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Solutions for Chapter 2 End-of-Chapter Problems
Problem 2.1.
Student answers will vary. This is an example. Use a pipette to collect samples from various
places throughout the mixture. Analyze each sample for properties like density, boiling point,
freezing point, or specific gravity. Compare the results to each other. If the solution is
homogeneous all the results should be the same (within the experimental uncertainty).
Problem 2.2.
Some examples of homogeneous solutions you might find at the supermarket or a pharmacy
might be: soft drinks, saline solutions for contact lenses, rubbing alcohol, liquid bleach, and
vinegar. Some examples of heterogeneous solutions (suspensions) might be: milk, Milk of
Magnesia, Peptobismol, calamine lotion, and paint.
Problem 2.3.
(a) An energy diagram for the change of a liquid (water) going to a gas is shown in Figure 1.35
and this problem is here to help recall the earlier introduction. The energy diagram for this
general case with the energy change labeled is:
(b) For gaseous solvent molecules solvating a solute to form a liquid solution, we can think
about the reaction occurring two steps, first condensing the gas to a liquid and then dissolving
the solute in the liquid. The first step is exothermic, because condensing a gas to a liquid always
releases energy. The second step, dissolution of the solute, may be either exothermic or
endothermic, although not endothermic enough to make the overall two-step process
endothermic, because the energy released in condensation of solvents is quite substantial. Thus,
the overall process will be exothermic, but the solvation step may by endothermic or
exothermic. If solvation by the liquid solvent is endothermic, we have:
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ACS Chemistry Chapter 2 suggested solutions
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Aqueous Solutions and Solubility
Chapter 2
If solvation by the liquid solvent is exothermic, we have:
Problem 2.4.
(a) The solution feels cool to the touch because the process is drawing needed thermal energy
from your hand, providing the energy needed for the overall solution process. This is a sketch of
the energy diagram showing this endothermic solution process:
(b) The solution feels warm to the touch because the process is adding thermal energy to your
hand, because the overall solution process releases energy. This is a sketch of the energy
diagram showing this exothermic solution process:
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Problem 2.5.
The energy diagram for the overall solution process for the exothermic dissolution of calcium
chloride, CaCl2, in water is:
Problem 2.6.
Whether you can predict that a substance will be soluble in water by looking at its line formula
depends to some extent on how the formula is actually written. For example, the line formula
C2H6O, does not tell you how the atoms are connected, so the best you can do is predict that the
molecule will probably be polar (because of the oxygen bonded to other less electronegative
atoms) and, since it is a small molecule is likely to be reasonably soluble in water. However, if
the line formula is written as C2H5OH, you can predict that this low molecular mass alcohol
should be very soluble (actually it is miscible – mixes in any ratio) in water. If the line formula
is written as CH3OCH3, you can predict that this low molecular mass ether should be somewhat
soluble in water, but probably not as soluble as the alcohol (which can act as donor of an H to
form H bonds as well as an acceptor of H from water to form other H bonds. A structural
formula will always show the connectivity of the atoms and will, therefore, always provide the
kind of information we just discussed for simple line formulas written to help us understand the
connectivity. The regions of more positive and more negative charge in the molecule will be
relatively easy to locate and an estimate of the polar and non-polar contributions to the
solubility (or insolubility) will help you predict whether the substance will dissolve in water or
not.
Problem 2.7.
The like-dissolves-like expression reflects the fact that attractions between solute molecules and
some attractions between solvent molecules must be replaced by solute-solvent attractions when
a solution forms. If the new attractions are similar to those replaced, we expect that a solution
will be more easily formed. A polar liquid, such as water, is generally the best solvent for polar
compounds, especially those with H bonding sites. Non-polar liquids, such as hexane are better
solvents for non-polar compounds like, for example, wax, whose molecules are mostly attracted
by dispersion forces.
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Chapter 2
Problem 2.8.
(a) Two representations of the testosterone structure are:
The dipoles are easier to show on the structure on the left (with fewer atoms shown explicitly):
(b) The acetylsalicylic acid (aspirin) structure, with polar bonds circled and direction of dipoles
shown, is:
(c) The methyl salicylate (oil of wintergreen) structure, with polar bonds circled and direction of
dipoles shown, is:
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ACS Chemistry Chapter 2 suggested solutions
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Problem 2.9.
(a) The dashed lines in this diagram represent the network of hydrogen bonds that can form
between ethanol and water.
Note that there are two H bonds between H atoms in water molecules and nonbonding electron
pairs on the oxygen of ethanol, and another H bond between the –OH group’s hydrogen and a
nonbonding electron pair on the oxygen in water. The water molecules have other nonbonding
pairs on oxygen and covalently bonded hydrogen, all of which are capable of extending the
network of hydrogen bonds. In Chapter 1, we found that there are, on the average, fewer than
four H bonds per water molecule in liquid water. The three H bonds formed by ethanol fits right
into this structure and the rather small ethyl group, –CH2CH3, probably does not perturb the
structure very much, so ethanol molecules fit well into the liquid water structure and thus
account for their miscibility with water.
(b) The alcohol group in pentanol, CH3CH2CH2CH2CH2OH, can also H bond with water, as
shown for ethanol in part (a). However, the solubility of pentanol is expected to be a good deal
lower than the solubility of ethanol because the nonpolar hydrocarbon part of the molecule is
larger and has a much larger effect on the structure of the liquid water. The solubility of
pentanol in water is about 0.3 M (about 27 g·L–1).
Problem 2.10.
(a) Cyclohexane, C6H12, is a nonpolar liquid while methanol, CH3OH, is a polar liquid. The
interactions among cyclohexane molecules in the liquid are induced dipole attractions
(dispersion forces). For methanol, the largest interactions among the molecules in the liquid are
H bonds; methanol acts much like water as a solvent. In a mixture of methanol and
cyclohexane, the molecules of the two liquids lose some of their freedom of movement to make
way for each other. This type of reorganization of “unlike” molecules is unfavorable and limits
their mutual solubility, so they are not miscible. They are, however, rather soluble in one
another: 100 mL of methanol dissolves 57 g of cyclohexane. Interestingly, ethanol (with its
slightly larger alkyl group) and cyclohexane are miscible with one another, so we have to be
careful not to push the “like dissolves like” (or “unlikes do not dissolve”) simplistic idea to
extremes.
(b) Naphthalene, C10H8, is a nonpolar solid and water is a polar, hydrogen-bonding liquid. The
reorganization of the H-bonding structure of liquid water in order to accommodate naphthalene
molecules is unfavorable. In this case water molecules would lose some of their freedom of
movement while making way for naphthalene molecules.
(c) Both naphthalene, C10H8(s), and benzene, C6H6(l), are nonpolar molecules. The
reorganization involved in mixing two nonpolar compounds, both of which interact mainly by
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Chapter 2
induced dipole attractions (dispersion forces), favors the mixed state, so the solid naphthalene
dissolves in the liquid benzene.
