Solutes and
Solubility
Mixture
• Mixtures: made from physically mixing two or
more substances together without a chemical
reaction occurring.
• Mixing ionic compounds with water forms
aqueous solutions of dissolved ions.
• The polar water molecules attract to the ions,
tearing them apart from other ions and
holding them away from other ions.
• This is called molecule-ion attraction.
+
Na
+
Na
O2
H
+
+
H
+
Cl-
Na
Cl-
Na
+
+
Cl-
Cl-
Na
Cl-
Na
Cl-
Na
Na
Cl-
Na
Cl-
Na
Cl-
+
+
+
+
+
+
Cl-
Na+
Na+
Cl-
Cl-
Na+
Na+
Na+
Cl-
Na+
Na+
Cl-
Cl-
Na+
Cl-
Na+
Cl-
Na+
Cl-
Na+
Cl-
Na+
Cl-
Na+
Cl-
Na+
Cl-
Na+
Cl-
Na+
Na+
Cl-
Na+
Cl-
Na+
Na+
Cl-
Na+
Cl-
Na+
+
Cl-
Cl-
Na
Na+
Cl-
+
Cl
Na
Cl
Cl-
Partially positive ends of
the water molecules
(hydrogen) attracts to
the negatively charged Na+
chloride ions.
MOLECULE – ION
ATTRACTION
Na+
Partially negative
ends of
molecules
(oxygen) attracts
to the positively
charged sodium Clions.
Solubility
• The quantity of solute that can be added to a
quantity of solvent to make a saturated
solution at a given temperature and pressure.
Solubility Curve
• Solubility Curve – a graphical representation of
the amount of substance that can dissolve into
100 g of water at a specific temperature (Celsius)
Substances:
Compound being
dissolved in
water (H2O)
Y-axis:
Solubility of
substance
(g/100 g H2O)
X-axis:
Temperature
(Celsius)
Interpreting a Solubility Curve
• Each point on the solubility curve shows how many
grams can be dissolved at a specific temperature:
Each line shows
how much
substance can
dissolve as a
function of the
temperature of
the solution.
Using a Solubility Curve
How many grams of potassium bromide (KBr) can dissolve in
100 grams of water at 20°C?
70g
Answer: 70
grams of
KBr can
dissolve in
100g of
water at
20°C
Practice Using Solubility Curve
How many grams of potassium nitrate (KNO3) can dissolve in
100 g of water at 60°C?
130g
Answer: 130 g
of KNO3 can
dissolve in 100
g of H2O
Solution Concentration
• Saturated: The solution
holds as many dissolved
particles as it can
possibly hold
• Unsaturated: The
solution holds fewer
solute particles than can
theoretically be
dissolved, can add more
solute.
• Supersaturated: A very rare situation where the solution
holds more solute than is theoretically possible, unstable
situation where the excess will precipitate if the solution is
agitated
Saturated / Unsaturated / Supersaturated
• Saturated: solute = solubility
• Unsaturated: solute < solubility
• Supersaturated: solute > solubility
Saturated
NaNO3
Unsaturated
Supersaturated
Factors Affecting Solubility
1) Temperature
•
•
For solid and liquid solutes, solubility in water
increases as temperature increases
For gaseous solutes, solubility in water decreases
as temperature increases
2) Pressure
•
•
For gaseous solutes, solubility increases as
pressure increases
Pressure does not affect solid or liquid solutes
Factors Affecting Solubility
3) Nature of Solute and Solvent
(Like Dissolves Like)
• Polar solutes dissolve in polar solvents
• Nonpolar solutes dissolve in nonpolar solvents
Checking for understanding
1. Explain the difference between
saturated, unsaturated, and supersaturated
solutions
2. Explain how pressure and temperature affect
solubility
Concentration
Concentration and molarity
• a measure of the amount of solute dissolved in
a given quantity of solvent
• A concentrated solution has a large amount of
solute
• A dilute solution has a small amount of solute
1. Grams of solute/100 mL of solvent
Represents
Grams of solute
100mL of solvent
2. Parts Per Million (ppm)
• Number of grams of solute in
1 million gram of solvent
• One unit of concentration
used in pollution
measurements that involve
very low concentration
g solute
g solvent
X 1,000,000 = ppm
• A chemical analysis shows that there are 2.2
mg of lead in exactly 500 g of water. Convert
this measurement to parts per million.
mass of solute: 2.2 mg
mass of solvent: 500 g
parts per million = ?
