Chemistry I –
Chapters 15 & 16
Chemistry I HD –
Chapter 15
Solutions
ICP – Chapter 22
Why does a raw egg swell or shrink when
placed in different solutions?
1
Some Definitions
A solution is a
HOMOGENEOUS
mixture of 2 or more
substances in a
single phase.
One constituent is
usually regarded as
the SOLVENT and
the others as
SOLUTES.
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3
Parts of a Solution
• SOLUTE – the
part of a solution
that is being
dissolved (usually
the lesser
amount)
• SOLVENT – the
part of a solution
that dissolves the
solute (usually
the greater
amount)
• Solute + Solvent =
Solution
Solute Solvent
Example
solid
solid
Alloys (brass, steel)
solid
liquid
Salt water
gas
solid
Air bubbles in ice
cubes
liquid
liquid
“suicides” (mixed
drinks)
gas
liquid
Soft drinks
gas
gas
Air
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Definitions
Solutions can be classified as
saturated or unsaturated.
A saturated solution contains
the maximum quantity of
solute that dissolves at that
temperature.
An unsaturated solution
contains less than the
maximum amount of solute
that can dissolve at a
particular temperature
Example: Saturated and Unsaturated Fats
Saturated fats are
called saturated
because all of the
bonds between the
carbon atoms in a fat
are single bonds.
Thus, all the bonds
on the carbon are
occupied or
“saturated” with
hydrogen. These are
stable and hard to
decompose. The
body can only use
these for energy, and
so the excess is
stored. Thus, these
should be avoided in
diets. These are
usually obtained from
sheep and cattle fats.
Butter and coconut
oil are mostly
saturated fats.
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Unsaturated fats have at least one double bond
between carbon atoms; monounsaturated means
there is one double bond, polysaturated means
there are more than one double bond. Thus, there
are some bonds that can be broken, chemically
changed, and used for a variety of purposes.
These are REQUIRED to carry out many functions
in the body. Fish oils (fats) are usually
unsaturated. Game animals (chicken, deer) are
usually less saturated, but not as much as fish.
Olive and canola oil are monounsaturated.
Definitions
SUPERSATURATED SOLUTIONS
contain more solute than is
possible to be dissolved
Supersaturated solutions are
unstable. The supersaturation is
only temporary, and usually
accomplished in one of two ways:
1. Warm the solvent so that it will
dissolve more, then cool the
solution
2. Evaporate some of the solvent
carefully so that the solute does
not solidify and come out of
solution.
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7
Supersaturated
Sodium Acetate
• One application
of a
supersaturated
solution is the
sodium acetate
“heat pack.”
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C. Solubility
• Solubility Curve
– shows the dependence of
solubility on temperature
C. Johannesson
9
C. Solubility
• Solubility Curve
– shows the dependence of
solubility on temperature
– The line representssaturation
– Below the lineunsaturated
– Above the lineSupersaturated
C. Johannesson
• Some
• questions to
• answer
• What mass of solute will
dissolve in 100ml of
water at the following
temperatures. Which is
most soluble
• KNO3 at 70C
• NaCl at 100C
• NH4CL at 90C
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11
C. Solubility
• 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
C. Johannesson
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Henry’s Law
• At a given temperature the solubility of a
gasin a liquid (S) is directly proportional to
the pressure of the gas above the liquid (P).
• In other words if the pressure increases the
solubility increases
• S1 = S2
• P1
P2
13
IONIC COMPOUNDS
Compounds in Aqueous Solution
Many reactions involve ionic
compounds, especially reactions in
water — aqueous solutions.
KMnO4 in water
K+(aq) + MnO4-(aq)
Aqueous Solutions
How do we know ions are
present in aqueous
solutions?
The solutions conduct
electricity!
They are called
ELECTROLYTES
HCl, MgCl2, and NaCl are
strong electrolytes.
They dissociate
completely (or nearly so)
into ions.
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Aqueous
Solutions
Some compounds
dissolve in water but
do not conduct
electricity. They are
called nonelectrolytes.
Examples include:
sugar
ethanol
ethylene glycol
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16
It’s Time to Play Everyone’s
Favorite Game Show… Electrolyte
or Nonelectrolyte!
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Electrolytes in the Body
 Carry messages to
and from the brain
as electrical signals
 Maintain cellular
function with the
correct
concentrations
electrolytes
Make your own
50-70 g sugar
One liter of warm water
Pinch of salt
200ml of sugar free fruit
squash
Mix, cool and drink
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Concentration of Solute
The amount of solute in a solution
is given by its concentration.
Molarity (M) =
moles solute
liters of solution
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1.0 L of
water was
used to
make 1.0 L
of solution.
Notice the
water left
over.
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Steps to make a solution from a
SOLID
Step 1: Weigh out the amount of solid needed
Step 2: place the weighed solid into a VOLUMETRIC FLASK
Step 3: while mixing add solvent to bring the level up to
Mark on the flask neck.

