Chapter 12: Solutions I

1

• Chapter 12 Solutions I
• Solutions
• 1. A solution is a __________________ mixture (1
  phase). Fig. 1 pg. 401
• 2. A solution is composed of the dissolving
  substance (_____________) and the dissolved
  substance (___________). See fig. 2 and Table 1
  pg. 402 for some solute/solvent combinations.
• 3. The process of a solute dissolving in water is
  called dissolution.
• Note Alloys (Metal solutions) a mixture of ___
  or more metals to obtain desirable properties
  such as strength and resistance to corrosion.
• Example Carat gold pure gold is ___ carat. 14
  carat is (14/24)100 58 gold. Fig. 3 pg. 403
• Suspensions
• 1. A suspension has large particles that
  _________ out unless the mixture is agitated or
  stirred. Example muddy water.
• 2. Gravity pulls on these larger particles to
  cause them to settle and these particles can be
  separated from the mixture by ________________.
• Colloids
• 1. Mixtures with particles sizes ___________
  solutions and suspensions. Table 2 pg. 403
• 2. Many colloids behave as solutions but the
  particles are large enough to _________ light,
  known as the Tyndall Effect. See fig. 4 pg. 403

homogeneous
solvent
solute
2
24
settle
filtration
between
scatter

• Electrolytes vs. Nonelectrolytes
• When a solute dissolved in water ______________
  electricity it is called an electrolyte (if the
  dissolved solute does not conduct electricity it
  is called a nonelectrolyte).
• Soluble ionic compounds (like NaCl) and strong
  acids (like HCl) are __________ electrolytes
  (most soluble covalent compounds are
  nonelectroytes (like sugar).

conducts
strong
2

• Solubility
• 1. Solubility is a measure of the __________ of
  solute that will dissolve in a given amount of
  solvent at a certain temperature, for a saturated
  solution (a solution that contains all the solute
  it can possibly hold). See fig. 8 pg. 409.
• 2. A saturated solution has attained equilibrium
  between the rates of dissolution (dissolving of
  solute) and crystallization are _________.
• 3. If a saturated solution is heated it will
  often hold ______ solute and if cooled
  undisturbed the extra solute will remain in
  solution. At this point the solution is called a
  supersaturated solution. If a small seed crystal
  is placed in a supersaturated solution, _________
  crystallization will ensue.
• 4. Polar solutes dissolve in ___________ solvents
  and nonpolar solutes dissolve in nonpolar
  solvents (general rule of thumb-like dissolves
  like).
• 5. The solubility of most solids in liquids
  _________________ with increasing temperature
  (see fig. 15 pg. 414).
• 6. The solubility of non-polar gases
  _______________ with increasing temperature (see
  fig. 14 pg. 414).
• 7. Gases tend to have low solubility in water,
  the solubility can be increased by the
  application of _________________ (Henrys Law).
  Fig. 13 pg. 413
• 8. The dissolving of a solute in water may give
  off heat or be exothermic (____ ?H) or may
  require energy or be endothermic (____ ?H). Table
  5 pg. 416.
• Homework pg. 426 1a 2 4 5 7 8 9a 10a,b
  11a 12a 13d, f

amount
equal
more
rapid
polar
increases
increases
decreases
pressure
-

3

• Concentration of Solutions
• Solutions are prepared utilizing volumetric
  glassware which is sensitive to changes in
  _____________ (long and skinny). Fig. 17 pg.
  419, fig. 18 pg422.
• Percent Solutions
• a. mass / mass solutions mass solute (g) .
  100 commonly used for chemists when
• mass of solution (g)
  solids, liquids, or gases are dissolved in
  liquids
• (chemical reagent solutions)
• b. mass / volume solutions mass solute (g)
  . 100 most common method of describing solids
• volume of solution (mL)
  dissolved in liquids in the real world
• (sugar solutions)
• c. volume / volume solutions volume solute
  (mL) .100 commonly used when 2 liquids are
  mixed
• volume of
  solution (mL) (fruit juice
  solutions)
• d. proof twice the (volume / volume)
  percentage (100 proof 50 (v/v))
• A percent solution describes the parts of solute
  per ______ parts of solution.

volume
100
25
100
4

• Example Problems
• What is the percent (v/v) of a solution prepared
  by
• dissolving 45 ml of ethanol in water
  to make
• 125 ml of solution?

