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2. 
Solution - homogeneous mixture
Solute - substance
being dissolved
Solvent - present in
greater amount (usually
liquid, like H20)

3. Solute - KMnO4
Solvent - H2O

4. 
Solvation – the process of dissolving
solute particles are surrounded by
solvent particles
solute particles are separated and
pulled into solution

5. -
-
+
sugar
-
+
+
salt
acetic acid
NonElectrolyte
Weak
Electrolyte
Strong
Electrolyte
solute exists as
molecules
only
solute exists as
ions and
molecules
solute exists as
ions only
View animation online.
DISSOCIATION
IONIZATION

6. 
Dissociation
› separation of an ionic
solid into aqueous ions
NaCl(s)  Na+(aq) + Cl–(aq)

7. 
Ionization
› breaking apart of
some polar molecules
into aqueous ions
HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)

8. “Like Dissolves Like”
NONPOLAR
POLAR
NONPOLAR
POLAR

9. 
Soap/Detergent
› polar “head” with long nonpolar “tail”
› dissolves nonpolar grease in polar water

10. UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
concentration
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form

11. 
Solubility
› maximum grams of solute that will dissolve in
100 g of solvent at a given temperature
› varies with temp
› based on a saturated solution

12. 
Solubility Curve
› shows the
dependence of
solubility on
temperature

13. 
Solids are more soluble at...
› high temperatures.

Gases are more soluble at...
 low
temperatures &
 high pressures (Henry’s
Law).
 EX: nitrogen narcosis, the
“bends,” soda

15. 
The amount of solute in a solution.

How can we describe or communicate
concentration?
› Qualitatively through words
(“black” coffee vs. “a light amber shade”)
› Quantitatively through numbers
(2 sugars & 1 cream)

16. With Percent (%)
 % by mass
› Medicated creams
› Proactiv contains 2.5%
benzoyl peroxide by
mass

% by volume
With Moles*
 Molarity (M)
 Molality (m)
 Mole Fraction (X)
** Most often used by chemists **
› Usually used when both
solute & solvent are
liquids
› Gasoline can contain
up to 10% ethanol by
volume
Parts
› ppm, ppb - water
contaminants
(don’t talk about in here)

17. moles of solute
molarity (M) 
L of solution
% m/v 
mass of solute (grams)
volume of solution (mL)
% v/v 
volum e of solute
total volum e of solution
% m/m 
mass of solute (grams)
mass of solution (grams)
Remember to multiply the % formulas by 100 OR move decimal two spots left
Moles of solute
Mole Fraction 
Moles of solution

18. moles of solute
molality (m) 
kg of solvent
0.25 mol
0.25m 
1 kg
mass of solvent only
1 kg water = 1 L water
M 1V1  M 2V2

19. 
Find the molality of a solution containing 75
g of MgCl2 in 250 mL of water.
75 g MgCl2
1 mol MgCl2
95.21 g MgCl2 0.25 kg water
mol
m
kg
= 3.2m MgCl2

20. 
How many grams of NaCl are req’d to make
a 1.54m solution using 0.500 kg of water?
0.500 kg water
1.54 mol NaCl
58.44 g NaCl
1 kg water
1 mol NaCl
1.5 mol
1.5m 
1 kg
= 45.0 g NaCl

21. 
At a restaurant I saw a man put 7 packets
of sugar into his cup of coffee. What is the
percent by mass of sugar in this poor man’s
coffee?
› 1 packet of sugar = 4 grams
› 1 cup of coffee = 180 grams
28 g sugar
(180 + 28) g
= 13.5% sugar

22. 
Find the mole fraction of sugar in the coffee
of that now comatose man who has just
come down off his 7 sugar packet high.
› 0.08 moles sugar
› 10 moles coffee
Xsugar =
0.08 moles sugar
(10 + 0.08) total moles
= 0.0079
Xcoffee = 1 – Xsugar = 1 – 0.0079 = 0.9921

23. What is the mole fraction of each component in a solution
in which 3.57 g of sodium chloride, NaCl, is dissolved in
25.0 g of water?
First, convert from mass of NaCl to moles of NaCl.
3.57 g NaCl x 1 mole NaCl
=
0.0610857139 mole NaCl
58.44247 g NaCl
Next, convert from mass of water to moles of water.
25.0 g H2O x 1 mole H2O
=
1.387710877 mole H2O
18.01528 g H2O
Substitute these two quantities into the defining equation for mole fraction.
XNaCl =
0.0610857139 mole NaCl
= 0.0421630713 ~ 0.042
(0.0610857139 + 1.387710877) mol solution
Xwater =
1.387710877 mole H20
= 0.9578369287 ~ 0.96
(0.0610857139 + 1.387710877) mol solution
** Note the sum of the mole fractions for a solution will equal 1. **

24. Preparation of a desired solution by
adding water to a concentrate.
 Moles of solute remain the same.

M 1V1  M 2V2

25. 
What volume of 15.8M HNO3 is required
to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3

26. 
1.54m NaCl in
0.500 kg of water


mass 45.0 g of NaCl
add 0.500 kg of water

500 mL of 1.54M NaCl


mass 45.0 g of NaCl
add water until total
volume is 500 mL
500 mL
water
45.0 g
NaCl
500 mL
mark
500 mL
volumetric
flask

27. 
250 mL of 6.0M HNO3
by dilution
95 mL of
15.8M HNO3
› measure 95 mL
of 15.8M HNO3

combine with water until total
volume is 250 mL

Safety: “Do as you oughtta,
add the acid to the watta!”
250 mL
mark
water
for
safety

28. 
Turn in one paper per team.

Complete the following steps:
A) Show the necessary calculations.
B) Write out directions for preparing the solution.
C) Prepare the solution.

