Pre-AP Chemistry
1
• Solution – homogeneous mixture
• Solute – substance being dissolved
• Solvent – present in greater amount
2
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Make-up
Laundry detergent
Motor oil
Gasoline
Food preservatives
Deodorant
Lawn fertilizers & weed killers
Shampoo
Air fresheners
Furniture polish
Toothpaste and mouthwash
Oven cleaner
Glass cleaner
etc…
3
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Alloy – a solid solution of metals
• Bronze = Cu + Sn
• Brass = Cu + Zn
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Soluble – “will dissolve in”
Insoluble – “will not dissolve in”
Miscible – refers to two gases or two liquids that form a
solution
• More specific than “soluble”
• e.g. food coloring and water
•
•
Immiscible – refers to two gases or liquids that will not
form a solution
Suspension – appears uniform while being stirred, but
settles over time
4
Solute
Solvent
Solution
Gaseous Solutions
gas
Liquid
gas
gas
air (nitrogen, oxygen, argon
gases)
humid air (water vapor in air)
Liquid Solutions
gas
liquid
solid
liquid
liquid
liquid
carbonated drinks (CO2 in
water)
vinegar (CH3COOH in water)
salt water (NaCl in water)
Solid Solutions
liquid
solid
solid
solid
dental amalgam (Hg in Ag)
sterling silver (Cu in Ag)
5
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Solute – a substance in a smaller amount dissolved in a larger
amount of another substance (the solvent)
Concentration – the number of moles present in a certain
volume of solution
• Expressed as the amount of solute dissolved in a given
amount of solution.
• An intensive quantity
Molarity (M) expresses the concentration in units of moles of
solute per liter of solution
• Can be used as a conversion factor between volume of
solution and amount (mol) of solute
mol solute
M
L solution
6
A. mass % = mass of solute
mass of solution
B. parts per million (ppm) (also, ppb and ppt)
• commonly used for minerals or contaminants in water
supplies
C. molarity (M) = moles of solute
L of solution
• used most often in this class
D. molality (m) = moles of solute
kg of solvent
7
Concentration =
# of moles
volume (L)
V = 250 mL
n = 8 moles
[ ] = 32 molar
V = 1000 mL
n = 2 moles
[ ] = 2 molar
V = 1000 mL
V = 5000 mL
n = 4 moles
[ ] = 4 molar
n = 20 moles
[ ] = 4 molar
8
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
6.02  1023
MOLES
particles/mol
NUMBER
OF
PARTICLES
Molarity (mol/L)
LITERS
OF
SOLUTION
9
1.
Hydrobromic acid (HBr) is a solution of hydrogen
bromide gas in water. Calculate the molarity of
hydrobromic acid solution if 0.455 L contains 1.80 mol
of hydrogen bromide.
2.
How many moles of KI are in 84 mL of 0.50 M KI?
10
3.
How many grams of solute are in 1.75 L of 0.460 M
sodium hydrogen phosphate (Na2HPO4)?
4.
How many liters of 3.30 M sucrose (C12H22O11)
contain 135 g of solute?
11
5.
You have 10.8 g potassium nitrate. How many mL of
solution will make this a 0.14 M solution?
12
1.
Given the reaction
Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq),
what volume of 4.0 M KI solution is required to yield 89 g PbI2?
2.
How many mL of a 0.500 M CuSO4 solution will react w/excess
Al to produce 11.0 g Cu? Aluminum sulfate is also produced.
13
3.
Given the unbalanced reaction
Cu + AgNO3  Ag + Cu(NO3)2 ,
how many grams of Cu are required to react with 1.5 L of
0.10M AgNO3?
4.
79.1 g of zinc react with 0.90 L of 2.5M HCl to form zinc
chloride and hydrogen gas. Identify the limiting and excess
reactants. How many liters of hydrogen are formed at STP?
14
5.
A common antacid contains magnesium hydroxide, which reacts with HCl to form
water and magnesium chloride solution. As a government chemist testing
commercial antacids, you use 0.10 M HCl to simulate the acid concentration in the
human stomach. How many liters of stomach acid react with a tablet containing
0.10 g of magnesium hydroxide?
