SOLUTIONS TO SELECTED PROBLEMS IN CHAP. 7 & 8
Physical States of Solutions (Section 7.1)
7.1
Many solutions are found in the home. Some are listed below, with the
composition as printed on the label. When no percentage is indicated,
components are usually given in order of decreasing amount. When water is
present, it is often not mentioned on the label or it is listed as inert ingredients.
Identify the solvent and solutes of the following solutions:
a) Antiseptic mouthwash: alcohol 25 %, thymol, eucalyptol, methyl salicylate,
menthol, benzoic acid, boric acid
b) Paregoric: alcohol 45 %, opium 0.4 %
c) Baby oil: mineral oil, lanolin (there happens to be no water in this solutionwhy?)
d) Distilled vinegar: acetic acid 5%
Solution
a) Water is implied as the solvent. The alcohol and others are solutes.
b) Water is the solvent. Alcohol and opium are the solutes.
c) Mineral oil is the solvent. Lanolin is the solute. Water is immiscible.
d) Water is the solvent. Acetic acid is the solute.
7.3
Classify the following as being a solution or not a solution. Explain you reasons
when you classify one as not a solution. For the ones classified as solutions,
identify the solvent and solute(s).
a) maple syrup b) milk
c) eye drops
d) tomato juice e) tap water
Solution
a) Maple syrup is a solution. It is colored but is not cloudy. Water is the
solvent. Various sugars are the solutes.
b) Milk is not a solution. It is a homogenous colloid.
c) Eye drops is a solution. Water is the solvent. All other components are
solutes.
d) Tomato juice is not a solution. It is not homogeneous.
e) Tap water is a solution. Water is the solvent. Other ingredients are
solutes.
Solubility (Section 7.2)
7.5
Use the terms soluble, insoluble, or immiscible to describe the behavior of the
following pairs of substances when shaken together:
a) 25 mL of water and 1 g of salt-the resulting mixture is clear and colorless.
b) 25 mL of water and 1 g of solid silver chloride-the resulting mixture is cloudy
and the solid settles out.
c) 25 mL of water and 5 mL of mineral oil-the resulting mixture is cloudy and
gradually separates into two layers.
Solution
a) soluble
b) insoluble
127
c) immiscible
Chapter Seven.
7.11
Classify each of the following solutes into the approximate solubility categories of
Table 7.3. The numbers in parentheses are the grams of solute that will dissolve
in 100 g of water at the temperature indicated.
a) boric acid, H3 BO3 (6.35 g at 30°C)
b) calcium hydroxide, Ca(OH)2 (5.35 g at 30°C)
c) antimony(III) sulfide, Sb2S3 (1.75 x 10-4 g at 18°C)
d) copper(II) chloride, CuC12 (70.6 g at 0°C)
e) iron(II) bromide, FeBr2 (10 9 g at 10°C)
Solution
a) soluble
b) soluble
c) insoluble d) very soluble e) very soluble
The Solution Process (Section 7.3)
7.13
What is the difference between a nonhydrated ion and a hydrated ion? Draw a
sketch using the Cl- ion to help illustrate your answer.
Solution
The hydrated ion is the ion surrounded by water molecules, having the dipole
of the water oriented to spread the ionic charge. Figure 7.4 in the text
illustrates what the sketch should look like.
7.15
Ground up limestone (CaCO3) is used as a gentle abrasive in some powdered
cleansers. Why is this a better choice than ground up soda ash (Na2CO3)?
Solution
CaCO3 is not soluble in water. It forms a paste when mixed with a little bit of
water. The CaCO3 particles in the paste can act as an abrasive. In contrast,
Na2CO3 is soluble in water. It would dissolve and not make such a paste in
water.
7.17
Indicate which of the following substances (with geometries shown in the text)
would be soluble in water (a polar solvent) and in benzene (a nonpolar solvent):
H
O
O
a) H
S
b) H
HCl
c) H
H
d) N
N
Solution
a) H2 S is a polar molecule. It is soluble in water and insoluble in benzene.
b) HCl is a polar molecule. It is soluble in water and insoluble in benzene.
c) H2 O2 is a polar molecule. It is soluble in water and insoluble in benzene.
d) N2 is a nonpolar molecule. It is insoluble in water and soluble in benzene.
7.19
Suppose you put a piece of a solid into a beaker that contains water and stir the
mixture briefly. You find that the solid does not immediately dissolve completely.
Describe three things you might do to try to get the solid to dissolve.
Solution
Solutions and Colloids.
A solid would dissolve faster (1) at a higher temperature; (2) if the solid were
crushed into smaller particles; or (3) if the solution were stirred.
Solution Concentrations (Section 7.4)
7.23
Calculate the molarity of the following solutions:
a) A sample of solid Na2SO4 weighing 0.150 g is dissolved in enough water to
make 10.0 mL of solution.
b) A 6.50 g sample of glucose (C6 H12O 6) is dissolved in enough water to give
150 mL of solution
c) A 43.5 g sample of K2SO 4 is dissolved in a quantity of water, and the solution
is stirred well. A 25.0 mL sample of the resulting solution is evaporated to
dryness and leaves behind 2.18 g of solid K2SO 4.