For each of the following solutions:
1) 100.0 mL of 0.50M NaCl
2) 0.25m NaCl in 100.0 mL of water
3) 100.0 mL of 3.0M HCl from 12.1M concentrate.

30. 
Colligative Property
› property that depends on the concentration
of solute particles, not their identity

31. 
Freezing Point Depression (tf)
› f.p. of a solution is lower than f.p. of the pure
solvent

Boiling Point Elevation (tb)
› b.p. of a solution is higher than b.p. of the
pure solvent

32. Freezing Point Depression
View Flash animation.

33. Boiling Point Elevation
Solute particles weaken IMF in the solvent.

34. 
Applications
› salting icy roads
› making ice cream
› antifreeze
 cars (-64°C to 136°C)
 fish & insects

35. t = k · m · n
t: change in temperature ( C)
k: constant based on the solvent ( C·kg/mol)
m: molality (m)
n: # of particles

36. 
# of Particles
› Nonelectrolytes (covalent)
 remain intact when dissolved
 1 particle
› Electrolytes (ionic)
 dissociate into ions when dissolved
 2 or more particles

37. 
At what temperature will a solution that is
composed of 0.73 moles of glucose in 225 g
of phenol boil?
GIVEN:
WORK:
b.p. = ?
m = 0.73mol ÷ 0.225kg
tb = ?
tb = (3.60°C·kg/mol)(3.2m)(1)
kb = 3.60°C·kg/mol
tb = 12°C
m = 3.2m
b.p. = 181.8°C + 12°C
n=1
b.p. = 194°C
tb = kb · m · n

38. 
Find the freezing point of a saturated solution
of NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
tf = ?
kf = 1.86°C·kg/mol
WORK:
m = 0.48mol ÷ 0.100kg
tf = (1.86°C·kg/mol)(4.8m)(2)
m = 4.8m
n=2
tf = kf · m · n
f.p. = 0.00°C - 18°C
tf = 18°C
f.p. = -18°C

40. 




Once you write the molecular equation (synthesis,
decomposition, etc.), you should check for
reactants and products that are soluble or
insoluble (determine states of matter, which stay
the same throughout reaction).
Use a solubility table to tell us what compounds are
aqueous (dissolved in water), solids, liquids, and/or
gases
We usually assume the reaction is in water
If the compound is soluble (does dissolve in water),
the the compound splits into its ions
If the compound is insoluble (does NOT dissolve in
water), then it remains as a compound

41. 1.
2.
3.
Write a balanced chemical (molecular) equation
Consult the solubility rules (along with strong acids and
strong bases) and assign the correct state of matter
symbol annotation
Write the Total Ionic Equation (T.I.E)
a.
4.
Eliminate spectator ions
a.
b.
5.
All compounds that are annotated (aq) break up into
individual cations and anions in that order
Spectator ions-those ions that do not participate in the
chemical reaction but are present in the reaction mixture
Spectator ions are in the same form on each side of the
equation arrow
Write the Net Ionic Equation (N.I.E)
a.
The convention is to write the cation first followed by the anion
on the reactants side

42. 
Strong Acids
› Hdrochloric acid - HCl
› Nitric acid - HNO3
› Sulfuric acid - H2SO4
› Hydrobromic acid HBr
› Hydroiodic acid - HI
› Chloric acid - HClO3
› Perchloric acid HClO4
A strong acid is an acid
that ionizes completely

Strong Bases
› Lithium Hydroxide – Li(OH)
› Sodium Hydroxide –
Na(OH)
› Potassium Hydroxide –
K(OH)
› Calcium Hydroxide –
Ca(OH)2
› Rubidium Hydroxide –
Rb(OH)
› Strontium Hydroxide –
Sr(OH)2
› Cesium Hydroxide –
Cs(OH)

44. Solubility Rules

45. Potassium chromate mixes with lead (II) nitrate
Molecular Equation:
K2CrO4 + Pb(NO3)2 
Soluble
Soluble
PbCrO4 + 2 KNO3
Insoluble
Total Ionic Equation:
2 K+ + CrO4 -2 + Pb+2 + 2 NO3- 
PbCrO4 (s) + 2 K+ + 2 NO3Net Ionic Equation:
CrO4 -2 + Pb+2  PbCrO4 (s)
Soluble

46. Balanced Chemical Equation:
Pb(NO3)2(aq) + 2NaI(aq)  PbI2(s) + 2NaNO3(aq)
“Complete Ionic” Equation:
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq)+ 2I-(aq)  PbI2(s) + 2Na+(aq) + 2NO3(aq)
Cancel the “spectator ions” that appear on both sides of the
arrow
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq)+ 2I-(aq)  PbI2(s) + 2Na+(aq) + 2NO3(aq)
“Net Ionic” Equation:
Pb2+(aq) + 2I-(aq)  PbI2(s)

47. States of matter assigned in the
molecular equation stay the same
throughout the T.I.E. and the N.I.E.
 Solids, liquids, and gases DO NOT ionize
 Gases keep their subscript (it doesn’t
become a coefficient like other
compounds)

› For example: H2, Br2, Cl2, etc. stay like this
 They don’t become 2H-1, 2Br -1, 2Cl -1