(d) Molecules of water can form hydrogen bonds with 1-propanol, CH3CH2CH2OH, like those
shown for methanol in the solution to Problem 2.9(a). The nonpolar part of the molecule, like
those of methanol and ethanol, apparently does not disturb the liquid water structure enough to
make the interactions unfavorable. Thus mixing of the two liquids, water and 1-propanol, is not
impaired and they mix in all proportions (are miscible).
Problem 2.11.
Since gasoline, C8H18, is a nonpolar molecule, we predict that it will be pretty insoluble in
water. Its interactions with water molecules will be unfavorable as the freedom of movement of
the water molecules is impaired.
Problem 2.12.
(a) The Lewis structures for 1-hexanol, CH3(CH2)5OH, and 1,6-hexanediol, HO(CH2)6OH, with
their regions for hydrogen bonding identified, are:
1-hexanol
1,6–hexanediol, HO(CH2)6OH
(b) 1,6-hexanediol is predicted to be more soluble in water than 1-hexanol, because compounds
with multiple polar groups present more opportunity for hydrogen bonding with water
molecules.
Problem 2.13.
Recall (from Section 2.2) that, for alcohols, the solubilizing and hydrogen bonding effects of the
polar hydroxy (–OH) group becomes less and less important as the hydrocarbon portion of the
alcohol increases in size. To a first approximation, the solubility is related directly to the ratio of
carbon atoms to polar groups. In part (a), the polar group is an alcohol and in part (b), the polar
group is an amine, –NH2, which can also hydrogen bond with water. The ratios and rank orders
of solubilities are shown (#1 is the most soluble) here:
carbon/polar grp ratio
solubility rank
(a) CH3-CH2-CH2-CH2-OH
4/1
3
CH3-CH2-CH2-CH2-CH2-OH
5/1
4
4/2 = 2/1
2
HO-CH2-CH2-CH2-CH2-OH
HO- CH2-CH2-CH2-OH
3/2 = 1.5/1
1
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carbon/polar grp ratio
solubility rank
4/1
5/1
6/1
8/1
1
2
3
4
(b) CH3CH2CH2CH2NH2
(CH3)2CHCH2CH2NH2
(CH3)3CCH2CH2NH2
(CH3 CH2)3CCH2NH2
Problem 2.14.
[There are many books (especially beginning biochemistry texts) and web resources that have
the structures of these sugars. We are not interested here in the exact stereochemistry of these
molecules (for which most students are unprepared at this stage), but in the connectivity that
shows the several alcohol groups in each one. The structures here are simply modeled after that
for glucose in Figure 2.6 without showing the nonbonding electrons.]
The structure of fructose might be found in either the pyranose (six-membered ring) or furanose
(five-membered ring) form:
CH2OH
HO
HO C O H
C
C
H
H
C C
HO
OH
H H
HOJ2C
O CH2OH
C H
HO C
C C
HO
OH
H H
In both structures, there are five alcohol (–OH) groups that can H bond with water, as well as a
ring oxygen atom with nonbonding electrons that can also donate electron pairs to H bonds with
water. Looking, as in the solution to Problem 2.13, at the ratio of number of carbons to number
of alcohol groups, we have 6/5 = 1.2/1. This ratio is comparable to that for methanol (1/1) and
ethanol (2/1), both of which are miscible with water. The high solubility of fructose is expected.
The structures of lactose and sucrose are a bit more complicated, because they are disaccharides
combinations of two simpler sugars, two glucose molecules in lactose and glucose and fructose
in sucrose:
CH2 OH
CH2 OH
H
H
C O OH
HO C O O
C
C
C
C
H C C HH C C H
HO H OHH HO H OHH
lactose
CH2 OH
H
CH 2OH
HO C O O C O CH2 OH
C
C
C
H C C H C
H
C
HO H OHH H OH OH
H
sucrose
Lactose and sucrose both have eight alcohol (–OH) groups that can H bond with water, as well
as the two ring oxygen atoms and the oxygen atom bonding the rings together which have
nonbonding electrons that can also donate electron pairs to H bonds with water. (One alcohol
group from each of the simple sugars has been lost in forming the disaccharides via a reaction
we can write as R1–OH + R2–OH R1–O–R2 + H2O.) Looking again at the ratio of number of
carbons to number of alcohol groups, we have 12/8 = 1.5/1. Again, this ratio is comparable to
that for the simplest alcohols and helps to explain the high solubility of these sugars.
ACS Chemistry Chapter 2 suggested solutions
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Problem 2.15.
The more polar groups present in a molecule the more soluble it will be in water. Conversely,
the fewer the polar groups and the larger the nonpolar portion of a molecule the less soluble it
will be in water and the more soluble it will be in nonpolar solvents.
H
H H
H
H
H H H H H H
H C
H
C H H
H C
C
H
H
C
C
C
C
C
C
C
C
C
C
C
C
O
H
H
H
C
C
C
H H
H
H
H
vitamin A
H
H
O
H
O
H
H H C C
C O
C
H
C
O
O
C
H
H
O
H
vitamin C
Vitamin A has a single polar alcohol group and a substantial nonpolar part, so it will be
insoluble in water, but relatively soluble in nonpolar solvents like the fats in our body. Vitamin
A is a fat-soluble vitamin.
Vitamin C has three polar alcohol groups that can H bond with water as well as two other
oxygen atoms with nonbonding electrons that can also donate electron pairs to H bonds with
water. Using our ratio of number of carbon atoms to number of alcohol groups (as in the
solutions to Problems 2.13 and 2.14), we have 5/4 = 1.25/1, which indicates that vitamin C
should be quite soluble in water. Vitamin C is a water-soluble vitamin.
Problem 2.16.
(a) Vitamins that are soluble in the fatty tissues in our bodies will tend to stay in the body for a
substantial period of time. Vitamins that are soluble in water will dissolve in fluids like our
blood plasma and can be transported to the kidneys, removed from the blood stream, and
excreted. The fat-soluble vitamins D, E, and K can be stored in your body, although they are
slowly being used and lost and need to be part of your diet (which is part of the definition of
what constitutes a vitamin), but not in large amounts. Water soluble, vitamin B is not stored and
needs to be included in your daily diet at a larger dosage, in order to maintain an appropriate
level in your body.
(b) Vitamin A can be stored in your body, while vitamin C should be included in your daily diet
in relatively large amounts.
(c) Since only fat-soluble vitamins (vitamins A, D, E, and K) can be stored in your body, true
hypervitaminosis has been observed only for these vitamins. This does not mean that you are
entirely safe ingesting excessive doses of water-soluble vitamins. Vitamin C is a case in point.
One fad is to take massive doses of vitamin C, in order to prevent colds. The problems with this
regimen occur when the dosage is lowered. Your body becomes used to the high level and, if it
is reduced to the normal level, you begin to show the signs of scurvy, a vitamin C deficiency
disease.
Problem 2.17.
In order for the light bulb to glow when testing a solution for electrical conductivity, as in
Investigate This 2.10, ions must be present in the solution.
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Problem 2.18.