Solution
1mol K2 SO 4
174.3 gK 2 SO 4
= 0.500 M
0.0250L of soln
2.18 g K2 SO 4 x
c)
7.25
M=
Use the data that relates the
volume and the mass of the
same sample
Calculate:
a) How many moles of solute are contained in 1.75 L of a 0.215 M solution?
b) How many moles of solute are contained in 250 mL of a 0.750 M solution?
c) What volume of a 0.415 M solution contains 0.250 mol of solute?
Solution
7.31
Calculate the concentration in % (w/w) of the following solutions. Assume water
has a density of 1.00 g/mL.
a) 5.20 g of CaCl2 is dissolved in 125 mL of water
b) 0.200 mol of solid KBr is dissolved in 200 mL of water
c) 50.0 g of solid is dissolved in 250 ml of water
d) 10.0 mL of ethyl alcohol (density= 0.789 g/mL) is mixed with 10.0 mL of
ethylene glycol (density = 1.11 g/mL).
Chapter Seven.
Solution
5.20 g CaCl2
a) % (w/w)= 125 gH2 O + 5.20 g CaCl2 x 100 = 4.00% (w/w)CaCl2
119 g KBr
0.200 molKBr x
1molKBr
% (w/w)=
x 100 = 10.6% (w/w)KBr
119 g KBr
200 gH2O + (0.200 mol KBr x
)
1molKBr
b)
50 g solid
% (w/w)=
x 100 = 16.7% (w/w) solid
c)
250 gH2 O + 50 g solid
d) wt alcohol = 10.0 mL x 0.789 g/mL = 7.89 g alcohol
wt ethylene glycol = 10.0 mL x 1.11 g/mL = 11.1 g ethylene glycol
wt soln = 11.1 g ethylene glycol + 7.89 g alcohol = 19.0 g soln
7.89 g alcohol
% (w/w)=
x 100 = 41.5% (w/w) alcohol
19.0 g soln
7.33
Calculate the concentration in % (w/w) of the following solutions:
a) 424 g of solute is dissolved in water to give 1.00 L of solution. The density of
the resulting solution is 1.18 g/mL
b) A 50.0 mL solution sample with a density of 0.898 g/mL leaves 12.6 g of solid
residue when evaporated
c) A 25.0 g sample of solution upon evaporation leaves a 2.32 g residue of
NH4Cl
Solution
a) wt soln = 1.00L x 1000 mL/L x 1.18 g/mL = 1180 g soln
424 g solute
% (w/w)=
x 100 = 35.9% (w/w)
1180 g soln
b) wt soln = 50.0 mL soln x 0.898 g soln/mL soln = 44.9 g soln
12.6 g solute
% (w/w)=
x 100 = 28.1% (w/w)
44.9 g soln
2.32 g solute
% (w/w)=
x 100 = 9.28% (w/w)
c)
25.0 g soln
7.37
The blood serum acetone level for a person is determined to be 1.8 mg of
acetone per 100 mL of serum. Express this concentration as % (v/v) if liquid
acetone has a density of 0.79 g/mL.
Solution
%(v/v)=
7.39
1.8 mg acetone
1gacetone
1mL acetone
x
x
x100=0.0023%acetone(v/v)
100mLblood 1000 mgacetone 0.79 g acetone
Calculate the concentration in % (w/v) of the following solutions:
a) 28.0 g of solute is dissolved in 200 mL of water to give a solution with a
density of 1.10 g/mL
b) A 25.0 mL solution sample upon evaporation leaves a solid residue of 0.38 g.
c) Upon analysis for total protein, a blood serum sample of 15.0 mL is found to
contain 1.02 g of total protein.
Solutions and Colloids.
Solution
weight of soln = 28.0 g solute + (200 mlH2 O x
volume of soln =
1.00 g H2 O
) = 228 gsoln
1mlH2 O
228 g soln
= 207 mL soln
1.10 g soln/mL soln
28.0 g solute
a) % (w/v) = 207 mL soln x 100 =13.5% (w/v)
0.38 g solute
b) % (w/v) = 25.0 mL soln x100 = 1.52% (w/v)
1.02 g protein
c) % (w/v) = 15.0 mL soln x100 = 6.80% protein (w/v)
7.41
Assume the density of the solution prepared in Exercise 7.40 is 1.18 g/mL and
express the concentration in % (w/v).
Solution
From Figure 7.3, the solubility of KBr is about 65 g/100 g H2O weight of soln =
65 g KBr + 100 g H2O = 165 g soln
165 g soln
volume of soln =
= 140 mL soln
1.18 g soln/mL soln
65 g KBr
% (w/v) =
x100 = 46%KBr (w/v)
140 mL soln
Solution Preparation (Section 7.5)
7.43
Explain how you would prepare the following solutions using pure solute and
water. Assume water has a density of 1.00 g/mL.
a) 500 mL of 2.00 M NaOH solution
b) 250 mL of 40.0 % (v/v) alcohol solution (C2 H5 OH)
c) 100 mL of 10.0 % (w/v) glycerol solution. Glycerol is a liquid with a density of
1.26 g/mL. Describe two ways to measure out the amount of glycerol
d) approximately 100 mL of normal saline solution, 0.89 % (w/w) NaCl.
