Chapter 8A
Solutions
1
CHAPTER OUTLINE







Type of Solutions
Electrolytes & Non-electrolytes
Equivalents of Electrolytes
Solubility & Saturation
Soluble & Insoluble Salts
Formation of a Solid
Precipitation Reactions
2
TYPE OF
SOLUTIONS
 A solution is a homogeneous mixture of two
substances:
Solute:
substance being dissolved
present in smaller amount
Solvent:
substance doing the dissolving
present in larger amount
Solutes and solvents may be of any form of matter:
solid, liquid or gas.
3
TYPE OF
SOLUTIONS
Type
Example
Solute
Solvent
Gas in gas
Air
Oxygen
Nitrogen
CO2
Water
Gas in liq.
Soda water
Liq. in liq.
Vinegar
Acetic acid
Water
Solid in liq.
Seawater
Salt
Water
4
TYPE OF
SOLUTIONS
Type
Example
Solute
Solvent
Liquid in
solid
Dental
amalgam
Mercury
Silver
Solid in
solid
Brass
Zinc
Copper
5
SOLUBILITY
 Solutions form between solute and solvent
molecules because of similarities between them.
Like dissolves Like
Ionic solids dissolve in water because the charged ions
(polar) are attracted to the polar water molecules.
Non-polar molecules such as oil and grease dissolve in
non-polar solvents such as kerosene.
6
SOLUBILITY
Water (polar)
CH2Cl2 (non-polar)
Ni(NO3 )2
(polar)
I2 (non-polar)
7
ELECTROLYTES &
NON-ELECTROLYTES
 Solutions can be characterized by their ability to
conduct an electric current.
 Solutions containing ions are conductors of electricity
and those that contain molecules are non-conductors.
 Substances that dissolve in water to
form ions are called electrolytes.
 The ions formed from these
substances conduct electric current
in solution, and can be tested using
a conductivity apparatus.
8
STRONG
ELECTROLYTES
 Electrolytes are further classified as strong
electrolytes and weak electrolytes.
 In water, a strong electrolyte exists only as ions.
Only ions
 Strong electrolytes make
the light
present after
bulb on the conductivity
solution
apparatus glow brightly.
 Ionic substances such as NaCl
are strong electrolytes.
NaCl (s)
Na+ (aq) + Cl (aq)
9
WEAK
ELECTROLYTES
 Solutions containing weak electrolytes contain only
a few ions.
 These solutions make the light
bulb on the conductivity
Few ions
apparatus glow dimly.
present after
 Weak acids and basessolution
that
dissolve in water and produce few
ions are weak electrolytes.
HF (aq)
H+ (aq) + F (aq)
10
NONELECTROLYTES
 Substances that do not form any ions in solution
are called non-electrolytes. No ions
present
 With these solutions the bulb
on theafter
conductivity
solution
apparatus does not glow.
 Covalent molecules that dissolve in water but do
not form ions, such as sugar, are non-electrolytes.
C12H22O11 (s)
C12H22O11 (aq)
11
ELECTROLYTES &
NON-ELECTROLYTES
12
Example 1:
Identify the predominant particles in each of the following
solutions and write the equation for the formation of the
solution:
NH4Br
NH4Br (s)
Strong electrolyte
(only ions)
NH4+ (aq) + Br (aq)
13
Example 1:
Identify the predominant particles in each of the following
solutions and write the equation for the formation of the
solution:
CH4 N2O
Non-electrolyte
CH4N2O (s)
(only molecules)
CH4N2O (aq)
14
Example 1:
Identify the predominant particles in each of the following
solutions and write the equation for the formation of the
solution:
HClO
Weak electrolyte
HClO (aq)
(few ions)
H+ (aq) + ClO (aq)
15
EQUIVALENTS OF
ELECTROLYTES
 Body fluids typically contain a mixture of several
electrolytes, such as Na+, Cl–, K+ and Ca2+.
 Each individual ion is measured in terms of an
equivalent (Eq), which is the amount of that ion
equal to 1 mole of positive or negative electrical
charge.
 For example, 1 mole of Na+ ions and 1 mole of Cl–
ions are each 1 equivalent (or 1000 mEq) because
they each contain 1 mole of charge.
16
EQUIVALENTS OF
ELECTROLYTES
 An
Some
ionexamples
with a charge
of ionsofand
2+ or
their
2– contains
equivalents
2 are
equivalents
shown
below:
per mole.
Ion
Electrical
Charge
No. of Equivalents
in 1 Mole
Na+
1+
1 Eq
Ca2+
2+
2 Eq
Fe3+
3+
3 Eq
Cl–
1–
1 Eq
SO42–
2–
2 Eq
17
EQUIVALENTS OF
ELECTROLYTES
 In body, the charge of the positive ion is always
balanced by the charge of the negative ion.
