Transcript Solutions

Solutions
Chapter 18
Recall…..
• Solutions are
– homogeneous mixtures
– that may be solid, liquid, or gas
Examples
Solution of oxygen and several other gases.
In this solution aquarium salt is the
solute, water is the solvent.
6.1
Solutions and Other Mixtures
• Suspension – a mixture that looks
uniform when stirred or shaken that
separates into different layers when it
is no longer agitated
• Colloid – a mixture of very tiny
particles of pure substances that are
dispersed in another substance but do
not settle out of the substance
• Emulsion – any mixture of immiscible
liquids in which the liquids are spread
throughout one another
Suspensions
• The pulp in orange
juice is spread
throughout the
mixture right after
the juice is shaken
– over time, the
pulp does not stay
mixed with the
water molecules
and settles to the
bottom of the
container.
Examples of suspensions
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•
Water and coffee
Oil and vinegar
Sand and water
Water and oil
Blood
Examples of colloids
• Egg whites
• Paint
• Whipped cream (dispersing gas
in a gas)
• Marshmallows (dispersing gas in
a solid)
• Fog (water dispersed in air)
• Smoke (solid particles dispersed
in air)
Emulsions
• Examples:
– Ice cream
– Milk
– Mayonnaise
– Emulsifying agents like soap and
detergent allow the formation of
colloidal dispersions between
liquids that do not ordinarily mix
Factors determining solubility
• Temperature
• Stirring (agitation)
• Surface area
Properties of Solubility
• Miscible – two liquids are miscible IF
they dissolve in each other
– EXAMPLE: Water and alcohol
• Immiscible – liquids that are
insoluble in each other
– EXAMPLE: Oil and vinegar
Factors affecting solubility
• Solubility of most solid substances
increases as temperature of solvent
increases
• Solubility of gases is greater in cold
water than hot water
The Dissolving Process
More solid will dissolve as
temperature increases
Less gas will dissolve as
temperature increases
Henry’s Law
• …at a given temperature the
solubility (S) of a gas in a liquid is
directly proportional to the pressure
(P) of the gas above the liquid
– (simply put, as pressure of gas above
the liquid increases, the solubility
increases; similarly, as pressure of the
gas decreases, the solubility of the gas
decreases)
Solubility Curve Graphs
What mass of sodium nitrate
will dissolve in 100cm3 of water at 50°C?
About 112g
How much oxygen will dissolve in 100g
water at 80C°?
0.5 mg
How much oxygen will dissolve in
100g water at 10C°? 1.3 mg
Henry’s Law - Equation
S1
S2
------ = ------
P1
P2
Solubility # 1 divided by pressure # 1 is equal to
Solubility # 2 divided by Pressure #2
Henry’s Law - Example
• If the solubility of a gas in water is 0.77 g/L at 3.5
atm of pressure what is its solubility (in g/L) at 1.0
atm of pressure? (The temperature is held
constant at 250C)
• 1. ANALYZE List the knowns and the unknown.
Knowns:
P1 = 3.5 atm
S1 = 0.77 g/L
P2 = 1.0 atm
Henry’s Law: S1/P1 = S2/P2
Unknown:
S2 = ? g/L
Example - continued
• 2. Calculate Solve for the unknown.
– Solving Henry’s law for S2 yields
S2 = S1 X P2
P1
Substituting the known values and
calculating yields
S2 = 0.77 g/L X 1.0 atm
= 0.22 g/L
3.5 atm
Example continued
• 3. Evaluate Does the result make sense?
– The pressure is reduced (from 3.5 atm to 1.0
atm, so the solubility of the gas should
decrease, which it is shown to do. Because
the new pressure is approximately one-third of
the original pressure, the new solubility should
be approximately one-third of the original,
which it is. The answer is correctly expressed
to two significant figures.
Concentrations of Solutions
• Concentration of a solution is a
measure of the amount of solute that
is dissolved in a given quantity of
solvent
• A dilute solution is one that contains
only a low concentration of solute
• A concentrated solution contains a
high concentration of solute
Molarity (M)
• THE most important unit of
concentration in chemistry……
• Molarity (M) is the number of moles
of a solute dissolved per liter of
solution
Molarity
• Molarity = moles of solute or M = mol
•
liters of solution
L
For example: to calculate the molarity
when 2 mol of glucose is dissolved in
5 L of solution, divide the number
of moles by the volume in liters.