2.2 mg
1g
1000 mg
2.2 x 10-3 g
500 g
g solute
g solvent
= 2.2 x 10-3 g
X 1,000,000
= 4.4 ppm
X 1,000,000 = ppm
• Helium gas, 3.0 x 10-4 g is dissolved in 200.0g
of water. Express this concentration in ppm.
mass of solute: 3.0 x 10-4 g
mass of solvent: 200.0 g
parts per million = ?
g solute
g solvent
X 1,000,000 = ppm
3.0 x 10-4 g
X 1,000,000 = 1.5ppm
200 g
3. % by mass
g of solute
% mass =
x 100 %
g of solution
4. % by volume
mL of solute
% volume =
x 100 %
mL of solution
• Percent by volume is often used to describe the
concentration of alcohol in alcoholic beverages
or in medications containing alcohol.
What is the percentage by mass of a solution
made by dissolving 0.49 g of potassium sulfate
in 12.70 g of water?
g of solute
% mass =
g of solution
= 0.49 g of K2SO4
12.70g water + 0.49 g K2SO4
x 100 %
=3.7% K2SO4 by mass
=
0.49 K2SO4
12.70g water
• A 50.0 mL sample of an aqueous ethanol
solution is distilled to yield 33.2 mL of ethanol.
What is the percent by volume of ethanol in
this solution?
mL of solute
x 100 %
% volume=
mL of solution
=
33.2 mL of ethanol
50.0 mL solution
x 100 %
=66.4% ethanol by volume
5. Molarity (M)
• Molarity (M) is a concentration unit of a
solution expressed as moles of solute
dissolved per liter of solution.
M
Moles of solute
= Liters of solution
•
What is the molarity of a potassium chloride
solution that has a volume of 400.0 mL and
contains 85.0 g KCl?
solute = 85 g KCl
1 mol KCl
74.55 g KCl
solution = 400 mL
1L
= 1.14 mol KCl
= 0.4 L
1000 mL
M=
Moles
L
M=
1.14 mol
0.4 L
= 2.85 M
• An aqueous solution of sodium carbonate, Na2CO3,
contains 53 g of solute in 215 mL of solution. What is it
concentration (M)?
solute = 53 g Na2CO3 1 mol Na2CO3
105.99 g Na2CO3
solution = 215 mL
1L
= 0.5 mol Na2CO3
= 0.215 L
1000 mL
M=
Moles
L
M=
0.5 mol
0.215 L
= 2.3 M
• How many moles of sugar are dissolved in 202 mL of a
0.150 M solution?
solution= 202ml
1L
1000ml
Molarity = 0.150 M
M=
Moles
L
mol = M · L = (0.150 mol/L)(0.202 L)
= 0.0303 mol C12H22O11
= 0.202 L
•
A mass of 98 g of sulfuric acid, H2SO4, is
dissolved in water to prepare a 0.50 M solution.
What is the volume of the solution in liters ?
solute = 98 g H2SO4 1 mol H2SO4
98.08 g H2SO4
= 1.0 mol H2SO4
solute = 1.0mol H2SO4
Molarity = 0.50 M
M=
Moles
L
mol = M · L
L=
Moles
M
=
1.0mol
0.50 M
= 2.0 L H2SO4
5. Molality (m or molal)
• Solute in moles and the mass of solvent in
kilograms;
m=
For
water
only!
Moles of solute
Kg of solvent
Density of water = 1 g/ 1 mL
1 mL = 1 g
1 L = 1 Kg
1000 g = 1 Kg
What is the molality of a solution with 9.3 mole
of NaCl in 450 g of water?
mols of solute = 9.3 mol
solution= 450g
1Kg
1000g
m=
m=
Moles
kg
9.3 mol
0.45 Kg
= 21 m
= 0.45kg
• Determine the molality of a solution of
560. g of acetone, CH3COCH3, in 620 g of
water.
solute = 560g CH3COCH3 1 mol
solution= 620g
= 9.6 mol CH3COCH3
58.09g
1Kg
= 0.62kg
1000g
m=
Moles
kg
=
9.6 mol
= 15.5 m
0.62 Kg
Checking for understanding
Formula
g of solute/100mL of
solvent
Parts per million
% by mass
% by volume
Molarity
Molality
Resulting units
Dilution
Concentration of Solution
• Dilution is the process of preparing a less
concentrated solution from a more concentrated one.
moles of solute before dilution = moles of solute after dilution
MiVi = MfVf
remove
sample
moles of
solute
initial solution
mix
Making a Dilute
Solution
diluted solution
Timberlake, Chemistry 7th Edition, page 344
same number of
moles of solute
in a larger volume
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
Mi = 4.00 mol/L Mf = 0.200mol/L
Vf = 0.06 L
Vi = ? L
MiVi = MfVf
Vi =
MfVf
Mi
(0.200mol/L ) ( 0.06L)
= 0.003 L = 3 mL HNO3
=
(4.00 mol/L)
3 mL of HNO3 + 57 mL of water
= 60 mL of solution
How to mix a Standard Solution
Wash bottle
Volume marker
(calibration mark)
Weighed
amount
of solute
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480
Process of Making a Standard Solution
from Liquids
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 483
In an experiment, a student needs 250.0 mL of a 0.100 M
CuCl2 solution. A stock solution of 2.00 M CuCl2 is
available.
