Solutions
Chapter 14
1
Solutions
• Solutions are homogeneous mixtures
• Solute is the dissolved substance
– Seems to “disappear” or “Takes on the state” of the solvent
• Solvent is the substance the solute dissolves in
– Does not appear to change state
– When both solute and solvent have the same state, the solvent
is the component present in the highest percentage
• Solutions in which the solvent is water are called
aqueous solutions
– Water is often called the universal solvent
• Solutions that contain metal solutes and solvent are
called alloys
2
The Solution Process Ionic Compounds
• When ionic compounds dissolve in water
they dissociate into ions
– ions become surrounded by water molecules hydrated
• When solute particles are surrounded by
solvent molecules we say they are solvated
3
Figure 14.1: When
solid sodium
chloride dissolves,
the ions are
dispersed randomly
throughout the
solution
4
Figure 14.2: Polar water molecules interact with the
positive and negative ions of a salt
5
The Solution Process
Covalent Molecules
• Covalent molecules that are small and have “polar”
groups tend to be soluble in water
• The ability to H-bond with water enhances solubility
H
C
O
6
Figure 14.3: Polar O—H bond similar to those in the
water molecule
7
Solubility
• When one substance (solute) dissolves in another
(solvent) it is said to be soluble
– Salt is soluble in Water,
– Bromine is soluble in methylene chloride
• When one substance does not dissolve in another they
are said to be insoluble
– Oil is insoluble in Water
• There is usually a limit to the solubility of one substance
in another
– Gases are always soluble in each other
– Some liquids are always mutually soluble
8
Solutions & Solubility
• Molecules that are similar in structure tend to form
solutions
– Like dissolves like
• The solubility of the solute in the solvent depends on the
temperature
– Higher Temp = Larger solubility of solid in liquid
– Lower Temp =Larger solubility of gas in liquid
• The solubility of gases depends on the pressure
– Higher pressure = Larger solubility
9
Figure 14.4: The structure of common table sugar
(called sucrose)
10
Figure 14.5: A molecule typical of those found in
petroleum
11
Figure 14.6: An oil layer floating on water
12
Describing Solutions - Qualitatively
• A concentrated solution has a high proportion of solute
to solution
• A dilute solution has a low proportion of solute to
solution
• A saturated solution has the maximum amount of solute
that will dissolve in the solvent
– Depends on temp
• An unsaturated solution has less than the saturation limit
• A supersaturated solution has more than the saturation
limit
– Unstable
13
Describing Solutions Quantitatively
• Solutions have variable composition
• To describe a solution accurately, you need
to describe the components and their
relative amounts
• Concentration = amount of solute in a
given amount of solution
– Occasionally amount of solvent
14
Figure 14.7: Preparation of a standard aqueous
solution
15
Solution Concentration
Percentage
• Parts Per Hundred
• % = grams of solute per 100 g of solution
– Mass Percent or Percent by Mass
– 5.0% NaCl has 5.0 g of NaCl in every 100 g of
solution
• Mass of Solution = Mass of Solute + Mass of Solvent
• Divide the mass of solute by the mass of solution
and multiply by 100%
16
Solution Concentration
Molarity
• moles of solute per 1 liter of solution
• used because it describes how many
molecules of solute in each liter of solution
• If a sugar solution concentration is 2.0 M ,
1 liter of solution contains 2.0 moles of
sugar, 2 liters = 4.0 moles sugar, 0.5 liters
= 1.0 mole sugar, etc.
moles of solute
molarity =
liters of solution
17
Molarity & Dissociation
• the molarity of the ionic compound allows you to determine the
molarity of the dissolved ions
• CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq)
• A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each
liter of solution
– 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2, 0.5 L = 0.5
moles CaCl2
• Because each CaCl2 dissociates to one Ca+2, 1.0 M CaCl2 = 1.0 M
Ca+2
– 1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2, 0.5 L = 0.5 moles
Ca+2
• Because each CaCl2 dissociates to 2 Cl-1, 1.0 M CaCl2 = 2.0 M Cl-1
– 11 L = 2.0 moles Cl-1, 2 L = 4.0 moles Cl-1, 0.5 L = 1.0 moles Cl18
Dilution
• Dilution is adding solvent to decrease the
concentration of a solution
• The amount of solute stays the same, but the
concentration decreases
• Dilution Formula
M1 x V1 = M2 x V2
• Concentrations and Volumes can be most units as
long as consistent
19
Figure 14.8: Acetic acid dilution
20
Solution Stoichiometry