6.
Mercury (II) nitrate reacts with sodium sulfide solution to produce solid mercury (II)
sulfide and sodium nitrate solution. In a laboratory simulation, 0.050 L of 0.010 M
mercury (II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. How many grams
of mercury (II) sulfide form?
15
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The volume of a solution is made up of solute and
solvent.
To prepare a solution of a specific molarity:
• Weigh the solid (solute) needed.
• Transfer the solid to a volumetric flask that contains
about half the final volume of solvent.
• Dissolve the solid thoroughly by swirling.
• Add solvent until the solution reaches its final volume.
• To dilute the solution to a lesser molarity, add only
solvent to increase the volume.
18
19
1.
Assume that 1 L of 3 M NaOH solution is needed for a
class lab. Determine the mass of sodium hydroxide
required to create the solution.
2.
What mass of salt is required to prepare 500 mL of
1.54 M NaCl solution?
20
•
Concentration – a measure of solute to solvent
ratio
• Concentrated – lots of solute
• Dilute – not much solute (“watery”)
•
•
To dilute solutions, add more solvent (often water)
to solution.
Moles of solute remain the same.
21
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Dilution of solutions  Acids (and sometimes
bases) are purchased in concentrated form
(concentrate) and are easily diluted to any
desired concentration.
• Safety Tip: When diluting, add acid or base to
water (A&W!)
•
Dilution Equation:
• MCVC=MDVD
• C  “concentrate”
• D  “dilute”
23
1.
Concentrated H3PO4 is 14.8 M. What volume of
concentrate is required to make 25.00 L of 0.500 M
H3PO4?
2.
What volume of 15.8M HNO3 is required to make
250 mL of a 6.0M solution?
24
3.
Isotonic saline is a 0.15 M aqueous solution of NaCl
that simulates the total concentration of ions found in
many cellular fluids. Its uses range from a cleansing
rinse for contact lenses to a washing medium for red
blood cells. How would you prepare 0.80 L of
isotonic saline from a 6.0 M stock solution?
25
4.
If 25.0 mL of 7.50 M sulfuric acid are diluted to
exactly 500. mL, what is the mass of sulfuric acid
per milliliter?
26
•
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Beer’s Law relates the absorption of light to the
properties of the material through which the light is
traveling.
There is a dependence between the absorbance of
light by a substance and the product of the molar
absorptivity coefficient, a, the distance the light travels
through the material, b, and the molar concentration,
c.
•
•
A  abc
Also called the Beer-Lambert Law, the more
concentrated a solution is, the more light it will absorb
and the darker it will appear.
27
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Beer’s law is typically used in a process called
spectroscopy.
In this process, a beam of light is sent through a
small sample of a solution. A detector on the other
side of the solution records the amount of light that
passed through the solution.
28
1.
A red light is passed through a blue solution as shown.
The molar absorptivity constant of the solution is
1.3x106 M-1 cm-1. The absorbance reading is 0.52.
Determine the concentration of the solution.
0.52
1 cm
2.
If water is added to the solution in question #1, what
do you think will happen to the concentration and the
absorbance reading?
29
3.
A green light is passed through a purple solution as
shown. The molar absorptivity constant of the solution
is 1.8x104 M-1 cm-1. The absorbance reading is 0.18.
Determine the concentration of the solution.
0.18
1 cm
4.
List at least two laboratory procedures you could do to
increase the absorbance reading for the purple
solution in #3.
30
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Experiments with Beer’s Law typically create a “calibration
curve” by recording absorbance values for solutions with
known concentrations.
Then, the absorbance for a solution of unknown concentration
can be measured and the concentration may be calculated
using the equation from the calibration curve.
An example of a calibration curve is shown below.
Calibration Curve for Crystal Violet
1.2
y = 1.0096x + 0.0869
1
Absorbance
•
R² = 0.9771
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
Concentration (M)
31
•
Covalent bonds between H and O have
unequal sharing.
• Electrons spend more time closer to O, creating a
slightly negative pole.
•
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Water molecule has bent shape; atoms form
angle.
The distribution of its bonding electrons and
its overall shape makes water an ionizing
solvent.