Solution
a) mol NaOH = (2.00 mol NaOH/L soln x 0.500 L soln) = 1.00 mol NaOH
g NaOH = (1.00 mol NaOH x 40.0 g NaOH/mol Na0H) = 40.0 g NaOH
Dissolve 40.0 g NaOH in enough water to make 500 mL solution.
% (v/v) x mL soln 40.0% x 250 mL soln
=
= 100 mL alcohol
b) mL alcohol =
100
100
Dissolve 100 mL alcohol in enough water to give 250 mL solution.
% (w/v) x mL soln 10.0% x 100 mL soln
=
= 10.0 g glycerol
100
100
10.0 g glycerol
volume glycerol
= 7.94 mL glycerol
1.26 g glycerol/mL
Weigh out 10.0 g glycerol or measure 7.94 mL glycerol, and dissolve in
enough water to make 100 mL of solution.
c) g glycerol =
Chapter Seven.
d) Assuming the solution has a density of approximately 1.00 g/mL, the 100
mL (approx) weighs approximately 100 g.
% (w/w) x wt soln 0.89% x 100 g
wt NaCl =
=
= 0.89 g NaCl
100
100
Dissolve 0.89 g NaCl in slightly less than 100 mL (approx. 99 mL) of
water.
7.49
What is the molarity of the solution prepared by diluting 25.0 mL of 0.412 M
Mg(NO3 )2 to each of the following final volumes?
a) 40.0 mL
b) 100 mL
c) 1.10 L
d) 350 mL
Solution
Rearrange Equation 7.9 to get:
Md =
Mc x Vc
Vd
Solution Stoichiometry (Section 7.6)
7.53
How many milliliters of 0.250 M HCl would be needed to react exactly with 10.5 g
of solid NaHCO3? NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2 (g) + H2O(l)
Solution
1molNaHCO3
1molHCl
HCl1Lsoln 1000mL
x
x
= 500mLHCl
84.0gNaHCO3 1molNaHCO3 0.250molHCl 1L soln
How many milliliters of 0.124 M NaOH solution will exactly react with 35.0 mL of
0.210 M H3PO 4 solution? 3NaOH(aq) + H3 PO4 (aq)  Na3PO 4(aq) + 3H 2O(l)
25.0 gNaHCO3 x
7.57
Solution
7.61
An ingredient found in some antacids is magnesium hydroxide, Mg(OH)2 .
Calculate the number of grams of Mg(OH)2 needed to react exactly with 250 mL
of stomach acid (see Exercise 7.60). Mg(OH) 2(s) + 2HCl(aq)  MgCl2(aq) + 2H2 O(l)
Solution
250 mL x
1L
0.10 molHCL 1molMg (OH)2 58.3gMg(OH)2
x
x
x
= 0.73gMg(OH)2
1000 mL
1L
2 molHCL
1molMg(OH)2
Solution Properties (Section 7.7)
7.63
If you look at the labels of automotive products used to prevent radiator freezing
(antifreeze) and radiator boiling, you will find the same ingredient listed, ethylene
Solutions and Colloids.
glycol. Use the idea of colligative solution properties to explain how the same
material can prevent an automobile cooling system from freezing and boiling.
Solution
Since pure ethylene glycol has a higher boiling than water, a solution of
ethylene glycol in water has a higher boiling point than pure water. The
freezing point of a water solution is lower than that of pure water. Thus, the
solution prevents freezing of the coolant, and can get hotter without boiling.
7.65
Calculate the boiling and freezing points of water solutions that are 1.00 M in the
following solutes:
a) KBr, a strong electrolyte
b) ethylene glycol, a nonelectrolyte
c) (NH4 )2 CO3, a strong electrolyte
d) Al2 (SO4)3 , a strong electrolyte
Solution
Use equations 7.13 and 7.14. M = 1.00;
Kb = 0.52;
Kf = 1.86
a) For KBr, n = 2
∆tb = 2 x 0.52 x 1.00 = 1.04°C higher; Boiling point = 101.04°C
∆tf = 2 x 1.86 x 1.00 = 3.72°C lower; Freezing point = -3.72°C
b) For ethylene glycol, n = 1
∆tb = 1 x 0.52 x 1.00 = 0.52°C higher; Boiling point = 100.52°C
∆tf = 1 x 1.86 x 1.00 = 1.86°C lower; Freezing point = -1.86°C
c) For (NH4 )2CO3, n = 3
∆tb = 3 x 0.52 x 1.00 = 1.56°C higher; Boiling point = 101.56°C
∆tf = 3 x 1.86 x 1.00 = 5.58°C lower; Freezing point = -5.58°C
d) For Al2(SO4 )3, n = 5
∆tb = 5 x 0.52 x 1.00 = 2.6°C higher; Boiling point = 102.6°C
∆tf = 5 x 1.86 x 1.00 = 9.30°C lower; Freezing point = -9.30°C
7.67
Calculate the boiling and freezing points of the following solutions. Water is the
solvent unless otherwise indicated.