2. Assuming volumes are additive, what is the
percent (v/v) concentration of a solution
prepared by dissolving 45 ml of ethanol in 125 ml
of water?
(v/v) mL solute . 100 mL
solution
(v/v) 45 mL . 100 125 mL
When liquids are mixed, the solute is the liquid
of smallest volume (ethanol in this case)
(v/v) mL solute . 100 mL solution
(v/v) 36
(v/v) 45 mL . 100 (45mL 125 mL)
What is the proof concentration of this solution?
(v/v) 45 mL . 100 170. mL
Proof 2 ( (v.v))
Proof 2 (36.00)
(v/v) 26
Proof 72
5

• Example Problems
• 3. If the density of water is 0.997 g/cm3 and the
  density, of ethanol is 0.789 g/cm3, what would be
  the percent (m/m) concentration of the solution
  prepared in problem 2?

(m/m) g solute . 100 g
solution
?g ethanol 45 mL . 0.789 g 35.51 g ethanol
mL
\ \
(m/m) 35.51 g . 100 (35.51 g
124.63 g)
?g water 125 mL . 0.997 g 124.63 g water
mL
\ \
(m/m) 35.51 g . 100 160.14 g
(m/m) 22
4. Describe the preparation of 2.50 liters of a
2.00 (m/v) glucose sugar (C6H12O6) solution.
dissolve 50.0 g C6H12O6 in water to make 2.50
L of solution
?g C6H12O6 2.50 L sln
1000 mL sln L sln
2.00 g C6H12O6 100 mL sln
\ \
\ \
6
Homework 1. Describe the preparation of 250. ml
of a 15.0 (m/v) sugar solution. 2. A solution
is prepared by dissolving 2.75 g of copper (II)
sulfate in 75.0 ml of solution, determine the
percent (m/v) of the solution. 3. If 10.0 ml of
acetic acid is diluted with water to a total
volume of 250.0 ml of solution, what is the
percent (v/v) of the solution? 4a. 25.0 ml of
ethanol is dissolved in 100. ml of water.
Assuming volumes are additive, what is the
percent (v/v) of the solution? b. Assuming the
density of water to be 1.00 g/cm3 and the density
of ethanol to be 0.789 g/cm3, determine the
(m/m) of the solution prepared in problem 4a. 5.
How many ml of ethanol are required to prepare
2.50 L of a 1.50 (v/v) ethanol
solution? Bonus Assuming the density of ethanol
to be 0.789 g/cm3, how many grams of ethanol are
required to prepare 1.50 liters of a 3.50 (v/v)
ethanol in water solution?
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Obtain the solution concentration in molarity
Why a volumetric flask?
What is Molarity?
Molarity is the moles of solute per liter of
solution
The long narrow neck makes for sensitive volume
measurements
M represents molarity
12.983 g of CoCl2 is dissolved in 250.00 mL
solution What is the molarity of the solution?
? mol CoCl2 L
12.983 g CoCl2 250.00 mL
. mol CoCl2 129.83 g CoCl2
. 1000 mL L
0.40000 M
8

• Molarity moles solute per liter of solution
• Sample A pg. 420
• Sample B pg. 420
• Sample C pg. 421
• Homework
• Pg. 421 1-3 pg. 427 19b 20b 21

? mol NaCl L
90.0 g NaCl .
1mol NaCl 58.443 g NaCl

0.440 M NaCl

3.50 L
0.8 L .
0.5 mol HCl L
0.4 mol HCl
? mol HCl
1mol K2CrO4 194.2 g K2CrO4
L K2CrO4 6.0 mol K2CrO4
? Lof 6.0 M K2CrO4
23.4 g K2CrO4 .
0.020 L of 6.0 M K2CrO4
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In this solution each purple dot represents 1
mol of particles and every volume marking is 0.5 L
2 L of 1 M solution needs a total of 2 moles
The concentrated solution contains 4 moles per
liter
What volume of 4 M solution is required to
prepare 2 L of 1 M solution?
?L of 4 M sln 2 L . 1 mol
L
L 4 mol
0.5 L of 4 M sln
What is the molarity? 4 mol for every L of
solution or ?mol 16 mol
4 M L sln 4 L
If 1 Lof the 4 M solution is dissolved in
enough water to make 2L of solution, what is the
new solution molarity?
?mol 1 L . 4 mol L
2 M
L sln
________ 2 L
10
In this solution each purple dot represents 1
mol of particles and every volume marking is 0.5 L
2 L of 1 M solution needs a total of 2 moles
The concentrated solution contains 4 moles per
liter
What volume of 4 M solution is required to
prepare 2 L of 1 M solution?
?L of 4 M sln 2 L . 1 mol
L
L 4 mol
0.5 L of 4 M sln
What is the molarity? 4 mol for every L of
solution or ?mol 16 mol .
4 M L sln 4 L
If 1 Lof the 4 M solution is dissolved in
enough water to make 2L of solution, what is the
new solution molarity?
?mol 1 L . 4 mol L
2 M
L sln 2L
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• Preparation of Dilute Solutions from Concentrated
  Solutions
• Sometimes it is inconvenient to prepare a series
  of solutions by obtaining a specific ______ of
  solute for every solution. When many solutions of
  a given solute are to be prepared, a concentrated
  solution (called a _________ solution) will be
  made. Volumes of the stock solution will be
  measured into a graduated cylinder (or pipet) to
  obtain the desired ___________ of solute. The
  number of moles of solute obtained will then be
  ___________ to the desired volume of solution to
  obtain the desired concentration of solution.
• Example1 Describe the preparation of 250.0 cm3
  of 2.0 M copper (II) sulfate.
• Example 2a Describe the preparation 50.0 ml of
  a 0.20 M solution from the stock solution above.
• For dilution problems determine the volume of
  the concentrated solution to be obtained.
• b. Determine the number of formula units of
  copper (II) sulfate present in the solution
  prepared in 2a.

