f is the freezing point depression constant
for the solvent (1.86°C·kg/mol for water) (can be
looked up on a table) ,m is molality (the number of
moles of solute in solution per kilogram) of solvent
for ALL particles.

Because the presence of a solute lowers the freezing
point, the Department of Highways puts salt on our
roads after a snowfall, to keep the melted snow
from refreezing.
Also, the antifreeze used in automobile heating and
cooling systems is a solution of water and ethylene
glycol (or propylene glycol); this solution has a lower
freezing point than either pure water or pure
ethylene glycol.
 ΔTf
= Kf x m
What is the freezing point of these solutions?
1.4 mol of Na2SO4 in 1750 of H2O
m = 1.4 mol
1.75 kg
= 0.8 m
3 particles are formed when Na2SO4 ionizes
3 x 0.8m = 2.4 m
Δ Tf = Kf x m
= 1.86 oC/m x 2.4 m
= 4.46 oC
0 oC – 4.5 oC = - 4.46 oC
 What
is the freezing point of 0.60 mol of
MgSO4 in 1300 g of H2O?
m = 0.60 mol
1.3 kg
= 0.46 m
2 particles x 0.46 m = 0.92 m
ΔTf = 1.86 oC x 0.92 m
= 1.71 oC
0oC – 1.71 oC = - 1.71 oC

The boiling point of a solution is higher than that
of the pure solvent. The actual boiling point
elevation is the difference in temperature
between the boiling points of a solution and the
pure substance. Accordingly, the use of a
solution, rather than a pure liquid, in antifreeze
serves to keep the mixture from boiling in a hot
automobile engine. As with freezing point
depression, the effect depends on the number of
solute particles present in a given amount of
solvent, but not the identity of those particles. If
10 grams of sodium chloride are dissolved in 100
grams of water, the boiling point of the solution
is 101.7°C (215.1°F; which is 1.7°C (3.1°F)
higher than the boiling point of pure water).

The formula used to calculate the change in
boiling point (Δ T b ) relative to the pure
solvent is similar to that used for freezing
point depression:
Δ T b= K bm ,
where K b is the boiling point elevation
constant for the solvent (0.52°C·kg/mol for
water), and m has the same meaning as in
the freezing point depression formula.
Note that Δ T b represents an increase in
the boiling point, whereas Δ T f represents a
decrease in the freezing point.
 Boiling
Point
Δ Tb = Kb x m
What is the boiling point of a solution that contains
1.2 mol of sodium chloride in 800 g of water?
m= 1.2 mol NaCl
0.8 kg
= 1.5 m
Total molality = 1.5 m x 2 = 3m
Δ Tb = Kb x m
= 0.512 oC/m x 3 m
= 1.54 oC
100 oC + 1.54 oC = 101.54 oC
 What
is the boiling point of a solution that
contains 1.25 mol of CaCl2 in 1400 g of
water?
m= 1.25 mol
1.4 kg
= 0.89 m
CaCl2 = 3 particles when it ionizes
3 x 0.89 m = 2.67 m
Δ Tb = Kb x m
= 0.512 oC/m x 2.67 m
= 1.37 oC
100 oC + 1.37 oC = 101.37 oC
A solution of 7.50 g of a nonvolatile compound in
22.60 g of water boils at 100.78 oC . What is the
molecular mass of the solute?
 Molality of the solution:
100.78 oC – 100 oC = 0.78 oC ( ΔTb)
If ΔTb = Kb x m then m = Δ Tb
Kb