a) a solution containing 50.0 g of H2SO 4, a strong electrolyte (both H's
dissociate), per 250 mL
b) a solution containing 200 g of table sugar (C12H22O11), a nonelectrolyte, per
250 mL
c) a solution containing 75.0 g of octanoic acid (C8 H16O 2), a nonelectrolyte, in
enough benzene to give 250 mL of solution
Solution
Use equations 7.13 and 7.14. Kb = 0.52; Kf = 1.86 for water
a) For H2SO4 , n = 3
molH2 SO4 = 50.0 gH2 SO4 x
M=
1molH2 SO4
= 0.510 molH2 SO4
98.1g H2SO4
0.510 molH2 SO4
= 2.04 M H2 SO4
0.250 L soln
∆tb = 3 x 0.52 x 2.04 = 3.18°C higher; Boiling point = 103.18°C
∆tf = 3 x 1.86 x 2.04 = 11.4°C lower; Freezing point = -11.4°C
Chapter Seven.
b) For sucrose, n = 1
mol C12H22O11= 200 g C12H22 O11 x
M=
1molC12H22 O11
= 0.585 mol C12H22 O11
342 g C12H22 O11
0.585 molC12H22 O11
= 2.34 M C12 H22O11
0.250 L soln
∆tb = 1 x 0.52 x 2.34 = 1.22°C higher; Boiling point = 101.22°C
∆tf = 1 x 1.86 x 2.34 = 4.35°C lower; Freezing point = -4.35°C
c) For benzene,
Kb = 2.53
Kf = 4.90;
normal boiling point = 80.1°C
normal freezing point = 5.5°C
For C8H16O2, n = 1
molC 8H16 O2 = 75.0 g C 8H16O 2 x
M=
1mol C8 H16 O2
= 0.521molC 8H16 O2
144 g C 8H16 O2
0.521mol C8 H16 O2
= 2.08 M C8H16 O2
0.250 L soln
∆tb = 1 x 2.53 x 2.08 = 5.26°C higher; BP = 80.1 + 50.26 = 85.4°C
∆tf = 1 x 4.90 x 2.08 = 10.2°C lower; FP = 5.5 – 10.2 = -4.7°C
7.71
Calculate the osmotic pressure of any solution with an osmolarity of 0.250.
Solution
= nMRT = 0.250 mol/L x 0.0821 L atm/mol K x 298 K = 6.12 atm
760 torr
6.12 atm=
= 4.65 x103 torr = 4.65 x103 mmHg
1atm
7.73
Calculate the osmotic pressure of a 0.200 M solution of methanol, a
nonelectrolyte.
Solution
7.75
Calculate the osmotic pressure of a solution that contains 1.20 mol of CaCl2 in
1500 mL.
Solution
= nMRT
120 mol CaCl2
n = 3; M = 1.500 L soln = 0.800 MCaCl2
= 3 x 0.800 mol/L x 0.0821 L atm/mol K x 298 K = 58.7 atm
760 torr
58.7 atm =
= 4.46 x104 torr = 4.46 x104 mmHg
1atm
7.77
Calculate the osmotic pressure of a solution that is 0.122 M in solute, and has a
boiling point 0.19°C above that of pure water.
Solutions and Colloids.
Solution
The osmolarity of that solution can be obtained from the ∆tb.
t 0.19
nM = b =
= 0.37
Kb 0.52
= nMRT = 0.365 mol/L x 0.0821 L atm/mol K x 298 K = 8.93 atm
760 torr
3
3
8.9 atm =
= 6.8 x10 torr = 6.8 x10 mmHg
1atm
7.79
Calculate the osmotic pressure of a solution that contains 245.0 g of ethylene
glycol (C2 H6 O2 ), a nonelectrolyte, per liter.
Solution
= nMRT; For ethylene glycol, n = 1
1mol C2 H6 O2
mol C 2H6 O2 = 245.0 g C 2H6 O2 x
= 3.95 mol C2 H6 O2
62.0 g C2 H6 O 2
3.95 mol C2 H6 O2
= 3.95 MC 2H6 O2
1L soln
= 1 x 3.95 mol/L x 0.0821 L atm/mol K x 298 K = 96.6 atm
760 torr
96.6 atm =
= 7.34 x104 torr = 7.34 x104 mmHg
1atm
M=
Colloids (Sections 7.8)
7.81
Explain how the following behave in a colloidal suspension: dispersing medium,
dispersed phase, and colloid emulsifying agent.
Solution
A dispersing medium is similar to the solvent of a solution. A dispersed
phase is similar to the solute of a solution. A colloidal emulsifying agent is
used to reduce the size of dispersed phase particles, making a more stable
colloid.