mass
stock
moles
diluted
. 2.0 mol CuSO4 . 1000 mL
159.60 g CuSO4 mol CuSO4
/ /
/ /
?g CuSO4 250.0 cm3
dissolve 80. g CuSO4 in enough water to make
250.0 ml sln
1000 mL sln 2.0 mol CuSO4
0.20 mol CuSO4 1000 mL sln
. / /
/ /
dilute 5.0 mL of 2.0 MCuSO4 in water to make
50.0 mL sln
?mL 2.0 M sln 50.0 mL
0.20 mol CuSO4 1000 mL
6.02.1023 for us CuSO4 mol CuSO4
/ /
/ /
?for. us CuSO4 50.0 mL
6.02.1021 for us CuSO4
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• Example 3 Describe the preparation of 100.0
  ml of 0.68 M solution from the same 2.0 M
  solution.
• Example 4 Describe the preparation of 100.0 ml
  of a 4.0 (m/v) solution of NaCl from a 10.0
  (m/v) NaCl solution. Remember that 4.0 (m/v)
  means ____ g NaCl for every _____ ml of solution
• Describe the preparation of 75.0 ml of 4.15
  (m/v) NaCl solution from a 5.25 M NaCl solution.
• If 15 mL of 2.00 M solution is diluted to a total
  volume of 55 mL, what is the new molarity of the
  solution?

0.68 mol CuSO4 1000 mL
1000 mL 2.0 mol CuSO4
/ /
/ /
dilute 34 mL of 2.0 M CuSO4 in water to make
100.0 mL sln
?mL 2.0 M sln 100.0 mL
4.0
100
100 mL 10.0 g NaCl
4.0 g NaCl 100 mL
/ /
/ /
dilute 40. mL of 10.0 NaCl in water
to make 100.0 mL sln
?mL 10.0 sln 100.0 mL
1000 mL NaCl 5.25 mol NaCl
4.15 g NaCl 100 mL sln
1 mol NaCl 58.44 g NaCl
. / /
/ /
/ /
dilute 10.1 mL of 5.25 M NaCl into water to
make 75.0 mL sln
?mL 5.25 M sln 75.0 mL
2.00 mol 1000 mL
?mol L
15 mL
?mol L
15 mL
2.00 mol 1000 mL
1000 mL L
_____ 55 mL
0.55 M
0.55 M
_____________ 55 mL L 1000 mL
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Homework 1a. Determine the molarity of a calcium
chloride solution when 52.0 g of calcium
chloride is dissolved in sufficient water to
prepare 250.0 mL of solution. b. Determine the
(m/v) of the solution prepared in problem 1a.
c. Describe the preparation of 100.0 ml of a 0.35
M solution from the solution prepared in
problem 1a. d. Describe the preparation of 50.0
cm3 of a 3.0 (m/v) solution form the solution
prepared in problem 1a. e. Describe the
preparation of 40.0 ml of 0.050 M calcium
chloride solution from the 0.35 M solution
prepared in problem 1c. 2a. Describe the
preparation of 2.50 L of a 10.0 M C6H12O6
(glucose) solution. b. Determine the number of
glucose molecules present in the solution made in
2a. c. Describe the preparation of 50.0 ml of
2.33 M glucose solution from the solution
prepared in problem 2a. d. Describe the
preparation of 0.500 L of 1.00 (m/v) solution
from the solution prepared in problem 2a.
e. Describe the preparation of 55.0 cm3 of 0.22
(m/v) solution from the solution 1.00
(m/v) solution prepared in problem 2d.
Bonus Describe the preparation of 1.2 gallons of
0.10 M ferrous phosphate from a 35 (m/v)
solution of ferrous phosphate. Day 2 pg.
901-903 328 332 335 341a 357 367 381
Bonus 377
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• Optional Writing Paper!!!!
• Topic Solutions
• Minimum 800 words
• A 100/100
• B 70/70
• C 40/40
• Due Monday

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• 1. What is a solution?