In the representations of the Fe3+(aq) and NO3–(aq) ions in the Web Companion, Chapter 2,
Section 2.3, page 1, movies, both ions have six water molecules around them in their hydration
layer. The positively charged Fe3+(aq) ions attract the negative (oxygen) ends of the water
molecules, so the water molecules are oriented with their oxygen atoms directed toward the ion.
The negatively charged NO3–(aq) ions attract the positive (hydrogen) ends of the water
molecules, so the water molecules are oriented with their hydrogen atoms directed toward the
ion.
Problem 2.19.
Solution A, ethanoyl chloride (acetyl chloride), CH3C(O)Cl, dissolved in water, must contain
ions, because it conducts and electric current. Solution B, 2-chloroethanol, ClCH2CH2OH,
dissolved in water, does not conduct an electric current, so does not contain ions. The identity of
the ions in solution A cannot be determined from these data.
Problem 2.20.
(a) Barium chloride, BaCl2(s), dissolves in water to give barium cations, Ba2+(aq), and chloride
anions, Cl–(aq).
(b) Potassium chloride, KCl(s), dissolves in water to give potassium cations, K+(aq), and
chloride anions, Cl–(aq).
(c) Sodium triphosphate, Na3PO4(s), dissolves in water to give sodium cations, Na+(aq), and
phosphate anions, PO43–(aq).
(e) Ammonium chloride, NH4Cl(s), dissolves in water to give ammonium cations, NH4+(aq),
and chloride anions, Cl–(aq).
(f) Sodium sulfide, Na2S(s), dissolves in water to give sodium cations, Na+(aq), and sulfide
anions, S2–(aq).
(g) Magnesium sulfate, MgSO4(s), dissolves in water to give magnesium cations, Mg2+(aq), and
sulfate anions, SO42–(aq).
Problem 2.21.
(a) If current flow through a conducting solution were simply caused by a flow of electrons
through the solution, we would not expect any chemical changes to occur in the solution.
However, in Investigate This 2.10, you probably observed bubbles of gas formed at the
electrodes when electric current passed through the conducting solutions. Formation of the
gases is a sign that chemistry occurs as a result of current passing through the solution. Some
mechanism for chemistry to occur, coupled with current flow, is necessary to explain the
observations. Movement of ions in the solution provides a mechanism that explains both current
flow and the observation that chemical reactions occur at the electrodes, presumably when the
ions pick up electrons from or deliver them to the electrodes. See Chapter 10, Section 10.1 for a
discussion of the chemistry that occurs at electrodes.
(b) The Web Companion, Chapter 2, Section 2.3, page 2, represents the motion of cations and
anions in a solution, so you can see how they are always in motion. You have to indicate which
ion moves which way when responding to this page, so the assumption is made that the ions
move and are the source of the electric current in the solution. If your friend is not convinced by
the argument in part (a), s/he may not be convinced by an illustration of the mechanism that is
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in the textbook. The movement of colored ions in a conducting medium, as in electrophoresis
(introduced in Chapter 8, Section 9.5), might be more convincing
Problem 2.22.
Neither solid NaCl nor solid HgCl2 will conduct electricity because any ions present are not free
to move and transport the charge. An aqueous solution of sodium chloride is a good conductor
because sodium cations, Na+(aq), and chloride anions, Cl–(aq), surrounded by polar water
molecules, are free to move about in the solution. Although HgCl2 is soluble in water, the
aqueous solution does not conduct electricity. Evidently no ions are present when this
compound goes into solution. The molecule itself stays together in solution as HgCl2(aq). This
is not the usual circumstance for what appears to be a salt, but mercury(II) chloride molecules
stay bonded together even in aqueous solution.
Problem 2.23.
This is a representation of a positively charged ion surrounded by polar molecules like our
simple ellipsoids with positive and negative ends in Chapter 1, Figures 1.15 and 1.16.
If you are the positive ion, you will feel a good deal of attraction from the partial negative
charges on the surrounding solvation sphere of polar molecules. Other positive or negative ions
elsewhere in the solution are farther away and have little influence on your behavior. They, too,
are surrounded by solvating polar solvent molecules and are little influenced by your charge.
Problem 2.24.
(a) MgBr2(s) dissolves in water to produce Mg2+(aq) and Br–(aq) ions, so the solution will
conduct an electric current (by movement of the ions).
(b) CH3OH(l) dissolves in water and forms H bonds with the water molecules, but produces no
ions, so the solution will not conduct an electric current (there are no ions to carry the current).
(c) NaOH(s) dissolves in water to produce Na+(aq) and OH–(aq) ions, so the solution will
conduct an electric current (by movement of the ions).
(d) CH3OCH3(g) dissolves in water and forms some H bonds with the water molecules, but
produces no ions, so the solution will not conduct an electric current (there are no ions to carry
the current).
(e) KNO3(s) dissolves in water to produce K+(aq) and NO3–(aq) ions, so the solution will
conduct an electric current (by movement of the ions).
(h) CH3CH2CH2CH3(g) dissolves to a very limited extent in water and produces no ions, so the
solution will not conduct an electric current (there are no ions to carry the current).
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ACS Chemistry Chapter 2 suggested solutions
Chapter 2
Aqueous Solutions and Solubility
Problem 2.25.
(a) Alkali metals (column I of the periodic table) lose their single valence electron to form
positive ions (cations) with a 1+ charge. Familiar examples are Na+ and K+ and members of a
family ( column of the table) have similar properties.
(b) Oxygen family elements (column VI of the periodic table) gain two electrons in their
valence shell to form negative ions (anions) with a 2– charge. The oxide, O2–, and sulfide, S2–,
anions are familiar examples and we expect other members of the family to have similar
properties.
(c)Alkaline earth metals (column II of the periodic table) lose their two valence electrons to
form positive ions (cations) with a 2+ charge. Familiar examples are Mg2+ and Ca2+ and we
expect members of the family to have similar properties.
(d) Halogens (elements in column VII of the periodic table) gain an electron in their valence
shell to form negative ions (anions) with a 1– charge. The chloride, Cl–, and bromide, Br–,
anions are familiar examples and we expect other members of the family to have similar
properties.
Problem 2.26.
(a) The magnesium cation, Mg2+, and bromide anion, Br–, combine to form the electrically
neutral ionic compound magnesium bromide, MgBr2.
(b) The calcium cation, Ca2+, and nitrate anion, NO3–, combine to form the electrically neutral
ionic compound calcium nitrate, Ca(NO3)2.
(c) The magnesium cation, Mg2+, and sulfate anion, SO42–, combine to form the electrically
neutral ionic compound magnesium sulfate, MgSO4.
(d) The potassium cation, K+, and oxide anion, O2–, combine to form the electrically neutral
ionic compound potassium oxide K2O.
Problem 2.27.
(a) Na2SO4 is sodium sulfate, an ionic compound of the Na+ cation and SO42– anion.
(b) MgCl2 is magnesium chloride, an ionic compound of the Mg2+ cation and Cl– anion.
(c) (NH4)2CO3 is ammonium carbonate, an ionic compound of the NH4+ cation and CO32– anion.