Dialysis (Sections 7.9)
7.83
Suppose you have a bag made of a membrane like that in Figure 7.18. Inside
the bag is a solution containing water and dissolved small molecules. Describe
the behavior of the system when the bag functions as an osmotic membrane and
when it functions as a dialysis membrane.
Solution
With osmosis, pore size is smaller than molecule size so water would migrate
into the bag. With dialysis, pore size is larger than molecule size so the
molecules would migrate out of the bag.
ADDITIONAL EXERCISES
7.85
When 5.0 mL of water and 5.0 mL of rubbing alcohol are mixed together, the
volume of the resulting solution is 9.7 mL. Explain in terms of the molecules why
Chapter Seven.
the final solution volume is not 10.0 mL.
Solution
Volumes are not additive, with resulting solutions having larger or smaller
volumes. In this case, a viable explanation is that there is a large attraction
between the molecules, causing them to be pulled closer together.
7.87
Ethylene glycol is mixed with water and used as an antifreeze in automobile
cooling systems. If a 50% (v/v) solution is made using pure ethylene glycol and
pure water, it will have a freezing point of about 65 o C. What will happen to the
freezing point of the mixture if ethylene glycol is added until the concentration is
70% (v/v)? Explain your answer.
Solution
The freezing point will increase since ethylene glycol is now the solvent. Its
freezing point would be the one depressed.
SOLUTIONS TO CHEMISTRY FOR THOUGHT
7.95 When a patient has blood cleansed by hemodialysis, the blood is circulated
through dialysis tubing submerged in a bath that contains the following solutes in
water: 0.6% NaCl, 0.04% KCL, 0.2% NaHCO3, and 0.72% glucose (all
percentages are w/v). Suggest one or more reasons why the dialysis tubing is
not submerged in pure water.
Solution
If the tubing were submerged in pure water, osmosis would move water
through the membrane into the blood. This would dilute the blood serum,
increasing the tendency for the blood cells to undergo hemolysis.
7.97
Refer to Figure 7.3 and propose a reason why fish sometimes die when the
temperature of the water in which they live increases.
Solution
Fish must be able to extract dissolved oxygen from the water. As the
temperature increases, the amount of dissolved gases, including oxygen,
decreases. If the oxygen level gets too low, the gills of the fish cannot extract
sufficient oxygen and the fish "suffocates."
7.98
Small, souvenir, salt-covered objects are made by forming the object out of wire
mesh and suspending the mesh object in a container of water from a salt lake
such as the Dead Sea or Great Salt Lake. As the water evaporates, the wire
mesh becomes coated with salt crystals. Describe this process using the key
terms introduced in Section 7.2.
Solution
As water evaporates from the salt water, the solution becomes saturated.
Continuing evaporation causes salt to crystallize out, coating the object with
salt crystals.
Solutions and Colloids.
Spontaneous and Nonspontaneous Changes (Section 8.1)
8.1
Classify the following processes as spontaneous or nonspontaneous. Explain
your answers in terms of whether energy must be continually supplied to keep
the process going.
a) Water is decomposed into hydrogen and oxygen gas by passing electricity
through the liquid.
b) An explosive detonates after being struck by a falling rock.
c) A coating of magnesium oxide forms on a clean piece of magnesium exposed
to air.
d) A light bulb emits light when an electric current is passed through it.
e) A cube of sugar dissolves in a cup of hot coffee.
Solution
a) Nonspontaneous; energy must be continually supplied.
b) Spontaneous; only an initial energy is needed to be supplied.
c) Spontaneous; no outside energy is required.
d) Nonspontaneous; energy must be continually supplied.
e) Spontaneous; no outside energy is required.
8.3
Classify the following processes as exergonic or endergonic. Explain your
answers.
a) any combustion process
b) perspiration evaporating from the skin
c) melted lead solidifying
d) an explosive detonating
e) an automobile being pushed up a slight hill (from point of view of automobile)
Solution
a) Exergonic; any combustion process liberates energy to its surroundings.
b) Endergonic; perspiration point of view; energy must be absorbed.
c) Exergonic; lead point of view; energy goes from lead to surroundings.
d) Exergonic; energy goes from the explosive to the surroundings.
e) Endergonic; energy goes from surroundings to the automobile.
8.7
Pick the example with the highest entropy from each of the following sets.
Explain your answers.
a) solid ice, liquid water, steam
b) leaves on a tree, fallen leaves blown about on the ground, fallen leaves raked
and placed in a basket
c) A stack of sheets of paper, a wastebasket containing sheets of paper, a
wastebasket containing torn and crumpled sheets of paper
d) A 0.10 M sugar solution, a 1.0 M sugar solution, a 10.0 M sugar solution
e) A banquet table set for dinner, a banquet table during dinner, a banquet table
immediately after dinner
Solution
a) Steam has the highest entropy (disorder).
b) Fallen leaves blown about have the highest entropy.
c) Wastebasket containing torn and crumpled paper has the higher entropy.
