Transcript Chapter 8 Solutions

Solutions
Properties of Water
Solutions
1
Predict the % water in the following
foods
2
Predict the % water in the following
foods
88% water
94% water
85% water
86% water
3
Water in the Body
water gain
liquids
1000 mL
food
1200 mL
cells
300 mL
water loss
urine
1500 mL
perspiring 300 mL
exhaling
600 mL
feces
100 mL
Calculate the total water gain and water loss
Total ______ mL
_____ mL
4
Water
 Most common solvent
 A polar molecule
O a hydrogen bond
H +
H +
5
Hydrogen Bonds Attract Polar
Water Molecules
6
Explore: Surface Tension HW
 Fill a glass to the brim with water
 How many pennies can you add to the
glass without causing any water to run
over?
Predict _________________
Actual _________________
 Explain your results
7
Explore
1. Place some water on a waxy surface. Why
do drops form?
2. Carefully place a needle on the surface of
water. Why does it float? What happens if
you push it through the water surface?
3. Sprinkle pepper on water. What does it
do? Add a drop of soap. What happens?
8
Surface Tension
 Water molecules within water hydrogen
bond in all directions
 Water molecules at surface cannot hydrogen
bond above the surface, pulled inward
 Water surface behaves like a thin, elastic
membrane or “skin”
 Surfactants (detergents) undo hydrogen
bonding
9
Solute and Solvent
Solutions are homogeneous mixtures of
two or more substances
Solute
The substance in the lesser amount
Solvent
The substance in the greater amount
10
Nature of Solutes in Solutions
 Spread evenly throughout the
solution
 Cannot be separated by filtration
 Can be separated by evaporation
 Not visible, solution appears
transparent
 May give a color to the solution
11
Types of Solutions
air
O2 gas and N2 gas
gas/gas
soda
CO2 gas in water
gas/liquid
seawater
NaCl in water
solid/liquid
brass
copper and zinc
solid/solid
12
Discussion
Give examples of some solutions
and explain why they are solutions.
13
Learning Check SF1
(1) element
(2) compound
(3) solution
A. water
1
2
3
B. sugar
1
2
3
C. salt water
1
2
3
D. air
1
2
3
E. tea
1
2
3
14
Solution SF1
(1) element
(2) compound
A. water
2
B. sugar
2
C. salt water
3
D. air
3
E. tea
3
(3) solution
15
Learning Check SF2
Identify the solute and the solvent.
A. brass: 20 g zinc + 50 g copper
solute
=
1) zinc
2) copper
solvent
=
1) zinc
2) copper
B. 100 g H2O + 5 g KCl
solute
=
1) KCl
2) H2O
solvent
=
1) KCl
2) H2O
16
Solution SF2
A. brass: 20 g zinc + 50 g copper
solute
solvent
=
=
1) zinc
2) copper
B. 100 g H2O + 5 g KCl
solute
=
1) KCl
solvent
=
2) H2O
17
Learning Check SF3
Identify the solute in each of the following
solutions:
A. 2 g sugar (1) + 100 mL water (2)
B. 60.0 mL ethyl alcohol(1) and 30.0 mL
of methyl alcohol (2)
C. 55.0 mL water (1) and 1.50 g NaCl (2)
D. Air: 200 mL O2 (1) + 800 mL N2 (2)
18
Solution SF3
Identify the solute in each of the following
solutions:
A. 2 g sugar (1)
B. 30.0 mL of methyl alcohol (2)
C. 1.5 g NaCl (2)
D. 200 mL O2 (1)
19
Like dissolves like
A ____________ solvent such as water is
needed to dissolve polar solutes such as
sugar and ionic solutes such as NaCl.
A ___________solvent such as hexane
(C6H14) is needed to dissolve nonpolar
solutes such as oil or grease.
20
Learning Check SF4
Which of the following solutes will
dissolve in water? Why?
1) Na2SO4
2) gasoline
3) I2
4) HCl
21
Solution SF4
Which of the following solutes will
dissolve in water? Why?