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PROBLEM: Dissolve 5.00 g of NiCl2•6
H2O in enough water to make 250 mL
of solution. Calculate the Molarity.
Step 1: Calculate moles
of NiCl2•6H2O
1 mol
5.00 g •
= 0.0210 mol
237.7 g
Step 2: Calculate Molarity
0.0210 mol
= 0.0841 M
0.250 L
[NiCl2•6 H2O ] = 0.0841 M
USING MOLARITY
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a 0.0500 M
solution?
moles = M•V
Step 1: Change mL to L.
250 mL * 1L/1000mL = 0.250 L
Step 2: Calculate.
Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles
Step 3: Convert moles to grams.
(0.0125 mol)(90.00 g/mol) =
1.13 g
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24
Learning Check
How many grams of NaOH are required
to prepare 400. mL of 3.0 M NaOH
solution?
1) 12 g
2) 48 g
3) 300 g
25
Solution
M = moles of solute
Liters of solution
M * V = moles
3.0 mol/L * 0.400 L = 1.2 mol NaOH
1.2 mole NaOH x 40.0 g NaOH
1 mole NaOH
= 48 g NaOH
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Molarity from Solubility curve
• Determine the Molarity of a saturated NaCl
solution at 25C
• Go back to solubility curve
Concentration Units
An IDEAL SOLUTION is
one where the properties
depend only on the
concentration of solute.
Need conc. units to tell us the
number of solute particles
per solvent particle.
The unit “molarity” does not
do this!
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Two Other Concentration Units
MOLALITY, m
mol solute
m of solution =
kilograms solvent
% by mass
% by mass =
grams solute
X100
grams solution
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Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol
in 250. g of H2O. Calculate molality and % by
mass of ethylene glycol.
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Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g
of H2O. Calculate m & % of ethylene glycol (by mass).
Calculate molality
1.00 mol glycol
conc (molality) =
 4.00 molal
0.250 kg H2O
Calculate weight %
62.1 g
%glycol =
x 100 = 19.9%
62.1 g + 250. g
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Learning Check
A solution contains 15 g Na2CO3 and 235 g of
H2O? What is the mass % of the solution?
1) 15% Na2CO3
2) 6.4% Na2CO3
3) 6.0% Na2CO3
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Solution
mass solute =
15 g Na2CO3
mass solution=
15 g + 235 g = 250 g
%(by mass) = 15 g Na2CO3
x
250 g solution
= 6.0% Na2CO3 solution
100
33
Using mass %
How many grams of NaCl are needed to
prepare 250 g of a 10.0% (by mass) NaCl
solution?
250 g NaCl soln x 10.0 g NaCl
100 g NaCl soln
= 25 g NaCl
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Try this molality problem
• 25.0 g of NaCl is dissolved in 5000. mL of
water. Find the molality (m) of the resulting
solution.
m = mol solute / kg solvent
25 g NaCl
1 mol NaCl
58.5 g NaCl
= 0.427 mol NaCl
Since the density of water is 1 g/mL,
5000 mL = 5000 g, which is 5 kg
0.427 mol NaCl
5 kg water
= 0.0854 m salt water
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Mole Fraction (✗)
• N is the number of moles of each substance
solute and solvent
✔ all mole fractions add up to = 1
36
Problems
If I add 1.65 L of water to 112 grams of sodium
nitrate
•
a)What is the molality of NaNO3 in this
solution?
b) What is the percent by mass of sodium
nitrate in this solution?
c) What is the mole fraction of water in this
solution?
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•
•
•
•
•
•
•
If I add 1.65 L of water to 112 grams of
sodium acetate…
a)
What is the molality of NaC2H3O2 in
this solution?
0.82 m
b)
What is the percent by mass of
sodium acetate in this solution?
6.36%
c) What is the mole fraction of water in this
solution?
0.985
Colligative Properties
On adding a solute to a solvent, the properties
of the solvent are modified.
• Vapor pressure
decreases
• Melting point
decreases
• Boiling point
increases
• Osmosis is possible (osmotic pressure)
These changes are called COLLIGATIVE
PROPERTIES.
They depend only on the NUMBER of solute
particles relative to solvent particles, not on
the KIND of solute particles.
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39
Change in Freezing Point
Pure water
Ethylene glycol/water
solution
The freezing point of a solution is LOWER
than that of the pure solvent
Change in Freezing Point
Common Applications
of Freezing Point
Depression
Propylene glycol
Ethylene
glycol –
deadly to
small
animals
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Change in Freezing Point
Common Applications
of Freezing Point
Depression
Which would you use for the streets of
Bloomington to lower the freezing point
of ice and why? Would the temperature
make any difference in your decision?
a)
sand, SiO2
b)
Rock salt, NaCl
c)
Ice Melt, CaCl2
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Change in Boiling Point
Common Applications
of Boiling Point
Elevation
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Boiling point elevation
• .
• The solute gets in the way of the molecules
escaping to become vapor.