32
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To be soluble, the attraction between each ion and
the water molecules must outweigh the attraction
of the oppositely charged ions.
• If the electrostatic attraction among ions in the compound
is greater than the attraction between ions and water
molecules, the substance is insoluble.
33
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Water interacts strongly with dissolved reactants and can affect
their bonds.
Electrolyte – a substance that conducts a current when
dissolved in water
Soluble ionic compounds dissociate completely into ions and
create a large current; called strong electrolytes.
Solvated ions are surrounded by solvent molecules.
Oppositely charged ions separate when dissolved in water,
become surrounded by water molecules, and spread randomly
throughout the solution.


KBr ( s) 
 K (aq)  Br (aq)
H 2O
•
The H2O above the arrow indicates that water is required but is
not a reactant in the usual sense.
34
Na+
ions
Water molecules
Clions
NaCl(s) + H2O  Na+(aq) + Cl-(aq)
35
Animation
36
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Many covalent compounds dissolve in water.
• Do not dissociate into ions
• Remain intact molecules
Do not conduct an electric current; called
nonelectrolytes.
Acids are H-containing covalent compounds that do
dissociate into ions in water.
37
•
Water interacts most strongly with the hydrogen
cation (H+)
• H+ is just a proton
•
H+ attracts the negative pole of surrounding
water molecules so strongly that it forms a
covalent bond to one of them
• H3O+ (hydronium ion)
38
1+
1+
1+
+
H+
H2O
hydrogen ion
(a proton)
water
H3O+
hydronium ion
39
1.
Nitric acid is a major chemical in the fertilizer and
explosives industries. In aqueous solution, each
molecule dissociates and the H becomes a solvated
H+ ion. What is the molarity of H+ (aq) in 1.4 M nitric
acid?
2.
How many moles of H+(aq) are present in 451 mL of
3.20 M hydrobromic acid?
40
3.
How many moles of H+(aq) are present in 585 mL
of 3.50 M H3PO4?
4.
What is the molarity of the H+ ion in 1.6 M sulfuric
acid?
41
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Dissociation: separation of an ionic solid into
aqueous ions
e.g. NaCl(s)  Na+(aq) + Cl-(aq)
Ionization: breaking apart of polar molecules into
aqueous ions
e.g. HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
Molecular Solvation: molecules (covalent)
dissolve, but remain intact
e.g. C6H12O6(s)  C6H12O6(aq)
42
Predict the ionization or dissociation for the following
chemical species:
a. Sodium hydroxide
b. Hydrochloric acid
c. Sulfuric Acid
d. Acetic Acid
e. Potassium fluoride
f.
Calcium nitrate
43
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Strong Electrolytes exhibit nearly 100% dissociation
• e.g.
NOT in water:
In aq. solution:
•
NOT in water:
In aq. solution:
•
Na+
1000
1
0
999
+
Cl–
0
999
Weak electrolytes exhibit little dissociation
• e.g.
•
NaCl
CH3COOH
1000
980
CH3COO –
0
20
+
H+
0
20
“Strong” or “Weak” is a property of the substance.
One cannot be changed into the other.
When written in a chemical equation, a strong
electrolyte is shown as individual ions while a weak
electrolyte is shown in molecular form.
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Electrolytes - solutions that carry an electric current
strong electrolyte
NaCl(aq)
Na+ + Cl-
weak electrolyte
HF(aq)
nonelectrolyte
H+ + F-
45
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Temperature: as T increases, rate
increases
•
Particle Size: as particle size
decreases, rate increases
•
Mixing: more mixing/stirring, rate
increases
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Nature of solvent or solute: identity
determines whether a substance will
dissolve and to what extent
46
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
increasing concentration
47
Solids dissolved in liquids
Gases dissolved in liquids
Sol.
Sol.