1) Na2SO4
Yes, polar (ionic)
2) gasoline
No, nonnpolar
3) I2
No, nonpolar
4) HCl
Yes, Polar
22
Formation of a Solution
H2O
Cl- Na+
Na+ Clsolute
Na+
H2O Cl-
Hydration
Dissolved
solute
23
Electrolyte and Non-electrolyte
• Electrolyte: a substance that conducts
electricity when dissolved in water.
– Acids, bases and soluble ionic solutions are
electrolytes.
• Non-electrolyte: a substance that does
not conduct electricity when dissolved in
water.
– Molecular compounds and insoluble ionic
compounds are non-electrolytes.
24
Electrolytes
• Some solutes can
dissociate into ions.
• Electric charge can
be carried.
25
Types of solutes
high conductivity
Strong Electrolyte 100% dissociation,
all ions in solution
Na+
Cl26
Types of solutes
slight conductivity
Weak Electrolyte partial dissociation,
molecules and ions in
solution
CH3COOH
H+
CH3COO27
Types of solutes
no conductivity
Non-electrolyte No dissociation,
all molecules in
solution
Sugar
C6H12O6
28
Types of Electrolytes
• Strong electrolyte dissociates
completely.
– Good electrical conduction.
• Weak electrolyte partially
dissociates.
– Fair conductor of electricity.
• Non-electrolyte does not
dissociate.
– Poor conductor of electricity.
29
Representation of Electrolytes
using Chemical Equations
A strong electrolyte:
MgCl2(s) → Mg2+(aq) + 2 Cl- (aq)
A weak electrolyte:
-(aq) +H+(aq)
CH3COOH(aq) →
CH
COO
← 3
A non-electrolyte:
CH3OH(aq)
30
Electrolytes in Action
31
Strong Electrolytes
Strong acids: HNO3, H2SO4, HCl, HClO4
Strong bases: MOH (M = Na, K, Cs, Rb etc)
Salts: All salts dissolving in water are completely ionized.
Stoichiometry & concentration
relationship
NaCl (s)  Na+ (aq) + Cl– (aq)
Ca(OH)2 (s)  Ca2+(aq) + 2 OH– (aq)
AlCl3 (s)  Al3+ (aq) + 3 Cl– (aq)
(NH4)2SO4 (s)  2 NH4 + (aq) + SO42– (aq)
32
Writing An Equation for a
Solution
When NaCl(s) dissolves in water, the
reaction can be written as
H 2O
NaCl(s)
solid
Na+ (aq) + Cl- (aq)
separation of ions in water
33
Learning Check SF5
Solid LiCl is added to some water. It
dissolves because
A. The Li+ ions are attracted to the
1) oxygen atom(-) of water
2) hydrogen atom(+) of water
B. The Cl- ions are attracted to the
1) oxygen atom(-) of water
2) hydrogen atom(+) of water
34
Solution SF5
Solid LiCl is added to some water. It
dissolves because
A. The Li+ ions are attracted to the
1) oxygen atom(-) of water
B. The Cl- ions are attracted to the
2) hydrogen atom(+) of water
35
Rate of Solution
You are making a chicken broth using a
bouillon cube. What are some things you
can do to make it dissolve faster?

Crush it

Use hot water (increase temperature)

Stir it
36
How do I get
sugar to dissolve
faster in my
iced tea?
Stir, and stir, and stir
Fresh solvent contact and interaction with solute
Add sugar to warm tea then add ice
Faster rate of dissolution at higher temperature
Grind the sugar to a powder
Greater surface area, more solute-solvent
interaction
37
Learning Check SF6
You need to dissolve some gelatin in
water. Indicate the effect of each of the
following on the rate at which the gelatin
dissolves as (1) increase, (2) decrease,
(3) no change
A. ___Heating the water
B. ___Using large pieces of gelatin
C. ___Stirring the solution
38
Learning Check SF6
You need to dissolve some gelatin in
water. Indicate the effect of each of the
following on the rate at which the gelatin
dissolves as (1) increase, (2) decrease,
(3) no change
A. 1 Heating the water
B. 2 Using large pieces of gelatin
C. 2 Stirring the solution
39
Solubility
Percent Concentration
Colloids and Suspensions
40
Solubility
The maximum amount of solute that
can dissolve in a specific amount of
solvent usually 100 g.
g of solute
100 g water
41
Saturated and Unsaturated
 A saturated solution contains the maximum
amount of solute that can dissolve.