(d) Al2S3 is aluminum sulfide, an ionic compound of the Al3+ cation and S2– anion.
Problem 2.28.
(a) The electrically neutral formula for barium nitrate, a combination of the Ba2+ cation and
NO– anion, is Ba(NO3)2.
(b) The electrically neutral formula for ammonium phosphate, a combination of the NH4+ cation
and PO43– anion, is (NH4)3PO4.
(c) The electrically neutral formula for calcium oxide, a combination of the Ca2+ cation and O2–
anion, is CaO.
(d) The electrically neutral formula for potassium sulfate, a combination of the K+ cation and
SO42– anion, is K2SO4.
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Aqueous Solutions and Solubility
Chapter 2
Problem 2.29.
(a) MgS is magnesium sulfide, an ionic compound of the Mg2+ cation and S2– anion.
(b) Na3PO4 is sodium phosphate (or sometimes trisodium phosphate), an ionic compound of the
Na+ cation and PO43– anion.
(c) NH4NO3 is ammonium nitrate, an ionic compound of the NH4+ cation and NO3– anion.
(d) LiOH is lithium hydroxide, an ionic compound of the Li+ cation and OH– anion.
Problem 2.30.
(a) The electrically neutral formula for calcium iodide, a combination of the Ca2+ cation and I–
anion, is CaI2.
(b) The electrically neutral formula for sodium fluoride, a combination of the Na+ cation and F–
anion, is NaF.
(c) The electrically neutral formula for potassium carbonate, a combination of the K+ cation and
CO32– anion, is K2CO3.
(d) The electrically neutral formula for barium hydroxide, a combination of the Ba2+ cation and
OH– anion, is Ba(OH)2.
Problem 2.31.
The completed grid for ionic compound formation by the cation/anion pairs is:
Mg2+
NH4+
Al3+
Na+
CO32–
PO43–
F–
MgCO3
magnesium
carbonate
(NH4)2CO3
ammonium
carbonate
Al2(CO3)3
aluminum
carbonate
Na2CO3
sodium
carbonate
Mg3(PO4)2
magnesium
phosphate
(NH4)3PO4
ammonium
phosphate
AlPO4
aluminum
phosphate
MgF2
magnesium
fluoride
NH4F
ammonium
fluoride
AlF3
aluminum
fluoride
Na3PO4
sodium
phosphate
NaF
sodium
fluoride
Problem 2.32.
(a) Milk of Magnesia® is a suspension of the sparingly soluble ionic compound Mg(OH)2,
magnesium hydroxide.
(b) Epsom salt is MgSO4, magnesium sulfate.
(c) Plaster of Paris is CaSO4, calcium sulfate.
(d) Caustic soda (often called lye) is NaOH, sodium hydroxide.
(e) Soda ash is Na2CO3, sodium carbonate.
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Aqueous Solutions and Solubility
Problem 2.33.
The reaction equations for formation of the common cations or anions of these elements are:
(a) potassium
K(g) K+(g) + e–
(b) calcium
Ca(g) Ca2+(g) + 2e–
(d) sulfur
S(g) + 2e– S2–(g)
(e) bromine
Br(g) + e– Br-(g)
Problem 2.34.
By definition, ionization always involves the separation of a negative charge (an electron) and a
positive charge (the cation that remains after the electron has departed). If energy is released
when opposite charges come together (energy has a negative value for the process), then the
reverse process requires the input of energy (energy has a positive value for the process).
Mathematically, the energy of coulombic attraction, E (Q1·Q2)/d, equation (2.2), expresses the
energy of attraction of opposite charges when Q1 and Q2 have opposite signs. This attraction
must be overcome (a reversal of the mathematical sign from negative to positive) in order to
separate the opposite charges.
Problem 2.35.
We are asked whether lattice energies, such as those in Table 2.3, are consistent with electrical
attraction energies characterized by equation (2.2), E (Q1·Q2)/d. If they are consistent, we
would expect the lattice energies to be roughly proportional to the product, |Q1·Q2| for a series
of ionic compounds. The distance separating the ions, d in equation (2.2), might also affect the
lattice energies, but simple cations are about the same size, as are simple anions, so the
distances of separation are probably not very different and we will focus on the effects of
charge. We are interested only in relative values, so let us assign values of Q as the charges we
write on the ions. For our comparisons, two of the ionic compounds from Table 2.3 are chosen
here (but any other of the 1:1 cation-to-anion compounds could be chosen):
Compound
NaBr
MgS
Cation
Na+
Mg+2
Q1
Anion
1+
2+
Br–
S2–
Q2
Q1·Q2
12-
1
4
Elattice, kJ·mol–
1
751
3406
We see here that the lattice energies are roughly proportional to Q1·Q2 . An increase by a
factor of four in the product of the charges is accompanied by an increase of about 4.5 in the
lattice energy. Thus, the lattice energy data are consistent with coulombic attraction energy.
Other factors, such as the distance of separation of the ions and the geometric arrangement of
the ions with respect to one another, also affect lattice energies, but charge is most important.
Problem 2.36.
(a) Based on coulombic electrical attraction, CaBr2(s) would have greater forces of attraction
and repulsion than KBr(s), if the distance separating the charges is the same in both crystals.
The double charge on Ca2+ will result in larger coulombic forces, because they are directly
related to the size of the charges interacting.
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Chapter 2
(b) The data in Table 2.3 supports our answer in part (a). The lattice energy for CaBr2, 2176
kJ·mol–1, is about three times the lattice energy of KBr, 689 kJ·mol–1.
Problem 2.37.
The problem statement gives us Elattice = 2176 kJ·mol–1 and Eion form = 966 kJ·mol–1 for
calcium bromide, CaBr2(s). Combining these values in an energy diagram, we have:
We equate the energies for the two pathways from separated atoms to the ionic crystal to find
Extal form:
Extal form = Eion form + (– Elattice) = (966 kJ·mol–1) + (–2176 kJ·mol–1)
Extal form = –1210 kJ·mol–1 (as shown on the energy diagram)
Problem 2.38.
This is the table of lattice energies (in kJ·mol–1) we are to use for this problem.
F–
Cl–
Br–
Li+
1046
861
818
Na+
929
787
751
K+
826
717
689
(a) As we go across any row of this table, the size of the anion increases, because the size of the
ions in a group (column) of the periodic table increases as we go down the group and that is
what we are doing going from F– to Br–. We note that the lattice energy decreases across each
row, indicating that the lattice energy decreases as the size of the anion increases (and the cation
remains constant). This makes sense, because the distance between the cations and anions in the
crystal increases and we see from equation 2.2 that the energy of attraction between unlike
charges decreases as the distance between them increases.
(b) Analysis of the data for any column of this table is exactly like that in part (a) with the roles
of the cation and anion reversed. As we go down a column the size of the cation increases while
the size of the anion remains constant. The lattice energy decreases as the size of the cation
increases, because the distance between the cations and anions in the crystal increases and the
energy of attraction between unlike charges decreases as the distance between them increases.