To
As To , solubility
To
As To , solubility
48
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Solubility: how much solute dissolves in a given amount
of solvent at a given temperature
Unsaturated: Solution can hold more solute; below line
Saturated: Solution has “just the right” amount of solute;
on line
Supersaturated: Solution has “too much” solute dissolved
in it; above the line
KNO3 (s)
KCl (s)
Solubility
(g/100 g H2O)
HCl (g)
Temp. (oC)
49
Solubility vs. Temperature
140
KI
130
Solubility (grams of solute/100 g H2O)
120
NaNO3
110
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0
10 20 30 40 50 60 70 80 90 100
Temperature °C
50
140
KI
130
Solubility (grams of solute/100 g H2O)
120
NaNO3
110
gases
Describe each situation below according
to saturation and appearance using the
given solubility table:
A. Per 100 g H2O,100 g NaNO3 @ 60°C.
solids
100
KNO3
90
B.
Per 100 g H2O, 20 g KClO3 @ 80°C.
C.
Cool solution (B) very slowly to 30°C
D.
Quench solution (B) in an ice bath to
30°C
E.
Per 100 g H2O, 70 g KNO3 @ 50°C
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0
10 20 30 40 50 60 70 80 90 100
Temperature °C
51
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In a gas, the energy of attraction is small relative
to the energy of motion.
• On average, the particles are far apart.
•
Large interparticle distance has several
macroscopic consequences:
• A gas moves randomly throughout its container and fills
it.
• Gases are highly compressible (can be squeezed and
shrunk).
• Gases flow and diffuse through one another easily.
52
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In a liquid, the attractions are stronger because
the particles are in contact. But their kinetic
energy allows them to tumble randomly over and
around each other.
• Therefore, a liquid conforms to the shape of its container
but has a surface.
•
•
With very little free space between the particles,
liquids compress only very slightly.
Liquids flow and diffuse but much more slowly
than gases.
53
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•
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In a solid, the attractions dominate the motion to
such an extent that the particles remain in position
relative to one another.
With the particles very close together and
positions fixed, a solid has a specific shape.
Solids compress even less than liquids.
Solids do not flow significantly.
54
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Phase changes are determined by the interplay between
kinetic energy and intermolecular forces.
•
As the temperature increases, the average kinetic energy
increases as well so the faster moving particles can
overcome the attractions more easily.
•
Lower temperatures allow the forces to draw the slower
moving particles together.
•
The process by which a gas changes into a liquid is called
condensation. A liquid changing to a gas is called
vaporization.
•
The process by which a liquid changes into a solid is called
freezing. A solid changing to a liquid is called melting or
fusion.
•
The process by which a solid becomes a gas (without
becoming a liquid first) is called sublimation. A gas
changing to a solid is called deposition.
55
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•
Phase diagrams are used to depict the phase changes of
a substance at various conditions of temperature and
pressure.
A phase diagram has these four features:
• Regions of the diagram: Each region corresponds to one phase of
the substance. A particular phase is stable for any combination of
pressure and temperature within its region. If any other phase is
placed under those conditions, it will change to the stable phase.
• Lines between regions: The lines separating the regions represent
the phase transition curves. Any point along a line shows the
pressure and temperature at which the two phases exist in
equilibrium.
• The critical point: The liquid-gas line ends at the critical point.
Beyond the critical temperature, a supercritical fluid exists rather
than separate liquid and gaseous phases.
• The triple point: The three phase transition curves meet at the
triple point: the pressure and temperature at which three phases are
in equilibrium.
56
57
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Most substances have a solid phase that is more
dense than the liquid phase, giving a positive slope to
the solid-liquid line. (Because the liquid occupies
slightly more space than the solid, an increase in
pressure favors the solid phase, in most cases.)
•
Unlike almost every other substance, solid water is
less dense than liquid water (due to hydrogen bonds
causing the solid phase to have a greater volume
because of large spaces between molecules). An
increase in pressure favors the phase that occupies
less space, which for water is the liquid phase.
Therefore, the solid-liquid line for water has a negative
slope.
58
H2O
CO2
59
•
Using the phase diagram for water, determine what phase
water is in at the following conditions:
a. 2 atm, 55°C
b. 1 atm, 75°C
c. 1 atm, 200°C
d. 5 atm, -100°C
60
•
Colligative properties refer to the physical
properties that are changed by the presence of
any solute particles.
• The chemical identity of the solute is not important. Only
the concentration of the solute particles is significant in
altering physical properties.