Undissolved solute remains.
 An unsaturated solution does not contain all
the solute that could dissolve
42
Solubility
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
increasing concentration
Factors Affecting Solid Solubility
Polarity
Temperature
Surface Area
Stirring
Factors Affecting Solubility
Polarity
Temperature
Pressure
Intramolecular Bonding
• Intramolecular bonding refers to the chemical
bonding that holds atoms together within a
molecule of a compound
Covalent bonding and ionic bonding are the
two main types of intramolecular bonding
 Covalent bonding involves the sharing
of valence electrons involves the sharing
of valence electrons between two atoms.
POLAR- unequal sharing of electrons
NON POLAR – equal sharing of electrons
 Ionic bonding involves the transference
of valence electrons
SOLUTE
POLAR
SOLVENT
NONPOLAR
SOLVENT
Ionic
Soluble
Insoluble
Polar
Soluble
Insoluble
Nonpolar
Insoluble
soluble
47
Learning Check S1
At 40C, the solubility of KBr is 80 g/100 g
H2O. Indicate if the following solutions are
(1) saturated or (2) unsaturated
A. ___60 g KBr in 100 g of water at 40C
B. ___200 g KBr in 200 g of water at 40C
C. ___25 KBr in 50 g of water at 40C
48
Solution S1
At 40C, the solubility of KBr is 80 g/100 g
H2O. Indicate if the following solutions are
(1) saturated or (2) unsaturated
A. 2 Less than 80 g/100 g H2O
B. 1 Same as 100 g KBr in 100 g of water
at 40C, which is greater than its solubility
C. 2 Same as 60 g KBr in 100 g of water,
which is less than its solubility
49
Temperature and Solubility of
Solids
Temperature
0°
20°C
50°C
100°C
Solubility (g/100 g H2O)
KCl(s)
NaNO3(s)
27.6
74
34.0
88
42.6
114
57.6
182
The solubility of most solids (decreases or
increases ) with an increase in the
temperature.
50
Temperature and Solubility of
Solids
Temperature
0°
20°C
50°C
100°C
Solubility (g/100 g H2O)
KCl(s)
NaNO3(s)
27.6
74
34.0
88
42.6
114
57.6
182
The solubility of most solids increases with
an increase in the temperature.
51
Temperature and Solubility of
Gases
Temperature
0°C
20°C
50°C
Solubility (g/100 g H2O)
CO2(g)
O2(g)
0.34
0.17
0.076
0.0070
0.0043
0.0026
The solubility of gases (decreases or
increases) with an increase in
temperature.
52
Temperature and Solubility of
Gases
Temperature
0°C
20°C
50°C
Solubility (g/100 g H2O)
CO2(g)
O2(g)
0.34
0.17
0.076
0.0070
0.0043
0.0026
The solubility of gases decreases with an
increase in temperature.
53
Learning Check S2
A. Why would a bottle of carbonated drink
possibly burst (explode) when it is left out
in the hot sun ?
B. Why would fish die in water that gets too
warm?
54
Solution S2
A. Gas in the bottle builds up as the gas
becomes less soluble in water at high
temperatures, which may cause the bottle
to explode.
B. Because O2 gas is less soluble in warm
water, the fish may not obtain the needed
amount of O2 for their survival.
55
Gas Solubility
CH4
2.0
Solubility (mM)
O2
CO
1.0
He
0
10
20
30
40
50
56
Solubility Curves
Show the conditions that affect states of the
solution: unsaturated, saturated,
supersaturated.
57
Solubility vs. Temperature for Solids
140
KI
130
Solubility
Table
shows the dependence
of solubility on temperature
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
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517
10 20 30 40 50 60 70 80 90 100
How to determine the solubility of a given
substance?
• Find out the mass of solute needed to make a
saturated solution in 100 cm3 of water for a
specific temperature(referred to as the
solubility).