2 mol glucose
5 L of solution =
0.4 mol/L = 0.4 M
mole
Molarity Calculations
• Rearrange the equation:
Moles of solute = molarity (M) X liters of solution (V)
Moles of solute = M1 X V1
M1 X V1 = M2 X V2
V1 = M2 X V2 / M1
Colligative Properties
• Colligative Properties – properties
that depend only on the number of
particles dissolved in a given mass
of solvent. Three important ones:
– Vapor-pressure lowering
– Boiling-point elevation
– Freezing-point depression
Colligative properties
• Vapor pressure – the pressure exerted by
a vapor that is in dynamic equilibrium with
its liquid in a closed system
• Boiling-point Elevation – the difference in
temperature between the boiling point of a
solution and that of the pure solvent
• Freezing-point depression – the difference
in temperature between the freezing point
of a solution and that of the pure solvent
Calculations involving Colligative
Properties
• Molality (m) – is the number of moles of solute
dissolved per kilogram (1000g) of solvent. (Also
known as molal concentration)
• Molality = moles of solute
kilogram of solvent
moles of solute
kilogram of solvent
=
moles of solute
1000 g of solvent
Percent composition
• Percent by volume (% v/v)
=
Volume of solute
Volume of solution
X 100
Percent (mass/volume) (% m/v) =
mass of solute (g)
volume of solution (mL)
X 100
Assignment
• Complete calculations….
– Practice Problems # 1,2 pg 507
– Practice Problems # 8,9,10,11 pg 511
– Practice Problems # 12,13 pg 513
– Practice Problems # 14,15 pg 514
– Practice Problems # 16,17 pg 515
– Practice Problems # 28,29 pg 521
Problems page 507
• 1. The solubility of a gas in water is
0.16g/L at 104 kPa of pressure. What is the
solubility when the pressure of the gas is
increased to 288 kPa? (Assume the
temperature is constant)
• 2. A gas has a solubility in water at 0 C of
3.6 g/L at a pressure of 1.0 atm. What
pressure is needed to produce an
aqueous solution containing 9.5 g/L of the
same gas at 0 C?
Problems page 511
• 8. A solution has a volume of 2.0 L
and contains 36.0 g of glucose. If the
molar mass of glucose is 180 g/mol,
what is the molarity of the solution?
• 9. A solution has a volume of 250and
contains 0.70 mol NaCl. What is its
molarity?
Problems page 511
• 10. How many moles of ammonium
nitrate are in 335 mL of 0.425M
NH4NO3?
• 11. How many moles of solute are in
250 mL of 2.0 M CaCl2? How many
grams of CaCl2 is this?
Problems page 513
• 12. How many milliliters of a stock
solution of 4.00M KI would you need to
prepare 250.0 mL of 0.760M KI?
•
• 13. Suppose you need 250 mL of 0.20M
NaCl, but the only supply of sodium
chloride you have is a solution of 1.0M
NaCl. How do you prepare the required
solution? Assume that you have the
appropriate volume-measuring devices on
hand.
Problems page 514
• 14. If 10 mL of pure acetone is diluted
with water to a total solution volume
of 200 mL, what is the percent by
volume of acetone of this solution?
• 15. A bottle of hydrogen peroxide
antiseptic is labeled 3.0% (v/v). How
many mL H2O2 are in a 400.0 mL
bottle of this solution?
Problems page 515
• 16. Calculate the grams of solute
required to make 250 mL of 0.10%
MgSO4 (m/v).
• 17. A solution contains 2.7g CuSO4 in
75 mL of solution. What is the
percent (mass/volume) of the
solution?
Problems page 521
• 28. How many grams of sodium
fluoride are needed to prepare a
0.400m NaCl solution that contains
750.0g of water?
• 29. Calculate the molality of a
solution prepared by dissolving
10.0g NaCl in 600g of water.