Announcements & Agenda (02/16/07)
Please pick up exams if you haven’t.
Mon: Movie in VDW 102– You must attend &
complete a worksheet!
Today
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Solution Concentrations (7.4, 7.5)
Solution Properties & Some Other Mixtures (7.7)
Osmosis (7.7)
1
30 Good Practice Problems (Ch 7)
7.01, 7.03, 7.11, 7.13, 7.15, 7.17,
7.25, 7.27, 7.29, 7.31, 7.35, 7.37,
7.41, 7.45, 7.49, 7.53, 7.55, 7.61,
7.63, 7.65, 7.69, 7.71, 7.75, 7.77,
7.83, 7.87, 7.89, 7.93, 7.97, 7.99
2
Last Time: Nature of Solutions
Solution – homogeneous mixture of two or
more substances
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Solvent – the main medium, present in the
largest quantity
Solute – material dissolved in the solvent
They don’t just have to be liquids:
3
Last Time: “Like Dissolves Like”
What does this mean?
Answer: compatibility of intermolecular
forces (last week)
Typically, compounds with similar polarity
will be soluble in each other.
Predicting trends based on properties…
4
Which of the following vitamins are fat-soluble
and which are water soluble?
Vitamins: organic
molecules required for
proper function but not
made by the body.
(Obtain by eating!)
2 categories:
fat-soluble (dissolve in
fatty hydrocarbon-like
tissues)
water-soluble
Why we need at least some oil & fat in our diet!
5
Last Time: Solutes & Ionic Charge
In water,
• strong electrolytes produce ions and conduct an
electric current.
• weak electrolytes produce a few ions.
• nonelectrolytes do not produce ions.
6
Solubility & Concentration (7.3-7.4)
• the maximum amount of solute that
dissolves in a specific amount of solvent.
• expressed as grams of solute in 100
grams of solvent water.
g of solute
100 g water
Unsaturated vs. Saturated Solutions
7
Learning Check
At 40C, the solubility of KBr is 80 g/100 g H2O.
Identify the following solutions as either
1) saturated or 2) unsaturated. Explain.
A. 60 g KBr added to 100 g of water at 40C.
B. 200 g KBr added to 200 g of water at 40C.
C. 25 g KBr added to 50 g of water at 40C.
What happens to the “leftover” solute???
8
Effect of Temperature on Solubility
• Depends on Temp!
• Solids: usually
increases as
temperature inc.
• Gases: usually
decreases as
temperature inc.
9
Solubility and Pressure
Henry’s Law:
• Gas solubility is
directly related to
gas pressure above
the liquid
• at higher pressures,
more gas
molecules dissolve
in the liquid.
Real life examples… soda, the bends, etc.
10
Percent Concentration (7.4)
The amount of solute dissolved in a specific
amount of solution.
amount of solute
amount of solution
Comes in all sorts of fantastic flavors!
•
Mass Percent
•
Volume Percent
•
Mass/Volume Percent
•
Molarity
11
Mass Percent (% m/m)
• concentration is the percent by mass of
solute in a solution.
mass percent =
g of solute
x 100
g of solute + g of solvent
• amount in g of solute in 100 g of solution.
mass percent =
g of solute
x 100
100 g of solution
12
Mass of Solution
8.00 g KCl
Add water to
give 50.00 g
solution
50.00 g KCl
solution
13
Calculating Mass Percent
Calculation of mass percent (% m/m) requires the
• grams of solute (g KCl) and
• grams of solution (g KCl solution).
g of KCl
=
8.00 g
g of solvent (water)
=
42.00 g
g of KCl solution
=
50.00 g
8.00 g KCl (solute)
x 100 = 16.0% (m/m)
50.00 g KCl solution
14
Volume Percent (% v/v)
• percent volume (mL) of solute (liquid) to volume
(mL) of solution.
volume % (v/v) =
mL of solute x 100
mL of solution
• solute (mL) in 100 mL of solution.
volume % (v/v) = mL of solute
100 mL of solution
15
Mass/Volume Percent (% m/v)
• percent mass (g) of solute to volume (mL) of
solution.
mass/volume % (m/v) = g of solute x 100
mL of solution
• solute (g) in 100 mL of solution.
mass/volume % (m/v) =
g of solute x 100
100 mL of solution
16
Percent Conversion Factors
Two conversion factors can be written for
each type of % value.
TABLE 7.7
5% (m/v) glucose
There are 5 g of glucose
in 100 mL of solution.