• This is repeated for each of the temperatures
from 0ºC to 100ºC. The data is then plotted on
a temperature/solubility graph, and the points
are connected. These connected points are
called a solubility curve.
How to use a solubility graph?
A.
IDENTIFYING A SUBSTANCE ( given
the solubility in g/100 cm3 of water and
the temperature)
• Look for the intersection of the
solubility and temperature.
Learning Check
SG1: What
substance has
a solubility of
90 g/100 cm3 in
water at a
temperature of
25ºC ?
Learning Check
SG2:
What substance has
a solubility of 200
g/100 cm3 of water
at a temperature of
90ºC ?
Look for the temperature or solubility
•Locate the solubility curve needed and
see for a given temperature, which
solubility it lines up with and visa versa.
Learning
Check SG3:
What is the
solubility of
potassium
nitrate at
80ºC ?
• What is the
solubility of
potassium
nitrate at
80ºC ?
Learning Check
SG4:
At what
temperature
will sodium
nitrate have a
solubility of 95
g/100 cm3 ?
Learning Check
SG4:
At what
temperature
will sodium
nitrate have a
solubility of 95
g/100 cm3 ?
Learning
Check SG5:
At what
temperature
will
potassium
iodide have a
solubility of
230 g/100
cm3 ?
Learning
Check SG5:
At what
temperature
will
potassium
iodide have a
solubility of
130 g/100
cm3 ?
Using Solubility
Curves:
What is the
solubility of
sodium chloride at
25ºC in 150 cm3
of water ?
From the
solubility graph
we see that
sodium chlorides
solubility is 36 g.
Place this in the proportion below and solve for
the unknown solubility. Solve for the unknown
quantity by cross multiplying.
Solubility in grams = unknown solubility in grams
100 cm3 of water
other volume of water
___36 grams____ = unknown solubility in grams
100 cm3 of water
150 cm3 water
The unknown solubility is 54 grams. You can use this
proportion to solve for the other volume of water
if you're given the other solubility.
C.
Determine if a solution is saturated,
unsaturated,or supersaturated.
• If the solubility for a given substance places
it anywhere on it's solubility curve line it is
saturated.
• If it lies above the solubility curve line, then
it's supersaturated,
• If it lies below the solubility curve line it's an
unsaturated solution. Remember though, if
the volume of water isn't 100 cm3 to use a
proportion first as shown above.
Solubility how much solute dissolves in a given amt.
of solvent at a given temp.
SOLUBILITY
CURVE
Solubility
(g/100 g
H2O)
KNO3 (s)
KCl (s)
HCl (g)
Temp.
(oC)
unsaturated: solution could hold more solute; below line
saturated: solution has “just right” amt. of solute; on line
supersaturated:solution has “too much” solute dissolved in it;
above the line
Solids dissolved in liquids
Sol.
Gases dissolved in liquids
Sol.
To
As To , solubility
To
As To , solubility
Sometimes you'll need to determine how much
additional solute needs to be added to a
unsaturated solution in order to make it
saturated.
For example,30 grams of potassium nitrate
has been added to 100 cm3 of water at a
temperature of 50ºC. How many additional
grams of solute must be added in order to
make it saturated?
How many
additional grams
of solute must
be added in
order to make it
saturated?
From the graph
you can see that
the solubility for
potassium
nitrate at 50ºC is
84 grams
If there are already 30 grams
of solute in the solution, all
you need to get to 84 grams is
54 more grams ( 84g-30g )
Solubility vs. Temperature for S
140
KI
130
Solubility
Table
shows the dependence
of solubility on temperature
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
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517
10 20 30 40 50 60 70 80 90 100
Solubility vs. Temperature f
140
KI
Classify as unsaturated,
saturated, or supersaturated.
per
100
g
H2O
45 g KCl @
60oC
=saturated
50 g NH3 @ 10oC
=unsaturated
70 g NH4Cl @
70oC
=supersaturated
120
Solubility (grams of solute/100 g H2O)
80 g NaNO3 @ 30oC
=unsaturated
130
110
NaNO3
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
Describe each situation below.
(A) Per 100 g H2O, 100 g
NaNO3 @ 50oC.
(B) Cool solution (A) very
slowly to 10oC.