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Boiling Point Elevation and
Freezing Point Depression
∆T = K•m•i
i = van’t Hoff factor = number of particles
produced per molecule/formula unit. For
covalent compounds, i = 1. For ionic
compounds, i = the number of ions
present (both + and -)
Compound
Theoretical Value of i
glycol
1
NaCl
2
CaCl2
3
Ca3(PO4)2
5
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Boiling Point Elevation and
Freezing Point Depression
∆T = K•m•i
m = molality
K = molal freezing
point/boiling point constant
Substance
Kf
benzene
5.12
camphor
40.
carbon tetrachloride 30.
Substance
Kb
benzene
2.53
camphor
5.95
carbon tetrachloride 5.03
ethyl ether
water
ethyl ether
water
1.79
1.86
2.02
0.52
Change in Boiling Point
Dissolve 62.1 g of glycol (1.00 mol) in 250. g
of water. What is the boiling point of the
solution?
Kb = 0.52 oC/molal for water (see Kb table).
Solution
∆TBP = Kb • m • i
1.
2.
Calculate solution molality = 4.00 m
∆TBP = Kb • m • i
∆TBP = 0.52 oC/molal (4.00 molal) (1)
∆TBP = 2.08 oC
BP = 100 + 2.08 = 102.08 oC
(water normally boils at 100)
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Freezing Point Depression
Calculate the Freezing Point of a 4.00 molal
glycol/water solution.
Kf = 1.86 oC/molal (See Kf table)
Solution
∆TFP = Kf • m • i
= (1.86 oC/molal)(4.00 m)(1)
∆TFP = 7.44
FP = 0 – 7.44 = -7.44 oC
(because water normally freezes at 0)
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Freezing Point Depression
At what temperature will a 5.4 molal solution
of NaCl freeze?
Solution
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
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49
Molecular mass determination
• A Solution of 7.50 g of a nonvolital compound
in 22.60 g of water boils at 100.78°C at 760
mm Hg. What is the molecular mass of the
solute. (assume i =1)
• ΔTb= Kb X m X i 
m = ΔT
•
Kb
The Kb for water is 0.512 °C/molal
m = 0.78°C
0.512 °C/molal
m =1.5
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Now calculate moles of solute
in solution
• 1.5 m X 22,6 g of water X 1kg/1000g =0.034
mol solute
• Finally use the number of moles of solute and its
mass to determine the molecular mass of the
solute.
• MM of solute = mass of solute/moles of solute
• =7.50 g/0.0344mol= 2.2X102 g/mol
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Problem
• The freezing point for water is lowered to0.390 C when 3.90 grames of a molecular
solute is dissolved in 475 g of water.
Calculate the molar mass of the solute.
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osmosis
Setup for titrating an acid with a base
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Titration
1. Add solution from the buret.
2. Reagent (base) reacts with
compound (acid) in solution
in the flask.
3. Indicator shows when exact
stoichiometric reaction has
occurred. (Acid = Base)
This is called
NEUTRALIZATION.
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LAB PROBLEM #1: Standardize a
solution of NaOH — i.e., accurately
determine its concentration.
35.62 mL of NaOH is
neutralized with 25.2 mL of
0.0998 M HCl by titration to
an equivalence point. What
is the concentration of the
NaOH?
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35.62 mL of NaOH is neutralized with 25.2 mL of
0.0998 M HCl by titration to an equivalence point.
What is the concentration of the NaOH?
Ma Va = Mb Vb
M x V = moles
Ma Va
Since the moles of the acid
and base are equal and do
not change:
= Mb
Vb
M x V (acid) = M x V (base)
(0.0998 M) (25.2 mL)
=
(35.62 mL)
0.0706 M
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Preparing Solutions
• Weigh out a solid
solute and dissolve in a
given quantity of
solvent.
• Dilute a concentrated
solution to give one
that is less
concentrated.
58
SAFETY
•Safety: “Do as you
oughtta, add the
acid to the watta!”
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PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH.
What do you do?
Add water to the 3.0 M solution to lower
its concentration to 0.50 M
Dilute the solution!
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?
Conclusion:
add 250 mL
of water to
50.0 mL of
3.0 M NaOH
to make 300
mL of 0.50 M
NaOH.
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Preparing Solutions by
Dilution
A shortcut
M1 • V1 = M2 • V2
You try this dilution problem
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• You have a stock bottle of hydrochloric acid,
which is 12.1 M. You need 400 mL of 0.10 M
HCl. How much of the acid and how much
water will you need?
M1 V1 = M2 V2
M1 = 12.1 M
V1 = ??? L
M2 = 0.10 M
V1 = 0.0031 L (or 3.1 mL HCl)
Then add enough water so that
the total volume is 400 mL.
It should be ABOUT 396.9 mL
V2 = 400 mL  0.400 L (400 – 3.1), but it will be off
slightly due to the density of
the HCl not being 1.00 g/mL