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(c) The lattice energy of a salt decreases as the size of its ions increases. The lattice energies are
largest when the ions are the smallest.
(d) For CsI, the cation, Cs+, is larger than any of the alkali metal cations in the table and the
anion, I–, is larger than any of the anions in the table. The lattice energy for CsI should be even
lower than the lowest value in the table (that for KBr) and certainly considerably smaller than
that of NaCl. The lattice energy for CsI is 604 kJ·mol–1.
Problem 2.39.
This is the energy diagram for the formation of one mole of ionic crystals of MgCl2 that we are
to use for this problem.
(a) The lattice energy, Elattice, for MgBr2 is the energy required to convert one mole of the
compound from its solid crystal to separated ions in the gas phase. On the diagram, we see that
2524 kJ·mol–1 is released when the gaseous ions come together to form the ionic solid.
Therefore, Elattice = 2524 kJ·mol–1, the amount of energy to get the ions apart.
(b) From the energy diagram, the energy required to change gaseous atoms to gaseous ions,
Eion form = 1490 kJ·mol–1.
(c) The energy of formation of ionic crystals from gaseous atoms, Extal form, is the difference
between the two energies in parts (a) and (b): Extal form = –1034 kJ·mol–1 as shown on the
diagram.
(d) This table shows the comparison between these energies for MgBr2 and those for NaCl from
Figure 2.14.
Ionic
compound
MgCl2
NaCl
Elattice
kJ·mol–1
2524
787
Eion form
kJ·mol–1
1490
145
Extal form
kJ·mol–1
1034
642
Note that the large difference between the lattice energies is largely a result of the much higher
energy required to remove two electrons from magnesium atoms to form the doubly charged
Mg2+(g) cations compared to removal of only one electron to form the Na+(g) cation. This large
amount of energy is about 60% of the energy released when the coulombic attraction of the
doubly-charged cations and singly-charged anions brings them together to form the ionic solid.
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Chapter 2
Problem 2.40.
(a) We can use the information that the energy required to remove electrons from gaseous
silver atoms to form gaseous silver cations is 731 kJ·mol–1 and that 296 kJ·mol–1 of energy is
released when gaseous iodine atoms gain electrons to form gaseous iodide anions, to construct
an energy diagram to find the energy change for the net reaction: Ag(g) + I(g) Ag+(g) + I–(g).
The loss of an electron by each Ag(s) atom and gain of an electron by each I(g) atom is
represented on the diagram by the slanting arrow showing that the electron lost by a silver atom
is gained by an iodine atom, so there is no net loss or gain of electrons in the overall reaction
The diagram shows that the net reaction energy is 435 kJ·mol–1.
(b) Use the result from part (a) and the lattice energy for AgI(s) crystals, Elattice = 887 kJ·mol–1,
to draw an energy diagram analogous to Figure 2.14 and use it to find Extal form for the
formation of ionic crystals of AgI(s) from the gaseous atoms.
Problem 2.41.
In the energy diagrams we have seen in the textbook or constructed for the formation of solid
ionic compounds, we have seen this relationship among the energies represented on the
diagrams:
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Extal form = Eion form + (– Elattice) = Eion form – Elattice
For KBr(s) crystals, the problem statement says that Elattice = 689 kJ·mol–1. The formation of
separate gaseous atoms of potassium, K(g), and bromine, Br(g), from the ionic crystal requires
594 kJ·mol–1, so Extal form (energy change for the reverse of process, forming the ionic crystal
from the gaseous atoms) = –594 kJ·mol–1. Substituting these values in the preceding equation,
gives:
–594 kJ·mol–1 = Eion form – (689 kJ·mol–1)
Eion form = 95 kJ·mol–1
An energy diagram incorporating these data is shown here (not quite to scale, so the numeric
values can be included with the arrows representing them):
Problem 2.42.
As in the solution for Problem 2.41, to find the lattice energy, Elattice, for magnesium fluoride,
MgF2(s), we use the energy relationship:
Extal form = Eion form + (– Elattice) = Eion form – Elattice
Substituting the values given in the problem statement, Extal form = –1424 kJ·mol–1 and Eion form
= 1533 kJ·mol–1, we get:
–1424 kJ·mol–1 = (1533 kJ·mol–1) – Elattice
Elattice = 2957 kJ·mol–1
An energy diagram incorporating these data is shown here:
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Chapter 2
Problem 2.43.
(a) When ammonium acetate, NH4C2H3O2 [= (NH4+)(C2H3O2–)], is dissolved in water, the
mixture becomes quite cold. This observation means that the dissolution process is taking
thermal energy from its surroundings, the molecules in the solution, the container, and your
hand, if you are holding the container. A reaction that requires an input of thermal energy,
Ereaction > 0, is endothermic.
(b) The energy change for the process of dissolving ionic solutes in water can be broken into
two parts. Lattice energy is required to break the coulombic electrical attractions between
cations and anions in the lattice while hydration energy is released as water molecules surround
these ions and are attracted to them by coulombic attractions. The difference between these two
energies determines whether thermal energy will be absorbed (endothermic) or released
(exothermic) by the dissolving process. For ammonium acetate, the dissolving process absorbs
energy, so breaking the lattice attractions must take more energy than is gained back by
hydration of the ions. An energy diagram that represents this case is:
[Note that water is shown separately with the solid and gaseous ions as a reminder that the
system as a whole contains the water into which the solid dissolves and which hydrates the ions.
Sometimes, as in Figures 2.15 and 2.16 in the text and the solution to Problem 2.44 below, the
water is omitted for simplicity, but this is probably not a good idea.]
(c) The ions present in a solution of ammonium acetate are the same as the ions in the solid
crystal, ammonium cation and acetate anion, which we write as NH4+(aq) and C2H3O2–(aq) to
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signify that the ions are hydrated, that is, surrounded by polar water molecules that are attracted
by the ionic charges.
(d) The molecular level interactions of hydrated ions are represented in Figure 2.9 by showing
the negative oxygen end of water molecules oriented toward the positive cations and the
positive hydrogen end of water molecules oriented toward the negative cations. We expect
similar orientations and interactions for the NH4+ and C2H3O2– ions, but there is an added factor
for these ions, because they can hydrogen bond with the water molecules and form even more
directed interactions. Some of these H bonds are illustrated here:
H
O H
H
H
H O
H
H N H
H
H
O H
H
O H
O
H
H
H
O
O
CH3 C O
H O
H
H
H O
Problem 2.44.
(a) Use the data in Table 2.3 for LiCl, Elattice = 861 kJ·mol–1 and Ehydration = –898 kJ·mol–1, to
sketch an energy diagram for the dissolution process (broken into two steps). Use the diagram to
find that Edissolve = –37 kJ·mol–1. Energy is released, so the dissolution reaction is exothermic.
(b) Use the data in Table 2.3 for KBr, Elattice = 689 kJ·mol–1 and Ehydration = –670 kJ·mol–1, to
sketch an energy diagram for the dissolution process (broken into two steps). Use the diagram to
find that Edissolve = 19 kJ·mol–1. Energy is required, so the dissolution reaction is endothermic.