(C) Quench solution (A) in
an ice bath to 10oC.
Unsaturated; all solute
dissolves; clear solution.
Supersaturated; extra
solute remains in solution;
still clear.
Saturated; extra solute
(20 g) can’t remain in
solution, becomes visible.
Soluble and Insoluble Salts
A soluble salt is an ionic compound that
dissolves in water.
An insoluble salt is an ionic compound that
does not dissolve in water
84
Aqueous Solutions
How do we know ions are
present in aqueous
solutions?
The solutions:_________
They are called
ELECTROLYTES
HCl, MgCl2, and NaCl are
strong electrolytes.
They dissociate
completely (or nearly so)
into ions.
85
Solubility Rules
1. A salt is soluble in water if it contains
any one of the following ions:
NH4+
Li+
Na+
K+ or
NO3-
Examples:
soluble salts
LiCl
Na2SO4
KBr
Ca(NO3)2
86
Cl- Salts
2. Salts with Cl- are soluble, but not if the
positive ion is Ag+, Pb2+, or Hg22+.
Examples:
soluble
not soluble(will not dissolve)
MgCl2
AgCl
PbCl2
87
SO42- Salts
3. Salts with SO42- are soluble, but not if
the positive ion is Ba2+, Pb2+, Hg2+ or
Ca2+.
Examples:
soluble
not soluble
MgSO4
BaSO4
PbSO4
88
Other Salts
4. Most salts containing CO32-, PO43-, S2and OH- are not soluble.
Examples:
soluble
not soluble
Na2CO3
CaCO3
K 2S
CuS
89
Learning Check S3
Indicate if each salt is (1)soluble or (2)not
soluble:
A. ______ Na2SO4
B. ______ MgCO3
C. ______ PbCl2
D. ______ MgCl2
90
Solution S3
Indicate if each salt is (1) soluble or
(2) not soluble:
A. _1_ Na2SO4
B. _2_ MgCO3
C. _2_ PbCl2
D. _1_ MgCl2
91
Solutions
Molarity
92
Molarity (M)
A concentration that expresses the
moles of solute in 1 L of solution
Molarity (M) =
moles of solute
1 liter solution
93
Units of Molarity
2.0 M HCl
6.0 M HCl
=
2.0 moles HCl
1 L HCl solution
=
6.0 moles HCl
1 L HCl solution
94
Molarity Calculation
NaOH is used to open stopped sinks, to treat
cellulose in the making of nylon, and to
remove potato peels commercially.
If 4.0 g NaOH are used to make 500. mL of
NaOH solution, what is the molarity (M) of the
solution?
95
Calculating Molarity
1) 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH
40.0 g NaOH
2) 500. mL x
1L_
1000 mL
3. 0.10 mole NaOH
0.500 L
= 0.500 L
= 0.20 mole NaOH
1L
= 0.20 M NaOH
96
Learning Check M1
A KOH solution with a volume of 400 mL
contains 2 mole KOH. What is the molarity
of the solution?
1) 8 M
2) 5 M
3) 2 M
97
Solution M1
A KOH solution with a volume of 400 mL
contains 2 moles of KOH. What is the
molarity of the solution?
2) 5 M
M = 2 mole KOH = 5 M
0.4 L
98
Learning Check M2
A glucose solution with a volume of 2.0 L
contains 72 g glucose (C6H12O6). If glucose
has a molar mass of 180. g/mole, what is
the molarity of the glucose solution?
1) 0.20 M
2) 5.0 M
3) 36 M
99
Solution M2
A glucose solution with a volume of 2.0 L
contains 72 g glucose (C6H12O6). If glucose
has a molar mass of 180. g/mole, what is
the molarity of the glucose solution?
1) 72 g
x
1 mole x
180. g
1
=
2.0 L
0.20 M
100
Molarity Conversion Factors
A solution is a 3.0 M NaOH.. Write the
molarity in the form of conversion factors.
3.0 moles NaOH and
1 L NaOH soln
1 L NaOH soln
3.0 moles NaOH
101
Learning Check M3
Stomach acid is a 0.10 M HCl solution. How
many moles of HCl are in 1500 mL of stomach
acid solution?