• Many reactions occur in solution, therefore you
need to be able to predict amounts of reactants and
products in terms of concentrations and volumes
as well as masses
• Basic strategy is the same
1. Balance the Equation
2. Change Given Amounts to Moles
3. Determine Limiting Reactant
4. Calculate Moles of Required Substance
5. Convert Moles of the Required Substance into the
Desired Unit
21
Example
Calculate the Mass of Solid NaCl required to
Precipitate all the Ag+1 ions from 1.50 L of a
0.100 M AgNO3 Solution
1. Write and Balance the Reaction
The reaction is a Precipitation Reaction. The reaction
involves Cl-1 ions from NaCl reacting with Ag+1 ions
from AgNO3 to form AgCl(s). Therefore we get
Ag+1(aq) + Cl-1(aq)  AgCl(s)
After Balancing we get
Ag+1(aq) + Cl-1(aq)  AgCl(s)
22
Example
Calculate the Mass of Solid NaCl required to
Precipitate all the Ag+1 ions from 1.50 L of a
0.100 M AgNO3 Solution
2. Change the Given Amounts to Moles
We are given 1.50 L of 0.100 M AgNO3
Since 1 AgNO3 dissociates into 1 Ag+1
The concentration of Ag+1 = 0.100 M
1 L Solution = 0.100 mol Ag+1
0.100 mol Ag 1
1.50 L x
 0.150 mol Ag 1
1 L Solution
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Example
Calculate the Mass of Solid NaCl required to
Precipitate all the Ag+1 ions from 1.50 L of a
0.100 M AgNO3 Solution
3. Determine the Limiting Reactant
Since we are going to precipitate all the Ag+1
by adding Cl-1 , the Ag+1 is the Limiting Reactant
4. Determine the Number of Moles of the Required Substance
We need to calculate the Moles of Cl-1 Required to
precipitate 0.150 moles of Ag+1
1
1
mol
Cl
1
0.150 mol Ag 1 x

0.150
mol
Cl
1
1 mol Ag
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Example
Calculate the Mass of Solid NaCl required to
Precipitate all the Ag+1 ions from 1.50 L of a
0.100 M AgNO3 Solution
5. Convert Moles of the Required Substance into
the Desired Unit
We need 0.150 moles of Cl-1
Since 1 NaCl dissociates into 1 Cl-1
The moles of NaCl needed = 0.150 moles
1 mol NaCl = 58.44 g NaCl
58.44 g NaCl
0.150 mol NaCl x
 8.76 g NaCl
1 mol NaCl
25
Neutralization Reactions
• Acid-Base reactions are also called
Neutralization Reactions
• Often we use neutralization reactions to
determine the concentration of an unknown
acid or base
• The procedure is called a titration. With
this procedure we can add just enough acid
solution to neutralize a known volume of a
base solution
– Or visa versa
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Normality
• Normality is a concentration unit used
mainly for acids and bases
• One equivalent of an acid is the amount of
acid that can furnish 1 mol of H+1
• One equivalent of a base is the amount of
base that can furnish 1 mol of OH-1
• The equivalent weight is the mass of 1
equivalent of an acid or base
27
Equivalents
• 1 mol HCl = 1 mol H+1 = 1 equivalent HCl
– Therefore the equivalent weight of HCl = the Molar Mass of
HCl = 36.45 g
• 1 mol H2SO4 = 2 mol H+1 = 2 equiv H2SO4
– Therefore the equivalent weight of H2SO4 = one-half the Molar
Mass of H2SO4 = ½(98.07 g) = 49.04 g
• 1 mol NaOH = 1 mol OH-1 = 1 equivalent NaOH
– Therefore the equivalent weight of NaOH = the Molar Mass of
NaOH = 40.00 g
• 1 mol KOH = 1 mol OH-1 = 1 equiv KOH
– Therefore the equivalent weight of KOH = the Molar Mass of
KOH = 56.11 g
28
Solution Concentration
Normality
• equivalents of solute per 1 liter of solution
• used because it describes how many H+1 or OH-1
in each liter of solution
• If an acid solution concentration is 2.0 N , 1 liter
of solution contains 2.0 equiv of acid which
means 2 mol H+1
– 2 liters = 4.0 equiv acid = 4.0 mol H+1
– 0.5 liters = 1.0 equiv acid = 1.0 mol H+1
29
Normality
equivalents of solute
Normality =
liters of solution
Liters x Normality = Equivalents of Solute
30
Normality and Neutralization
• One Equivalent of Acid exactly neutralizes
One Equivalent of Base
• Can be used to simplify neutralization
stoichiometry problems to the equation
Nacid x Vacid = Nbase x Vbase
• The volumes can be most any unit, as long
as they are consistent
31
Example
What volume of 0.075 N KOH is required to
Neutralize 0.135 L of 0.45 N H3PO4
 Determine the quantities and units in the problem
Normality
Volume
Acid Solution
0.45 N
0.135 L
Base Solution
0.075 N
?L
 Solve the Formula for the Unknown Quantity
Nacid x Vacid  Nbase x Vbase
Nacid x Vacid
Vbase 
Nbase
32
Example
What volume of 0.075 N KOH is required to
Neutralize 0.135 L of 0.45 N H3PO4
 Plug the Values into the Equation and Solve
Normality
Volume
Acid Solution
0.45 N
0.135 L
Base Solution
0.075 N
?L
Nacid x Vacid
Vbase 
Nbase
0.45 N x 0.135 L
Vbase 
 0.81 L
0.075 N
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