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Chapter 2
Problem 2.45.
We are asked to try to explain why lithium sulfate, Li2SO4, is quite soluble in water (261 g·L–1)
while calcium sulfate, CaSO4, is essentially insoluble (4.9 mg·L–1). To see what factors might be
responsible for such a difference, let’s compare the data for a pair calcium ionic compounds in
Table 2.3:
Compound
Elattice
kJ·mol–1
Ehydration
kJ·mol–1
CaCl2(s)
2260
–2337
CaCO3(s)
2804
–2817
In CaCl2(s) and CaCO3(s), we have the cation with a 2+ charge and anions with a 1– charge and
a 2– charge, respectively. CaCl2(s) is quite soluble in water, as you found in Investigate This
2.22, and CaCO3(s), marble or chalk, is quite insoluble. As a first approximation, these
compounds are rather like Li2SO4(s) and CaSO4(s), where we have the anion with a 2– charge
and cations with a 1+ charge and 2+ charge, respectively. The ion-ion interactions in the solids
and ion-polar solvent interactions in aqueous solution should be roughly the same for an ionic
compound with 1+ cations and a 2– anion as for a compound with a 2+ cation and 1– anions. If
this supposition is correct, we see from the comparison in this table that the hydration energy
should substantially outweigh the lattice energy, for a 2:1 ionic compound like Li2SO4(s)
compared to a 2:2 ionic compound like CaSO4(s). As a first suggestion about the factor
responsible for the high solubility of Li2SO4(s) compared to CaSO4(s), we would probably say
that the hydration energy (due to solvation of the ions) favors dissolution of Li2SO4(s),
Edissolve = –67 kJ·mol–1, more than CaSO4(s), Edissolve = –13 kJ·mol–1. This is a large factor in
this case, but we have to be careful making to much of this argument, since we know that there
are soluble ionic compounds whose dissolution is endothermic, so energetics cannot be the
whole picture.
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Problem 2.46.
We asked to describe or represent in three different ways what happens when sodium sulfate,
Na2SO4(s), dissolves in water. First we describe the dissolution in words. Water molecules are
attracted to the Na+ and SO42– ions in the solid crystal lattice, especially those at the edges and
corners of the crystals. The negative oxygen ends of the water molecules are attracted to the Na+
and positive hydrogen ends in water are attracted to SO42-. These ion-dipole attractions compete
with the ion-ion attractions, lattice energy, that holds Na+ and SO42- in the crystal. In the case of
a soluble ionic compound like Na2SO4(s), the ion-dipole attractions are finally successful in the
tug of war and the ions are broken away from the crystal. Once the ions are broken away from
the crystal, more water molecules surround each ion creating a hydration layer. These hydration
layers create a shield, making it difficult for hydrated ions that are oppositely charged to get too
close to each other. In our example, hydrated Na+ and hydrated SO42- do not interact much with
one another (until their concentrations become so high that they are forced close to one another
because there are so many of them).
An ionic equation that succinctly represents this process is usually written as:
Na2SO4(s) 2Na+(aq) + SO42-(aq)
Although water is obviously involved in the process, it is usually left out of such equations,
because its stoichiometry is not easily represented. If we just add H2O(l) to the reactant side of
the equation, it might be interpreted as one molecule of water for each formula unit of Na2SO4
dissolved, and that would be misleading. We just have to remember that water is an active
reactant in the dissolution process, not an inert by-stander.
A molecular level representation of the dissolution process is shown in this very rough sketch
showing a two-dimensional crystal interacting with solvent (water):
S
+
Na+ cation
2–
SO4 2– anion
S
solvent, water molecules
+
+
2–
2–
+
+
2–
+
S + S
S
S
+ S
S
2– S
+ S
+
+
S
S S
S
S
2–
S
2–
+
S
+
S
S
S
S
For simplicity, the many water molecules that are not interacting with the ions are omitted and
the polar ends of the water molecules are not represented. See Figure 2.9(b) for a representation
of the orientation of the water molecules with respect to the cations and anions.
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Chapter 2
Problem 2.47.
(a) The mixing process described in this problem is represented in this table, modeled after
Table 2.4 in the text:
Before Mixing
Positive ion(s)
Negative ion(s)
Conductivity?
Precipitate?
Na3PO4
solution
Na+ (aq)
PO4-(aq)
yes
CaBr2
solution
Ca2+(aq)
Br–(aq)
yes
After Mixing
Na3PO4 and CaBr2
Na+(aq)+ Ca2+(aq)
PO4- (aq)+ Br– (aq)
yes
yes
(b) After mixing, two new combinations of cations and anions are possible: NaBr and
Ca3(PO4)2. Our solubility rules say that ionic compounds of alkali metal cations and halide
anions are soluble, so sodium bromide, NaBr, is likely to be a soluble compound. Calcium
phosphate, Ca3(PO4)2, must be the precipitate and it fits our solubility rule that says that ionic
compounds of multiply-charged cations and anions are likely to be insoluble.
(c) This diagram is a simple molecular level representation of this mixing and reaction
(precipitate formation) with only enough of each ion shown to represent the stoichiometry of the
reaction:
Note that each of the individual solutions that are mixed contains ions and there are also ions in
the solution remaining after the solid has precipitated. The presence of the ions in all three
solutions explains why they are all conduct an electric current, as shown in the table above.
(d) The complete ionic equation that represents the precipitation reaction involves all four ions:
6Na+(aq) + 2PO4–(aq) + 3Ca2+(aq) + 6 Br–(aq) 6Na+(aq) + 6Br--(aq) + Ca3(PO4)2(s)
(e) The net ionic equation that represents the precipitation reaction involves only the two ions
that react to form the precipitate (the spectator ions are omitted):
3Ca2+(aq) + 2PO4–(aq) Ca3(PO4)2(s)
Problem 2.48.
(a) When aqueous solutions of potassium chloride, KCl, and sodium bromide, NaBr, are mixed,
the solution contains all four ions, Na+(aq), K+(aq), Cl–(aq), and Br–(aq), and no precipitate is
formed. We can conclude that both NaCl(s) and KBr(s) are water soluble, since the solids could
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have formed from this mixture, but did not. We also note that all of these ionic compounds are
combinations of an alkali metal cation and a halide anion, which our solubility rules predict will
be soluble.
(b) This molecular level representation, similar to Figure 2.17, illustrates the result when the
two solutions are mixed.
(c) The complete ionic equation representing what happens when the two solutions are mixed
involves all four ions:
K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq)
The ions are present in two separate solutions on the left-hand (reactant) side of this equation
and together in a single solution on the right-hand (product) side. The separation on the left is
not represented in this standard ionic equation, but we can amend it slightly to suggest the
separation by bracketing the separate solution components on the left:
[K+(aq) + Cl-(aq)] + [Na+(aq) + Br-(aq)] K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq)
Since there is no net reaction (no apparent reaction of any kind), there is no net ionic equation
for this mixing. In a sense, all the ions are spectator ions.