1) 15 moles HCl
2) 1.5 moles HCl
3) 0.15 moles HCl
102
Solution M3
3) 1500 mL x
1 L
=
1000 mL
1.5 L
1.5 L x 0.10 mole HCl = 0.15 mole HCl
1L
(Molarity factor)
103
Learning Check M4
How many grams of KCl are present in 2.5 L
of 0.50 M KCl?
1) 1.3 g
2) 5.0 g
3) 93 g
104
Solution M4
3)
2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl
1L
1 mole KCl
105
Learning Check M5
How many milliliters of stomach acid, which is
0.10 M HCl, contain 0.15 mole HCl?
1) 150 mL
2) 1500 mL
3) 5000 mL
106
Solution M5
2) 0.15 mole HCl x 1 L soln
x 1000 mL
0.10 mole HCl
1L
(Molarity inverted)
= 1500 mL HCl
107
Learning Check M6
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
108
Solution M6
2) 400. mL x 1 L
= 0.400 L
1000 mL
0.400 L x 3.0 mole NaOH x 40.0 g NaOH
1L
1 mole NaOH
(molar mass)
= 48 g NaOH
109
Solution
Percent Concentration
110
Percent Concentration
Describes the amount of solute
dissolved in 100 parts of solution
amount of solute
100 parts solution
111
Mass-Mass % Concentration
mass/mass % = g solute
x 100%
100 g solution
112
Mixing Solute and Solvent
+
Solute
4.0 g KCl
Solvent
46.0 g H2O
50.0 g KCl solution
113
Calculating Mass-Mass %
g of KCl
g of solvent
g of solution
=
=
=
4.0 g
46.0 g
50.0 g
%(m/m) = 4.0 g KCl (solute) x 100 = 8.0% KCl
50.0 g KCl solution
114
Learning Check PC1
A solution contains 15 g Na2CO3 and 235 g of
H2O? What is the mass % of the solution?
1) 15% (m/m) Na2CO3
2) 6.4% (m/m) Na2CO3
3) 6.0% (m/m) Na2CO3
115
Solution PC1
mass solute
=
15 g Na2CO3
mass solution
=
15 g + 235 g = 250 g
%(m/m) = 15 g Na2CO3
x
100
250 g solution
= 6.0% Na2CO3 solution
116
Mass-Volume %
mass/volume % = g solute
x 100%
100 mL solution
117
Learning Check PC2
An IV solution is prepared by
dissolving 25 g glucose
(C6H12O6) in water to make
500. mL solution. What is the
percent (m/v) of the glucose in
the IV solution?
1) 5.0%
2) 20.% 3) 50.%
118
Solution PC2
1) 5.0%
%(m/v)
=
25 g glucose x 100
500. mL solution
=
5.0 %(m/v) glucose solution
119
Writing Factors from %
 A physiological saline solution is a 0.85%
(m/v) NaCl solution.
 Two conversion factors can be written for
the % value.
0.85 g NaCl
and 100 mL NaCl soln
100 mL NaCl soln
0.85 g NaCl
120
% (m/m) Factors
Write the conversion factors for a 10 %(m/m)
NaOH solution
NaOH
and
NaOH soln
NaOH soln
NaOH
121
% (m/m) Factors
Write the conversion factors for a 10 %(m/m)
NaOH solution
10 g
100 g
NaOH
NaOH soln
and
100 g
10 g
NaOH soln
NaOH
122
Learning Check PC 3
Write two conversion factors for each of the
following solutions:
A. 8 %(m/v) NaOH
B. 12 %(v/v) ethyl alcohol
123
Solution PC 3
Write conversion factors for the following:
A. 8 %(m/v) NaOH
8 g NaOH and 100 mL
100 mL
8 g NaOH
B. 12 %(v/v) ethyl alcohol
12 mL alcohol and 100 mL
100 mL
12 mL alcohol
124
Using % Factors
How many grams of NaCl are needed to
prepare 250 g of a 10.0% (m/m) NaCl solution?