 For example, a solution containing 25 mEq/L of Na+
and 4 mEq/L of K+ must have 29 mEq/L of Cl– to
balance.
 Shown next are examples of some common
intravenous solutions and their ion concentrations.
18
EQUIVALENTS OF
ELECTROLYTES
19
Example 1:
Indicate the number of equivalents in each of the
following:
2 mol K+
1 Eq
2 mol x 1 mol = 2 Eq
Mg2+
2 Eq
0.5 mol x 1 mol = 1 Eq
3 mol CO32
2 Eq
3 mol x 1 mol = 6 Eq
0.5 mol
20
Example 2:
A typical concentration for Ca2+ in blood is 8.8 mEq/L.
How many moles of Ca2+ are present in 0.50 L of blood?
Liter
blood
8.8 mEq
0.50 L x
1L
moles
Ca2+
mEq
x
1 Eq
103 mEq
x
1 mol
2 Eq
= 0.0022 mol
21
Example 3:
An IV solution contains 155 mEq/L of Cl–. If a patient
received 1250 mL of the IV solution, how many moles of
chloride were given to him?
1L
1250 mL x
103 mL
x
155 mEq
1L
x
1 Eq
103 mEq
x
1 mol
1 Eq
= 0.194 mol
22
Example 4:
A sample of Ringer’s solution contains the following
concentrations (mEq/L) of cations: Na+ 147, K+ 4, and
Ca2+ 4. If Cl– is the only anion in the solution, what is the
concentration of Cl– in mEq/L?
Total cation mEq/L = 147 + 4 + 4
= 155 mEq/L
Total cation mEq/L = Total anion mEq/L
Total Cl mEq/L = 155 mEq/L
23
SOLUBILITY
 Solubility refers to the maximum amount of solute
that can be dissolved in a given amount of solvent.
 Many factors affect the solubility of a solute in a
solution.
Type of solute
Type of solvent
Temperature
Solubility is measured in grams of solute per 100 grams
of solvent at a given temperature.
24
SATURATION
 A solution that does not contain the maximum
amount of solute in it, at a given temperature, is
called an unsaturated solution.
 A solution that contains the
maximum amount of solute in
it, at a given temperature, is
called a saturated solution.
Un-dissolved
solid in solution
25
SOLUBILITY
 Solubility of most solids
in water increases as
temperature increases.
 Using a solubility chart,
the solubility of a solute
at a given temperature
can be determined.
 For example, KNO3 has
a solubility of 80 g/100 g
H2O (80%) at 40 C.
26
SOLUBILITY
OF GASES
 Solubility of gases in water decreases as temperature
increases.
 At higher temperatures more gas molecules have the
energy to escape from solution.
 Henry’s law states that the solubility of a gas is
directly proportional to the pressure above the
liquid.
 For example, a can of soda is carbonated at high
pressures in order to increase the solubility of CO2.
Once the can is opened, the pressure is reduced and
the excess gas escapes from the solution.
27
SOLUBLE &
INSOLUBLE SALTS
 Many ionic solids dissolve in water and are called
soluble.
 However, some ionic salts do not dissolve in water
and do not form ions in solution. These salts are
called insoluble salts and remain solid in solution.
 Chemists use a set of solubility rules to predict
whether a salt is soluble or insoluble.
 These rules are summarized next.
28
SOLUBILITY
RULES
S
O
L
U
B
L
E
NO3
No exceptions
Na+, K+
NH4+
No exceptions
Cl, Br, I
Except those containing Ag+
, Pb2+
SO4
2
Except those containing
Ba2+ , Pb2+, Ca2+
29
SOLUBILITY
RULES
I
N
S
O
L
U
B
L
E
S2, CO32
PO43
Except those containing
Na+ , K+, NH4+
OH
Except those containing
Na+ , K+, Ca2+, NH4+
30
Example 1:
Use the solubility rules to determine if each of the
following salts are soluble or insoluble:
K3PO4
soluble
CaCO3
insoluble
(all salts of K are soluble)
(most carbonates are insoluble)
31
Example 2:
Using the solubility chart, determine if each of the
following solutions is saturated or unsaturated at 20C:
25 g NaCl in 100 g water
Solubility of NaCl
at 20 C is 40%
Solution is
unsaturated
32
Example 2:
Using the solubility chart, determine if each of the
following solutions is saturated or unsaturated at 20C:
11 g NaNO3 in 25 g water
Solubility of NaNO3
at 20 C is 85%
Solution is
unsaturated
33
Example 2:
Using the solubility chart, determine if each of the
following solutions is saturated or unsaturated at 20C:
400. g of glucose in 125 g water
Solubility of glucose
at 20 C is 80%
Solution is
saturated
34
FORMATION
OF A SOLID
 Solubility rules can be used to predict whether a
solid, called a precipitate, can be formed when
two solutions of ionic compounds are mixed.