Chapter Seven.
d) Since this mixture has two components, entropy must be considered for
each separately. The mixture with the most sugar (the 10.0 M solution,
with the least water) has the highest entropy for the water; but the mixture
with the most water (the 0.10 M solution, with the least sugar) produces
the highest disorder for the sugar.
e) The banquet table immediately after dinner will have the highest entropy,
assuming that everything is left on the table.
Reaction Rates (Section 8.2)
8.15
Consider the following reaction: A + B  C. Calculate the average rate of the
reaction on the basis of the following information:
a) Pure A, B, and C are mixed together at concentrations of A = B = 0.400 M; C
= 0.150 M. After 9.00 minutes, the concentration of C is 0.418 M.
b) Pure A and B are mixed together at the same concentration of 0.361 M. After
6.00 minutes, the concentration of A is found to be 0.0480 M.
Solution
8.17
A reaction generates hydrogen gas (H2 ) as a product. The reactants are mixed in
a sealed 250 mL vessel. After 15.0 minutes, 3.91 x 10-2 mol H2 have been
generated. Calculate the average rate of the reaction.
Solution
8.19
Suppose a small lake is contaminated with an insecticide that decomposes with
time. An analysis done in June shows the decomposition product concentration
to be 7.8 x 10-4 mol/L. An analysis done 35 days later shows the concentration of
decomposition product to be 9.9 x 10-4 mol/L. Assume the lake volume remains
constant and calculate the average rate of decomposition of the insecticide.
Solution
Co = 7.8 x 10-4 M; Ct = 9.9 x 10-4 M; ∆C = 2.1 x 10-4 M
- 4
rate = 2.1x 10 M = 6.0 x 10- 6 M/day
35days
Solutions and Colloids.
Energy Diagrams (Section 8.4)
8.23
Sketch energy diagrams to represent each of the following. Label the diagrams
completely and tell how they are similar to each other and how they are different.
a) exothermic (exergonic) reaction with activation energy
b) exothermic (exergonic) reaction without activation energy
Solution
Activation
Energy
reactants
reactants
E
E
products
products
Reaction progress
Reaction progress
a)
b)
The diagrams show reactants and products at the same energy levels, but the
diagrams have different activation energies.
8.25
Use energy diagrams to compare catalyzed and uncatalyzed reactions.
Solution
Activation
Energy
Activation
Energy
E
Uncatalyzed
Reaction
Energy (net)
Reaction progress
E
Catalyzed
Reaction
Energy (net)
Reaction progress
Factors That Influence Reaction Rates (Section 8.5)
8.27
8.29
The following reactions are proposed. Make a rough estimate of the rate of each
one: rapid, slow, won't react. Explain each answer.
a) H2O(l) + H +(aq)  H3O +(aq)
b) H3O +(aq) + H+  H4O 2+(aq)
c) 3H2(g) + N2 (g)  2NH3(g)
d) Ba2+(aq) + SO24 - (aq)  BaSO4 (s)
Solution
a) Rapid. Solution allows frequent collisions.
b) Won't react. The positive reactant ions repel each other and won't collide.
c) Slow. The bond strength of the N2 makes a very high activation energy.
d) Rapid. Oppositely charged ions in solution allow frequent collisions.
Which reaction mechanism assumption is most important in explaining the
following? Why? a) the influence of concentration on reaction rates. b) the
influence of catalysts on reaction rates.
Solution
a) #1. Reaction rate depends on frequency of collisions.
b) #2. Catalysts can lower the activation energy, improving the chance for an
effective collision; or #3. The orientation of the molecules is enhanced.
Chapter Seven.
8.31
A reaction is started by mixing reactants. As time passes, the rate decreases.
Explain this behavior that is characteristic of most reactions.
Solution
As the reaction proceeds, the reactants are used up. If the concentration of
the reactants decrease, the rate is slower.
8.33
What factor is more important than simply the amount of solid reactant present in
determining the rate of a reaction? Explain.
Solution
The surface area exposed. Reactions involving a solid can occur only on the
surface. Making more individual pieces of smaller size increases the total
area upon which the reaction may occur.
Chemical Equilibrium (Section 8.6)
8.35
Describe the establishment of equilibrium in a system represented by a shopper
walking up the "down" escalator.
Solution
When the rate of walking up is the same as the rate of the escalator moving
down, the shopper does not appear to be moving either direction.
8.39
Colorless N2O4 gas decomposes to form red-brown colored NO2 gas. Describe
how the concentrations of N2 O4 and NO2 would change (increases or decreases)
as equilibrium was established in a sealed container that was initially filled with
N2 O4. What observations would indicate that equilibrium was established?
Solution
Initially, only N2O 4 is present. To reach
equilibrium, its concentration will decrease
as NO2 is produced. This can be verified
by an increase in the red brown color of NO 2. When the color becomes
constant, equilibrium has been achieved.
Position of Equilibrium (Section 8.7)
8.41
Write an equilibrium expression for each of the following gaseous reactions.
Solution
Solutions and Colloids.
8.43
The following equilibria are established in water solutions. Write an equilibrium
expression for each reaction.