Problem 2.49.
The solubility rules indicate that multiple charged cations and anions tend to form insoluble
ionic compounds. Barium sulfate, BaSO4, falls into this category. The mixture that doctors use
to x-ray the gastrointestinal (GI) tract is a suspension of insoluble solid barium sulfate in water.
The solubility of the solid ionic compound is so low that the suspension contains only a tiny
amount of Ba2+ cation. Its concentration is below the toxic level of Ba2+, so the patient is not
harmed (except by having to swallow a substantial amount of chalky suspension).
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Chapter 2
Problem 2.50.
We are asked to suggest a sequence of selective precipitation reactions to separate Ag+, Ba2+,
and Fe3+ from solution in which all three cations are present. The information we have to work
with is in this table:
Cation
Test Solution
Ag+(aq)
Ba2+(aq)
Fe3+(aq)
NaCl
ppt
no ppt
no ppt
NaOH
ppt
no ppt
ppt
Na2SO4
no ppt
ppt
no ppt
The objective of our sequence of additions of the test solutions (which we might also call the
separation reagents) is to precipitate the cations one at a time without forming any mixtures of
precipitates that contain more than one cation. For example, if we were to add the NaOH test
solution to the original mixture of cations, both the Ag+(aq) and Fe3+(aq) cations would react to
form precipitates, so the solid product would be a mixture of solids and we would not have
separated the cations from one another. However, if the Ag+(aq) cation had already been
removed from the solution, addition of the NaOH test solution would result in a precipitate
containing the Fe3+(aq) cations and we would have the Fe3+(aq) cations as a separate precipitate.
Our strategy is to look for a test solution that will precipitate only one of the cations, use it to
precipitate that cation and then use a second test solution that will precipitate one or the other of
the remaining cations.
There are two ways to start our sequence. (1) We can add the NaCl test solution to precipitate
the Ag+(aq) cation as AgCl(s), leaving behind a solution containing the Ba2+(aq) and Fe3+(aq)
cations [as well as the Na+(aq) cations added with the test solution]. Or, (2) we can begin by
adding the Na2SO4 test solution to precipitate the Ba2+(aq) cations as BaSO4(s), leaving behind
a solution containing the Ag+(aq) and Fe3+(aq) cations [as well as the Na+(aq) cations added
with the test solution].
To pick the next step in sequence (1), consider this table that tells us how our test solutions react
with the remaining cations:
Cation
2+
Test Solution
Ba (aq)
Fe3+(aq)
NaOH
no ppt
ppt
Na2SO4
ppt
no ppt
We have two choices, both of which will separate these two cations from one another. In the
pathway that we will label (1a), we add NaOH test solution to the solution we get after
removing the AgCl(s) in order to precipitate the Fe3+(aq) as Fe(OH)3(s). The solution now
contains the Ba2+(aq) [and the Na+(aq) cations added with the test solutions]. The cations are
now separated as two precipitates, AgCl(s) and Fe(OH)3(s), with the remaining cation, Ba2+(aq),
in solution. If we need to have the Ba2+ separated from all the Na+(aq) also in the solution, we
can add the Na2SO4 test solution to precipitate it as BaSO4(s). The net ionic reactions for the
three steps of pathway/sequence (1a) [and the alternative (1b)] are:
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Pathway (1a)
Ag+(aq) + Cl–(aq)
3+
–
Fe (aq) + 3OH (aq)
AgCl(s)
Fe(OH)3(s)
Pathway (1b)
Ag+(aq) + Cl–(aq)
2+
2–
Ba (aq) + SO4 (aq)
AgCl(s)
BaSO4(s)
Ba2+(aq) + SO42–(aq) BaSO4(s)
Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)
The net ionic reactions for the three steps of pathway (2) are given here. You can reason
through the logic that leads to this sequence and the reason why there is no alternative
possibility for this sequence.
Pathway (2)
Ba2+(aq) + SO42–(aq) BaSO4(s)
Ag+(aq) + Cl–(aq)
3+
–
Fe (aq) + 3OH (aq)
AgCl(s)
Fe(OH)3(s)
Problem 2.51.
(a) When an aluminum nitrate, Al(NO3)3, solution is mixed with a sodium oxalate, Na2C2O4,
solution the solution formed contains aluminum, Al3+(aq), and sodium, Na+(aq), cations, and
nitrate, NO3–(aq), and oxalate, C2O42–(aq), anions. A precipitate forms from this mixed solution.
The possibilities for this solid ionic compound are Al2(C2O4)3(s) and NaNO3(s). Our solubility
rules indicate that ionic compounds with an alkali metal cation and nitrate anion are likely to be
soluble, so the precipitate is not NaNO3(s). The precipitate, Al2(C2O4)3(s), is an ionic compound
formed from a multiply-charged cation and a multiply-charged anion, which our rules suggest is
likely to be insoluble.
(b) The net ionic equation for the reaction that forms Al2(C2O4)3(s) has the aqueous ions with
appropriate stoichiometric coefficients as reactants and the solid as product:
2Al3+(aq) + 3C2O42-(aq) Al2(C2O4)3(s)
Problem 2.52.
When a solution of lithium nitrate, LiNO3, is mixed with a solution of sodium phosphate,
Na3PO4, the solution contains lithium, Li+(aq), and sodium, Na+(aq), cations and nitrate, NO3–
(aq), and phosphate, PO43–(aq), anions. A white precipitate is observed to form from this
mixture. The possibilities for this solid ionic compound are Li3PO4(s) and NaNO3(s). Our
solubility rules indicate that ionic compounds with an alkali metal cation and nitrate anion are
likely to be soluble, so the precipitate is not NaNO3(s). The precipitate must be Li3PO4(s).
Although ionic compounds with alkali metal cations are generally expected to be soluble, note
that, in Worked Example 2.34, lithium is discussed as an exception to the general rule.
(b) The net ionic equation for the precipitation reaction is:
3Li+(aq) + PO43–(aq) Li3PO4(s)
Problem 2.53.
[Note that there seems to be no symbol for the equilibrium double arrow that is common to both
PC and Macintosh computer platforms, so the symbol “ ” is used in these notes to make them
cross-platform compatible.]
When the reactants and products are separated by a forward arrow over a backward arrow, “ ”
in a reaction equation, it means that the reaction can or is going in both directions. That is, the
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reactant species are combining to yield products and the product species are combining to give
reactants and both processes are occurring simultaneously.
Problem 2.54.
(a) When a solution of cadmium chloride, CdCl2, is mixed with a solution of ammonium
sulfide, (NH4)2S, the solution contains cadmium, Cd2+(aq), and ammonium, NH4+(aq), cations
and chloride, Cl–(aq), and sulfide, S2–(aq), anions. A yellow-orange precipitate is observed to
form from this mixture. The possibilities for this solid ionic compound are CdS(s) and
NH4Cl(s). Our solubility rules indicate that ionic compounds with a singly-charged cation and
halide anion are likely to be soluble, so the precipitate is not NH4Cl(s). The precipitate must be
CdS(s). (Cadmium sulfide -- cadmium yellow -- is used as a pigment in paints.)