 A solid is formed when two ions of an insoluble
salt come in contact with one another.
 For example, when a solution of K2CrO4 is
mixed with a solution of Ba(NO3)2 a yellow
insoluble salt BaCrO4 is produced.
35
AQUEOUS
REACTIONS
K2CrO4 (aq) + Ba(NO3)2 (aq)
BaCrO4 (s) + 2 KNO3 (aq)
precipitate
Solid BaCrO4
forms
36
PRECIPITATION
REACTIONS
 Double replacement reactions in which a precipitate
is formed are called precipitation reactions.
 To predict a precipitate, follow the steps outlined
next.
37
PRECIPITATION
REACTIONS
1. Write the reactant ions that form after dissolution.
2. Write the product combinations possible when reactant
ions combine.
3. Use solubility rules to determine if any of the products are
insoluble.
4. If a precipitate forms, write the formula for the solid.
Write other ions that form soluble salts as ions. If no
precipitate forms, write “NO REACTION” after the
arrow.
5. Cancel ions that appear the same on both sides of the
equation (spectator ions), to form Net Ionic Equation.
38
PRECIPITATION
REACTIONS
The reaction of K2CrO4 and Ba(NO3)2 can be predicted
as shown below
 ???
Step 1:
2 K+ +CrO42 + Ba2+ + 2 NO3
Step 2:
2 K+ +CrO42 +Ba2+ + 2 NO3  BaCrO4(?) + 2K+NO3 (?)
Net Ionic
Step 3:

2 K+ +CrO42 +Ba2+ + 2 NO
Equation
 BaCrO4(s) + 2K+NO3 (aq)
3
Step 4:
2 K+ +CrO42 +Ba2+ + 2 NO3  BaCrO4(s) + 2K+ +2NO3
Step 5:
Ba2+ + CrO42  BaCrO4 (s)
spectator ions
precipitate
39
Example 1:
Predict the products, if any, for the reaction of AgNO3
and NaCl and write the net ionic equation.
Step 1:
Ag+ +NO3 + Na+ + Cl
Step 2:
Ag+ +NO3 +Na+ +ClNetIonic
Ag+Cl (?) + Na+ NO3 (?)
Step 3:
Ag+ +NO3 +Na+ +ClEquation
 Ag+Cl (s) + Na+ NO3 (aq)
Step 4:
Ag+ +NO3 +Na+ +Cl  AgCl (s) + Na+ + NO3 (aq)
Step 5:
 ???
+ +Cl
Agspectator
ions AgCl (s)
spectator ions
precipitate
40
Example 2:
Predict the products, if any, for the reaction of Na2SO4
and Pb(NO3)2 and write the net ionic equation.
Step 1:
2 Na+ +SO42 + Pb2+ + 2 NO3
Step 2:
2Na+ +SO42 +Pb2+ +2NO3  2Na+NO3(?) + Pb2+SO42(?)
Step 3:
2Na+ +SO42 +Pb2+ +2NO3  2Na+NO3(aq) + Pb2+SO42(s)
Step 4:
2Na+ +SO42 +Pb2+ +2NO3  2Na+ + 2NO3(aq) + PbSO4 (s)
Step 5:
Pb2+ +SO42
 ???
 PbSO4 (s)
(NIE)
precipitate
41
Example 3:
Predict the products, if any, for the reaction of PbCl2
and KI and write the net ionic equation.
Step 1:
Pb2+ + 2 Cl
Step 2:
Pb2+ + 2 Cl + K+ + I  Pb2+I2(?) + K+Cl(?)
Step 3:
Pb2+ + 2 Cl + 2 K+ +2 I  Pb2+I2(s) + 2 K+Cl(aq)
Step 4:
Pb2+ + 2 Cl + 2 K+ +2 I  PbI2 (s) + 2 K+ + 2 Cl
Step 5:
+ K+ + I  ???
Pb2+ + 2 I
 PbI2 (s)
(NIE)
precipitate
42
Example 4:
Predict the products, if any, for the reaction of NH4Cl
and KNO3 and write the net ionic equation.
Step 1:
NH4+ + Cl + K+ + NO3  ???
No Reaction
Step 2:
NH4+ + Cl + K+ + NO3  NH4+NO3(?) + K+Cl(?)
Step 3:
NH4+ + Cl + K+ + NO3  NH4+NO3(aq) + K+Cl(aq)
Step 4:
NH4+ + Cl + K+ + NO3  NH4+ + NO3 + K+ + Cl
43
THE END
44