2+
Ni(NH3)62a) Ni + 6 NH 3
Sn4+ + 2 Fe 2+
b) Sn 2+ + 2 Fe 3+
2 F + Cl2
c) F2 + 2 Cl
Solution
a) K =
8.45
[Ni (NH3)26 + ]
[Ni2 + ] [NH3 ]6
b) K=
[Sn4 + ] [Fe2 + ]2
[Sn2 + ] [Fe3 + ]2
c) K =
[F- ]2[NH3]4
[F2 ] [Cl- ]2
Write an equation that corresponds to each of the following equilibrium
expressions:
Solution
a) 3HF + 6H2O E PH3 + 3F2
b) 4NO2 + 6H2O E 7O2 + 4NH3
c) 2Cl2O 5 E 4ClO2 + O2
d) 2NO + 2H2 E N2 + 2H2O
8.47
At 600°C, gaseous CO and Cl2 are mixed together in a 1.00 L closed container.
At the instant they are mixed, their concentrations are CO = 0.79 mol/L and Cl2 =
0.69 mol/L. After equilibrium is established, their concentrations are CO = 0.25
mol/L and Cl2 = 0.15 mol/L. Evaluate the equilibrium constant for the reaction
CO(g) + Cl2 (g)  COCl2 (g)
Solution
The moles of COCl2 produced equal the moles of CO (and
[COCl2]
moles of Cl2) lost.
K=
[CO] [Cl2] CO mol lost = 0.79 mol - 0.25 mol
0.54
K=
= 14
(0.25) (0.15)
= 0.54 moles lost = 0.54 mol/L COCl2 at equilibrium
8.49
Consider the following equilibrium constants. Describe how you would expect the
equilibrium concentrations of reactants and products to compare with each other
(larger than, smaller than, etc.) for each case.
a) K = 2.1 x 10-6
b) K=0.15
c) K = 1.2 x 108
d) K = 0.00036
Solution
a) The products have a much smaller concentration than the reactants.
b) The products have a smaller concentration than the reactants.
c) The products have a much larger concentration than the reactants.
d) The products have a much smaller concentration than the reactants.
Chapter Seven.
Factors That Influence Equilibrium Position (Section 8.8)
8.51
Use Le Châtelier’s principle to predict the direction of equilibrium shift in the
following equilibria when the indicated stress is applied:
Solution:
a) shift left
8.55
b) shift right
c) shift left
Tell what would happen to each equilibrium concentration in the following when
the indicated stress is applied and a new equilibrium position established:
Solution
a) The [H+] will decrease; the [CO2 ] will increase; the [HCO-3 ] will be higher
than it was prior to adding, but less than the sum of the prior amount plus
the amount added. The [H2O] concentration is not changed significantly in
the water solution.
b) The [CO2] will decrease and the [H2 CO3] will increase. The H2 O
concentration is not decreased significantly in the water solution.
c) The [CO2] will decrease and the [H2 CO3] will increase. The H2 O
concentration is not decreased significantly in the water solution.
8.57
The gaseous reaction 2HBr (g) E H2 (g) + Br2 (g) is endothermic. Tell which
direction the equilibrium will shift for each of the following:
a) some H2 is removed
b) the temperature is decreased
c) some Br2 is added
d) a catalyst is added e) some HBr is added
f) the temperature is decreased and some HBr is removed
Solution
a) shift right
b) shift left
d) no effect
e) shift right
ADDITIONAL EXAMPLES
8.59
c) shift left
f) shift left
Assume the following reactions take place under identical conditions of
concentration and temperature. Which reaction would you expect to take place
faster? Explain your answer.
Solution
a) would proceed more rapidly. F2 is less massive than I 2 so at the same
temperature, it would move more rapidly. As a result, fluorine molecules will
interact with hydrogen molecules more often.
Solutions and Colloids.
8.61
Bacteria found in lakes use dissolved oxygen gas to metabolize organic
contaminates and use them as a food source, thus removing the contaminates
from the lakes. When lake temperatures increase significantly, the rate of this
chemical decontamination decreases. Remember your study of gases and
explain this apparent violation of the factors influencing reaction rates studied in
this chapter.
Solution
As temperature increases, the solubility of a gas in a liquid will decrease. In
this example, the amount of oxygen will decrease, reducing the rate of
metabolism.
ALLIED HEALTH EXAM CONNECTION
8.63
Which of the statements given as answers are true concerning the following
equilibrium:
a) An increase in temperature will shift the equilibrium to the right.
b) The addition of NH 3 will shift the equilibrium to the left.
c) The addition of H2 will shift the equilibrium to the right.
Solution
b) and c) are true.
8.65
Identify which of the following statements are not characteristic of exergonic
reactions:
a) They are “downhill” reactions.
b) They have a negative energy change.
c) They are “uphill” reactions
d) The products have less energy than the reactants.
Solution
c is not characteristic of an exergonic (exothermic) reaction.
SOLUTIONS TO CHEMISTRY FOR THOUGHT
8.68 Refer to Figure 8.4 and answer the question. How would the total energy
released as light for each light stick compare when they have both stopped
glowing?