A solution is a homogeneous mixture (mixture of
uniform composition) that consists of a solute
(the substance dissolved) and the solvent (the
substance that does the dissolving).
2. What is the difference between a solution,
suspension and a colloid?

• The solute particles of a solution are so small
  that they will not settle to
• the bottom of the container. When poured
  through a filter, the solute
• particles will not be trapped by the filter
  paper.

• The solute particles of a suspension are large
  enough that they will
• settle to the bottom of the container.
  When poured through a filter,
• the solute particles will be trapped by
  the filter paper.

• The solute particles of a colloid are so small
  that they will not settle to
• the bottom of the container, however they
  will scatter light (exhibit
• the Tyndall Effect) solute particles of a
  solution will not scatter light.

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3. What are some common solutions?

• Sugar in water (a solid dissolved in a liquid).
• Alcohol in water (a liquid dissolved in a
  liquid).
• Alloys (mixtures of metals).
• The air (mixtures of gases).
• Carbonated beverage (gas dissolved in liquid)

4. What differentiates a concentrated solution
from a dilute solution?
A concentrated solution has a larger amount of
solute dissolved in certain amount of solvent
than does a dilute solution.
5. In general, how does the solubility of a
solid in water change upon heating?
The solubility of most solids increases with an
increase in temperature
6. In general, how does the solubility of a gas
in water change upon heating?
The solubility of gases decreases with an
increase in temperature
17
7. How is it possible to increase the solubility
of a gas into water without altering temperature?
Increasing the pressure of a gas above a liquid
will increase the solubility of the gas in the
liquid (Henrys Law).
8. What determines whether something will
dissolve in water?
The polarity of the solute. If the solute is
polar it will dissolve in water. Non-polar
solutes will not be very soluble in water.
9. What are some common methods of
expressing solution concentration and what does
each mean?
m/v
Expresses the grams of solute dissolved in 100 ml
of solution.
Expresses the mL of solute dissolved in 100 ml of
solution.
v/v
m/m
Expresses the g of solute dissolved in 100 g of
solution.
molarity
Expresses the moles of solute dissolved in L of
solution.
proof
Twice the v/v
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10. What is the difference between mass of
solvent and mass of solution?
The mass of solution is the combined mass of
solute and solvent

• What is the difference between volume of solvent
  and volume of solution?

The volume of solution is the combined volume of
solute and solvent

• Describe any calculations and procedures required
  to prepare a
• solution of a certain molarity and volume if
  you are given a solid solute?

• You need to determine the number of moles of
  solute that will be placed
• into a volumetric flask (flask sensitive to
  volume change). The moles will
• be obtained from the product of the volume
  and molarity of the solution.
• For a solid solute the molar mass of the
  substance will be utilized to determine the
  number of grams of the solute that contains the
  certain
• number of moles (mass is obtained from the
  product of moles and molar
• mass and is measured on a balance).
• The calculated mass of solute will be placed into
  a volumetric flask,
• solvent will be added to this solute to
  until the total volume of solution
• is obtained.
• ? g solute L sln .
  mol solute . g solute
• 
  L sln mole

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13. Describe any calculations and procedures
required to prepare a solution of a certain
molarity and volume if you are given a more
concentrated solution of the desired solute?

• You need to determine the number of moles of
  solute that will be placed
• into a volumetric flask. The moles will be
  obtained from the product of
• the volume and molarity of the solution.
• Since a concentrated solution of a known molarity
  is given, the quotient
• of the moles of solute required and molarity
  of concentrated solution
• will give the desired volume of the
  concentrated solution to be placed
• into the volumetric flask.
• The calculated volume of concentrated solution
  will be measured into a
• graduated cylinder and added to a volumetric
  flask. Water will be added
• to the concentrated solution until the total
  volume of solution is obtained.
• ? mL L dilute sln . mol solute (dilute sln)
  . L concentrated sln
• (concentrated) L dilute
  sln mol solute (concentrated soln)