•
The physical properties that change when a solute
is present include vapor pressure, boiling point,
melting point, and osmotic pressure.
61
•
The lowering of vapor pressure caused by a
nonvolatile nonelectrolyte solute leads to an
increase in the boiling point and a decrease in the
freezing point of the solvent.
• Higher temperatures are needed to make the vapor
pressure of the system equal atmospheric pressure (boiling
point).
• Lower temperatures are needed to overcome the
interference with the crystallization process caused by the
attractive forces between the solute and solvent.
• The changes in the boiling and melting points of a solution
depend on the solvent and on the molal concentration of
the solute.
62
Boiling Point Elevation / Freezing Point Depression
•
The increase in the boiling point is given by
•
T  ik b m
• m is the molality of the solution
• kb is the boiling point elevation constant for the solvent
• i represents the number of particles produced by each part
of the solute
•
The decrease in the melting (freezing) point is given
by
T  ik f m
•
• The negative sign indicates a decrease in temperature
• kf is the freezing point depression constant
63
•
Chemists use the freezing point depression to make
cooling baths below 0°C by dissolving large quantities
of salt in water and then adding ice.
•
Antifreeze in a car’s engine lowers the freezing point
of water by adding a high concentration of ethylene
glycol. Antifreeze also increases the boiling point of
water to allow engines to run above 100°C without
boiling over.
•
A common cooking use is adding salt to water to
increase the boiling point, allowing the food to cook at
a higher temperature than otherwise possible.
•
The most important use is the determination of molar
masses of solutes.
64
•
Example #1: Which of the following, when added to 1.00
kg H2O, is expected to give the greatest increase in the
boiling point of water? (kb = 0.052°C/m)
a.
b.
c.
d.
e.
1.25 mol sucrose (C12H22O11)
0.25 mol iron (III) nitrate
0.50 mol ammonium chloride
0.60 mol calcium sulfate
1.00 mol acetic acid
65
•
•
•
•
•
Three types of equations:
• Molecular
• Total Ionic
• Net Ionic
Atoms and charges must balance in ionic equations.
Molecular equations show all reactants and products
as if they were intact, undissociated compounds
Total ionic equations show all the soluble ionic
substances dissociated into ions.
• Spectator ions appear in the same form on both
sides of the equation and are not involved in the
actual chemical change.
Net ionic equations eliminate the spectator ions and
show the actual chemical change taking place
66
•
Molecular Equation
2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s)+ 2NaNO3(aq)
•
Total Ionic Equation
2Na+(aq)+2Cl-(aq) + Pb2+(aq) + 2NO3-(aq)PbCl2(s) + 2Na+(aq)+2NO3-(aq)
Spectator Ions
•
Net Ionic Equation
2Cl-(aq) + Pb2+(aq)  PbCl2(s)
67
•
Molecular Equation
2 AgNO3 (aq)  Na2CrO4 (aq) 
 Ag 2CrO4 ( s)  2 NaNO3 (aq)
•
Total Ionic Equation

2

2 Ag  (aq)  2 NO3 (aq)  2 Na (aq)  CrO4 (aq) 
 Ag2CrO4 (s)  2 Na (aq)  2 NO3 (aq)
•
Net Ionic Equation

2
2 Ag (aq)  CrO4 (aq) 
 Ag 2CrO4 ( s)
68
Write the net ionic equations for the following
molecular equations and identify the spectator ions:
a.
Li2SO4(aq)+Ca(OH)2(aq)  CaSO4(s)+2LiOH(aq)
b.
2NaI(aq) + Pb(NO3)2(aq)  2NaNO3(aq) + PbI2(s)
69
Write the net ionic equations for the following
molecular equations and identify the spectator ions:
c.
AgNO3(aq) + KBr(aq)  AgBr(s) + KNO3(aq)
d.
H2SO4(aq) + NaOH(aq)  Na2SO4(aq) + H2O (l)
70
Write the net ionic equations for the following
molecular equations and identify the spectator ions:
e.
Al2(SO4)3(aq)+6NH4OH(aq)3(NH4)2SO4(aq) +2Al(OH)3(s)
71