Solution
Since there is a fixed amount of reactant(s) in the light stick, the one occurring
at a faster rate (the warmer one) will use up the reactant more quickly, but the
total amount of light energy would be the same in each.
8.69
A mixture of gases NOCl, NO, and Cl2 is allowed to come to equilibrium at 400°C
in a 1.50 L reaction container. Analysis shows the following number of moles of
each substance to be present at equilibrium: NOCl = 1.80 mol, NO = 0.70 mol,
and Cl2 = 0.35 mol. Calculate the value of the equilibrium constant for the
reaction at 400°C. The equation for the reaction is: 2NO Cl (g) E 2NO (g) + Cl2 (g)
Solution
Chapter Seven.
[NOCl] = 1.80 mol =1.20 M
[NO] = 0.70 mol = 0.47 M
1.50 L
1.50 L
0.35
mol
[Cl2] =
= 0.23M
1.50L
[Cl ] [NO]2 (0.23)(0.47)2
K= 2
=
= 0.035
[NOCl]2
(1.20)2
8.70
The equilibrium constant for the reaction PCl5 (g) E PCl3 (g) + Cl2 (g) is 0.0245 at
250°C. What molar concentration of PCl5 would be present at equilibrium if the
concentrations of PCl 3 and Cl2 were both 0.250 M?
Solution
[PCl3 ] [Cl2 ]
[PCl3 ] [Cl2 ]
for [PCl5 ] gives; [PCl5 ]=
[PCl5 ]
K
(0.250) (0.250)
[PCl5 ] =
= 2.55
0.0245
SolvingK =
8.71 At 448°C, the equilibrium constant for the H2 + I2 E 2HI is 50.5. What
concentration of I, would be found in an equilibrium mixture in which the
concentration of HI and H2 were 0.500 M and 0.050 M, respectively?
Solution
[HI]2
[HI]2
Solving K =
for [I2 ] gives; [I2 ]=
[H2 ][I2 ]
[H2 ]K
(0.500)2
[I2 ]=
= 0.099M
(0.050) (50.5)
8.72
Refer to Figure 8.10 and answer the questions. Would you expect a crushed
antacid tablet (like Alka-Seltzer) to dissolve faster or slower than a whole tablet?
Explain.
Solution
The other reactant for air combustion in air is molecules of oxygen, O2 . The
crushed tablet has much more surface area and would dissolve faster than a
whole tablet.
8.73
Refer to Figure 8.13 and answer the questions. Would the presence of a catalyst
in the tube influence the equilibrium concentrations of the two gases? Explain.
Solution
Since raising the temperature shifts the equilibrium to the NO 2 side, the
equation must be: N2 O 4 + heat E 2NO2 . It is endothermic from left to right.
Adding a catalyst would have no effect on the equilibrium position. The
catalyst would increase the rates of the forward and reverse reactions by the
same factor.
Solutions and Colloids.
8.74
In the blood, both oxygen (O2 ) and poisonous carbon monoxide (CO) can react
with hemoglobin (Hb). The respective compounds are HbO2 and HbCO. When
both gases are present in the blood, the following equilibrium is established:
HbCO + O2 E HbO2 + CO
Use Le Châtelier’s principle to explain why pure oxygen is often administered to
victims of CO poisoning.
Solution
Pure O2 would shift the equilibrium to the right, causing more CO to be
exhaled from the lungs, and more HbO2 in the blood.
8.75
Use the concept of reaction rates to explain why no smoking is allowed in
hospital areas where patients are being administered oxygen gas.
Solution
Increasing the O2 concentration make the rate of combustion increase. A
smoldering object, burning slowly in the air, will burst into an open flame in an
oxygen-enriched environment. The flame could easily spread to other
combustible materials, which would also burn faster in the oxygen-rich
atmosphere.
8.76
Suppose you have two identical unopened bottles of carbonated beverage. The
contents of both bottles appear to be perfectly clear. You loosen the cap of one
bottle and hear a hiss as gas escapes, and at the same time gas bubbles appear
in the liquid. The liquid in the unopened bottle still appears to be perfectly clear.
Explain these observations using the concept of equilibrium and Le Châtelier’s
principle. Remember, a carbonated beverage contains carbon dioxide gas
dissolved in a liquid under pressure.
Solution
The first bottle contains a dissolved gas. When the bottle is opened, the
pressure in the bottle is reduced, lowering the concentration of the gas above
the solution in the bottle. The equilibrium is shifted towards the ‘gas’ phase
causing the bubbles to appear.
8.77
Someone once suggested that it is impossible to unscramble a scrambled egg.
Describe an unscrambled and scrambled egg in terms of the concept of entropy.
Solution
Unscrambled eggs have a definite structure of shell, egg white and yoke. The
last two are liquid or gel like and contain proteins in their natural state (among
other substances). A scrambled egg (cooked) will have a broken shell, a
mixing of the egg white and yoke and now be a solid due to the denaturing of
the protein. All contribute towards an increased entropy.