5 g glucose
100 mL solution
and
100 mL solution
5 g glucose
17
How many grams of NaOH are needed to
prepare 75.0 g of 14.0% (m/m) NaOH solution?
1) 10.5 g NaOH
2) 75.0 g NaOH
3) 536 g NaOH
0%
0%
0%
1
2
3
4
5
18
Solution
1)10.5 g NaOH
75.0 g solution x 14.0 g NaOH = 10.5 g NaOH
100 g solution
14.0% (m/m) factor
19
How many milliliters of a 5.75% (v/v) ethanol
solution can be prepared from 2.25 mL ethanol?
0%
1) 2.56 mL
2) 12.9 mL
0%
3) 39.1 mL
0%
1
2
3
4
5
20
Solution
3) 39.1 mL
2.25 mL ethanol x 100 mL solution
5.75 mL ethanol
5.75% (v/v) inverted
= 39.1 mL solution
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Molarity and Dilution (7.5)
Molarity (M) is
• a chemist’s concentration term for solutions.
• gives the moles of solute in 1 L solution.
• moles of solute
liter of solution
22
Preparing a 1.0 Molar Solution
A 1.00 M NaCl solution is prepared
• by weighing out 58.5 g NaCl (1.00 mole)
and
• adding water to make 1.00 liter of solution.
23
Dilution
In a dilution
• water is added.
• volume increases.
• concentration decreases.
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Comparing Initial & Diluted Solutions
In the initial and diluted solution,
• the amount of solute is the same.
• the concentrations and volumes are related by
the following equations:
For percent concentration:
C1V1
= C2V2
initial
diluted
For molarity:
M1V1
= M2V2
initial
diluted
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Summary: Properties of Solutions
• contain small particles (ions or molecules).
• are transparent.
• do not separate*.
• cannot be filtered.
• do not scatter light.
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Colloids
• have medium size particles.
• cannot be filtered.
• can be separated by semipermeable
membranes.
• scatter light (Tyndall effect).
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Examples of Colloids
• Fog
• Whipped cream
• Milk
• Cheese
• Blood plasma
• Pearls
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Suspensions
• have very large particles.
• settle out.
• can be filtered.
• must be stirred to stay suspended.
Examples include blood platelets, muddy
water, and lotions.
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Osmosis
• water (solvent) flows
from the lower solute
concentration into the
higher solute
concentration.
• the level of the solution
with the higher
concentration rises.
• the concentrations of
the two solutions
become equal with
time.
30
Osmosis
Suppose a semipermeable membrane separates a 4%
starch solution from a 10% starch solution. Starch is a
colloid and cannot pass through the membrane, but
water can. What happens?
semi-permeable
membrane
4% starch
10% starch
H2O
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Water Flow Equalizes
• The 10% starch solution is diluted by the flow of
water out of the 4% and its volume increases.
• The 4% solution loses water and its volume
decreases.
• Eventually, the water flow between the two becomes
equal.
7% starch
7% starch
H2O
32
Osmotic Pressure
• produced by the solute particles dissolved in
a solution.
• equal to the pressure that would prevent the
flow of additional water into the more
concentrated solution.
• greater as the number of dissolved particles
in the solution increases.
33
Osmotic Pressure of the Blood
Red blood cells
• have cell walls that are semipermeable
membranes.
• maintain an osmotic pressure that cannot
change or damage occurs.
• must maintain an equal flow of water
between the red blood cell and its
surrounding environment.
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Isotonic Solutions
• exerts the same osmotic
pressure as red blood
cells.
• is known as a
“physiological solution”.
• of 5.0% glucose or
0.90% NaCl is used
medically because each
has a solute
concentration equal to
the osmotic pressure
equal to red blood cells.
H 2O
35
Hypotonic Solutions
• has a lower osmotic
pressure than red
blood cells.
• has a lower
concentration than
physiological solutions.
• causes water to flow
H2O
into red blood cells.
• causes hemolysis:
RBCs swell and may
burst.
36
Hypertonic Solutions
• has a higher osmotic
pressure than RBCs.
• has a higher
concentration than
physiological
solutions.
• causes water to flow
out of RBCs.
• cause crenation:
RBCs shrinks in
size.
H 2O
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Dialysis
In dialysis,
• solvent and small solute particles pass
through an artificial membrane.
• large particles are retained inside.
• waste particles such as urea from blood
are removed using hemodialysis (artificial
kidney).
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