How much of the stock solution is needed?
Mi = 2.00 mol/L Mf = 0.100mol/L
Vf = 250.0mL Vi = ? L
MiVi = MfVf
Vi =
MfVf
Mi
(0.100mol/L ) ( 2.50L)
= 0.0125 L
=
(2.00 mol/L)
= 12.5 mL CuCl2
To make the solution:
1) Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask.
2) Carefully dilute with water to the calibration mark.
Preparing Solutions
How to prepare 500 mL of
1.54 M NaCl solution
– mass 45.0 g of NaCl
– add water until total
volume is 500 mL
45.0 g NaCl
solute
500 mL
mark
500 mL
volumetric
flask
Preparing Solutions
molarity
molality
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
Colligative
Properties
Colligative Properties
• A property that depends only upon the
number of solute particles (concentration),
and not upon their identity.
• Three Important Colligative Properties of
Solutions.
– Vapor-pressure lowering
– Boiling-point elevation
– Freezing-point depression
Vapor-Pressure Lowering
• Vapor pressure: is the
pressure exerted by a
vapor that is in
dynamic equilibrium
with its liquid in a
closed system.
– A solution that contains
a solute that is
nonvolatile (not easily
vaporized) always has a
lower vapor pressure
that the pure solvent.
•
This is true because in a
solution, solute particles
reduce the number of free
solvent particles able to
escape the liquid.
Freezing-Point Depression
• Freezing-Point Depression: The difference in
temperature between the freezing point of a
solution and the freezing point of the pure
solvent (water).
– The presence of a solute in water disrupts the
formation of the orderly pattern of ice.
– Therefore more kinetic energy must be withdrawn
from a solution than from the pure solvent to
cause the solution to solidify.
– Solution containing solute causes decrease in
freezing point.
Freezing-Point Depression
Boiling-Point Elevation
• Boiling Point: The temperature
at which the vapor pressure of
the liquid phase equals
atmospheric pressure.
• Boiling-Point Elevation: The
difference in temperature
between the boiling point of a
solution and the boiling point of
the pure solvent.
Boiling-Point Elevation
• Because of the decrease in
vapor pressure, additional
kinetic energy must be
added to raise the vapor
pressure of the liquid
phase of the solution to
atmospheric pressure to
initiate boiling.
• The boiling point of a
solution is higher than the
boiling point of the pure
solvent.
1) The higher the concentration of solute is, the
higher the boiling point and the lower the
freezing point will be.
Which of the following solutions will boil at the
highest temperature?
a) 100 g NaCl in 1000 g of water
b) 100 g NaCl in 500 g water
c) 100 g NaCl in 250 g of water
d) 100 g NaCl in 125 g of water
The answer is D because it has the highest
concentration
2) The more particles that a solute ionizes into,
the higher the boiling point and the lower the
freezing point will be.
Which of the following solutions will boil at the
highest temperature?
a) 1 mole C6H12O6 in 500 g of water
b) 1 mole KBr in 500 g of water
c) 1 mole MgF2 in 500 g of water
d) 1 mole AlCl3 in 500 g of water
The answer is D because AlCl3 breaks into 4
particles, the most of any of the choices.
Electrolytes
vs.
Nonelectrolytes
Electrolyte
• Most ionic compounds and many acids dissolve
well in water.
• These are called electrolytes, because they cause
the solution to conduct electricity due to the
free-moving ions.
• They ionize 100% in water to yield ions in a
reaction that resembles a decomposition
reaction.
• The reaction is called dissociation, and it is a
physical change, not a chemical change.
• The more ions a solute breaks up into, the
higher the boiling point and the lower the
freezing point of the solution will be.
2 moles of
dissolved
ions total
Electrolytes
NaCl(s) Na+1(aq) + Cl-1(aq)
Free moving
ions will
conduct
electricity
Nonelectrolytes
• Substances formed from covalent bonding do not
dissolve into ions upon entering the water.
• These include polar molecules that dissolve, but
do not ionize.
• These include sugar (C6H12O6, C12H22O11),
antifreeze (CH2OHCH2OH ) and alcohol (C2H5OH).
• These have less impact on the melting and boiling
point of a solution than ionic compounds do,
because they do not break up any further.
Nonelectrolyte
C12H22O11 (s)  C12H22O11 (aq)
One mole of
sucrose
dissolves to
form one mole
of dissolved
sucrose.
No ions are
formed, so
no electricity
can be
conducted.
Checking for understanding
List at least 3 characteristics for each:
Electrolytes
Nonelectrolytes