m=
0.78 oC
0.512 oC/m
m = 1.5 m

1.5 m x 0.0226 kg = 0.034 mol
Molecular mass of solute = mass of solute
moles of solute
= 7.5 g
0.0344 mol
= 218 g/mol
 The
vapor pressure of a liquid is the
equilibrium pressure of gas molecules from
the liquid (evaporation) above the liquid
itself in a closed system. A glass of water
placed in an open room will evaporate
completely (and thus never reach
equilibrium); however, if a cover is placed on
the glass, the space above the liquid will
eventually contain a constant amount of
water vapor. How much water vapor is
present depends on the temperature, but not
on the amount of liquid that is present at
equilibrium.
If, instead of pure water, an aqueous solution is
placed in the glass, the equilibrium pressure will
be lower than it would be for pure water.
A solution that contains a nonvolatile solute (it
does not vaporize like sugar or salt) always has a
lower vapour pressure than the pure solvent. In
an aqueous solution of sodium chloride, there
are sodium ions and chloride ions throughout the
solution. The ions are surrounded by shells of
water of solvation. This reduces the number of
solvent molecules that have enough kinetic
energy to escape as vapour. The result is a
solution with a lower vapour pressure than the
pure solvent.
 The
decrease in the vapour pressure is
proportional to the number of particles the
solute makes in solution. A solute like sodium
chloride which dissociates into several
particles has a greater effect on the vapour
pressure than the same concentration of a
solute like glucose which does not dissociate.
 NaCl – 2 particles when dissociated
 Glucose – 1 molecule, no disassociation
 CaCl2 – 3 particles when dissociated
 Raoult’s
law states that the vapour pressure
of the solvent over the solution is
proportional to the fraction of solvent
molecules in the solution, if two-thirds of the
molecules are solvent molecules, the vapour
pressure due to the solvent is approximately
two-thirds of what it would be for pure
solvent at that temperature.
If the solute has a vapour pressure of its own,
then the total vapour pressure over the
solution would be:
vapour pressure =
(solution)
vapour pressure x mole fraction
(solvent)

Osmosis is the process whereby a solvent passes
through a semipermeable membrane from one
solution to another (or from a pure solvent into a
solution). A semipermeable membrane is a barrier
through which some substances may pass (e.g., the
solvent particles), and other species may not (e.g.,
the solute particles). Important examples of
semipermeable membranes are the cell walls in cells
of living things (plants and animals). Osmosis tends to
drive solvent molecules through the semipermeable
membrane from the low solute concentrations to the
high solute concentrations; thus, a "complete”
osmosis process would be one that ends with the
solute concentrations being equal on both sides of
the membrane. Osmotic pressure is the pressure that
must be applied on the high concentration side to
stop osmosis.
 Osmosis
is a very useful process. For
example, meats can be preserved by turning
them into jerky: The meat is soaked in a very
concentrated salt solution, resulting in
dehydration of the meat cells. Jerky does not
spoil as quickly as fresh meat, since bacteria
on the surface of the salty meat will fall
victim to osmosis, and shrivel up and die.
This process thus extends the life of the
meat without the use of refrigeration. There
are times when one wishes to prevent
osmosis when two solutions (or a pure
solvent and a solution) are on opposite sides
of a semipermeable membrane.
.
Osmosis can be prevented by applying
pressure to the more concentrated solution
equal to the osmotic pressure on the less
concentrated side. This can be accomplished
either physically, by applying force to one
side of the system, or chemically, by
modifying a solute concentration so that the
two solute concentrations are equal. (If one
applies a pressure greater than the osmotic
pressure to the higher concentration
solution, one can force solvent molecules
from the concentrated solution to the dilute
solution, or pure solvent. This process,
known as reverse osmosis, is often used to
purify water.)
A
hospital patient receiving fluids
intravenously receives an intravenous (IV)
solution that is isotonic with (i.e., at the
same solute concentration as) his or her
cells. If the IV solution is too concentrated,
osmosis will cause the cells to shrivel; too
dilute a solution can cause the cells to burst.
Similar problems would be experienced by
freshwater fish swimming in salt water, or
saltwater fish swimming in freshwater. The
osmotic pressure, like other colligative
properties, does not depend on the identity
of the solute, but an electrolyte solute will
contribute more particles per formula unit
than a nonelectrolyte solute.