Complete data:
____________ g solution
____________% or (______/_100 g_) solution
____________ g solute
125
Clculation Using % Factors
250
10.0%
?
g solution
or (10.0 g/100 g) solution
g solute
250 g NaCl soln x 10.0 g NaCl
= 25 g NaCl
100 g NaCl soln
126
Learning Check PC4
How many grams of NaOH do you need to
measure out to prepare 2.0 L of a 12%(m/v)
NaOH solution?
1) 24 g NaOH
2) 240 g NaOH
3) 2400 g NaOH
127
Solution PC4
2.0 L soln x 1000 mL = 2000 mL soln
1L
12 % (m/v) NaOH =
12 g NaOH
100 mL NaOH soln
2000 mL x 12 g NaOH
=
100 mL NaOH soln
240 g NaOH
128
Learning Check PC5
How many milliliters of 5 % (m/v) glucose
solution are given if a patient receives 150 g
of glucose?
1) 30 mL
2) 3000 mL
3) 7500 mL
129
Solution PC5
5% m/v factor
150 g glucose x
100 mL = 3000 mL
5 g glucose
130
Preparing a Solution by Dilution
131
Units of Concentrations
amount of solute per amount of solvent or solution
Percent (by mass) =
Molarity (M) =
g solute
g solution
x 100 =
g solute x 100
g solute + g solvent
moles of solute
volume in liters of solution
moles = M x VL
132
Solutions
Colloids and Suspensions
Osmosis and Dialysis
133
Solutions
Have small particles (ions or molecules)
Are transparent
Do not separate
Cannot be filtered
Do not scatter light.
134
Colloids
 Have medium size particles
 Cannot be filtered
 Separated with semipermeable membranes
 Scatter light (Tyndall effect)
135
Examples of Colloids

Fog

Whipped cream

Milk

Cheese

Blood plasma

Pearls
136
Suspensions
 Have very large particles
 Settle out
 Can be filtered
 Must stir to stay suspended
137
Examples of Suspensions
 Blood platelets
 Muddy water
 Calamine lotion
138
Osmosis
 In osmosis, the solvent water moves
through a semipermeable membrane
 Water flows from the side with the lower
solute concentration into the side with the
higher solute concentration
 Eventually, the concentrations of the two
solutions become equal.
139
Osmosis
4% starch
10% starch
H2O
semipermeable membrane
140
Equilibrium is reached.
7% starch
7% starch
H2OO
water flow becomes equal
141
Osmotic Pressure
 Produced by the number of solute particles
dissolved in a solution
 Equal to the pressure that would prevent
the flow of additional water into the more
concentrated solution
 Increases as the number of dissolved
particles increase
142
Osmotic Pressure of the Blood
 Cell walls are semipermeable membranes
 The osmotic pressure of blood cells
cannot change or damage occurs.
 The flow of water between a red blood
cell and its surrounding environment
must be equal
143
Isotonic solutions
•
Exert the same osmotic pressure as red
blood cells.
•
Medically 5% glucose and 0.9% NaCl are
used their solute concentrations provide
an osmotic pressure equal to that of red
blood cells
H2O
144
Hypotonic Solutions
 Lower osmotic pressure than red blood cells
 Lower concentration of particles than RBCs
 In a hypotonic solution, water flows into the
RBC
 The RBC undergoes hemolysis; it swells and
may burst.
H 2O
145
Hypertonic Solutions
 Has higher osmotic pressure than RBC
 Has a higher particle concentration
 In hypertonic solutions, water flows out
of the RBC
 The RBC shrinks in size (crenation)
H2O
146
Dialysis
 Occurs when solvent and small solute
particles pass through a semipermeable
membrane
 Large particles retained inside
 Hemodialysis is used medically (artificial
kidney) to remove waste particles such as
urea from blood
147
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
148
on the KIND of solute particles.
Change in Freezing Point
Pure water
Ethylene glycol/water
solution
The freezing point of a solution is
LOWER than that of the pure solvent
149
Change in Freezing Point
Common
Applications of
Freezing Point
Depression
Propylene glycol
Ethylene
glycol –
deadly to
small
animals
150
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
151
Change in Boiling Point
Common Applications of
Boiling Point Elevation
152