(b) The net ionic equation for the precipitation reaction is:
Cd2+(aq) + S2-(aq) CdS(s)
Problem 2.55.
The objective of this problem is to use our solubility rules, including the few exceptions we
have noted, to predict the products of mixing four pairs of aqueous solutions and to write
complete and net ionic equations for those reactions that produce a precipitate.
(a) barium chloride(aq) + sodium sulfate(aq)
The mixed solution contains barium, Ba2+(aq), and sodium, Na+(aq), cations and chloride, Cl–
(aq), and sulfate, SO42–(aq), anions. Our solubility rules indicate that an ionic compound with a
multiply-charged cation and anion is likely to be insoluble. Thus, we can predict that BaSO4(s)
is insoluble and write this complete and net ionic reaction equation:
Ba+2(aq) + 2Cl-(aq) + 2Na+(aq) + SO42-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq)
Ba+2(aq) + SO42-(aq)
BaSO4(s)
(b) silver nitrate(aq) + magnesium chloride(aq)
The mixed solution contains silver, Ag+(aq), and magnesium, Mg2+(aq), cations and nitrate,
NO3–(aq), and chloride, Cl–(aq), anions. Our solubility rules indicate that ionic compounds with
a halide or nitrate anion are likely to be soluble, so the initial instinct is to predict that no
precipitate will form. However, when we recall that the silver ion is a notable exception to the
rules, we predict that silver chloride, AgCl(s), is insoluble and write this complete and net ionic
reaction equation:
2Ag+(aq) + 2NO3–(aq) + Mg2+(aq) + 2Cl–(aq) 2AgCl(s) + Mg2+(aq) + 2NO3–(aq)
2Ag+(aq) + 2Cl–(aq)
2AgCl(s)
(c) strontium nitrate(aq) + potassium nitrate(aq)
The mixed solution contains strontium, Sr2+(aq), and potassium, K+(aq), cations and nitrate,
NO3–(aq), anions. No new products are possible, because there is only one anion that is
common to both soluble starting compounds. Thus we can write:
Sr2+(aq) + 3NO3-(aq) + K+(aq) + NO3-(aq) NO APPARENT REACTION
(d) ammonium phosphate(aq) + calcium bromide(aq)
The mixed solution contains barium, NH4+(aq), and calcium, Ca2+(aq), cations and phosphate,
PO43–(aq), and bromide, Br–(aq), anions. Our solubility rules indicate that an ionic compound
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with a multiply-charged cation and anion is likely to be insoluble. Thus, we can predict that
Ca3(PO4)2(s) is insoluble and write this complete and net ionic reaction equation:
6NH4+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Br-(aq) Ca3(PO4)2(s) + 6NH4+(aq) + 6Br-(aq)
3Ca2+(aq) + 2PO43-(aq)
Ca3(PO4)2(s)
Problem 2.56.
The objective of this problem is to use your knowledge of the solubility rules to propose a way
to prepare (synthesize) several solid ionic compounds by precipitation reactions from mixtures
of soluble ionic compounds.
(a) The net ionic reaction for preparing BaSO4(s) is:
Ba2+(aq) + SO42–(aq) BaSO4(s)
Mixing aqueous solutions of the soluble ionic compounds Ba(NO3)2 (nitrates are soluble) and
Na2SO4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction.
(b) The net ionic reaction for preparing AgCl(s) is:
Ag+(aq) + Cl–(aq) AgCl(s)
Mixing aqueous solutions of the soluble ionic compounds AgNO3 (nitrates are soluble) and
KCl (alkali metal halide salts are soluble) will provide the reactant ions necessary for this
reaction.
(c) The net ionic reaction for preparing Ca3(PO4)2(s) is:
3Ca2+(aq) + 2PO43- (aq) Ca3(PO4)2 (s)
Mixing aqueous solutions of the soluble ionic compounds CaCl2 (halides, except silver, are
soluble) and K3PO4 (alkali metal salts are soluble) will provide the reactant ions necessary for
this reaction.
(d) The net ionic reaction for preparing CaC2O4(s) is:
Ca2+ (aq) + C2O42-(aq) CaC2O4 (s)
Mixing aqueous solutions of the soluble ionic compounds CaCl2 (halides, except silver, are
soluble) and K2C2O4 (alkali metal salts are soluble) will provide the reactant ions necessary for
this reaction.
Problem 2.57.
Since the concentration of a solution (assuming it is well mixed after preparation) is uniform
throughout the entire volume, spilling some of it will not change the concentration of the
remaining solution. Concentration is an intensive variable (like density or temperature) that does
not depend upon the amount of solution you have.
Problem 2.58.
(a) If it is properly labeled, the solution in a bottle labeled, "0.5 M CaCl2" contains 0.5 moles of
CaCl2(aq) per liter of solution. There are 0.5 moles of Ca2+(aq) per liter of solution and 1 mole
of Cl–(aq) per liter of solution.
(b) If a 0.5-L bottle is about half full, then it contains approximately 0.25 L of 0.5 M CaCl2
solution. Use the definition of molarity, moles of solute per liter of solution, to determine the
number of moles of the solute in about one-quarter liter of solution:
ACS Chemistry Chapter 2 suggested solutions
27
Aqueous Solutions and Solubility
Chapter 2
mol CaCl2 in 0.25 L = (0.25 L)
0.5 mol
1L
= 0.13 mol CaCl2
(c) Use the molar formula mass of CaCl2, 111 g·mol–1, to convert the number of moles of CaCl2
from part (b) to mass:
0.13 mol CaCl2 = (0.13 mol CaCl2)
111 g
1 mol CaCl 2
= 14 g CaCl2
Problem 2.59.
(a) Counting the number of atoms of each element in the molecular structure for vitamin C in
Problem 2.15 gives the molecular formula C6H8O6.
(b) Find the molar mass of vitamin C by summing the masses of each element in a mole of the
compound:
12.01 g
(6 mol C)
= 72.06 g
1 mol C
1.008 g
= 8.06 g
(8 mol H)
1 mol H
16.00 g
= 96.00 g
(6 mol O)
1 mol O
molar mass vitamin C = 176.12 g = 176 g (accurate enough for the rest of the data)
(c) To find the number of moles of vitamin C in a tablet that contains 500-mg of the vitamin, we
use the molar mass to convert this mass to moles. The actual mass of vitamin in a vitamin tablet
is only accurate to a few percent, so we will assume that the mass of vitamin C in the tablet is
about 0.50 g [= (500 mg) 1 g 1000 mg ], with an implied accuracy of 1 part in 50 or about 2%.
0.50 g vit C = (0.50 g vit C)
1 mol vit C
176 g
= 0.0028 mol vit C = 2.8
10–3 mol vit C
(d) To find the number of molecules of vitamin C in a tablet that contains 500-mg of the
vitamin, we use Avogadro’s number to convert number of moles from part (c) to number of
molecules:
2.8
–3
10 mol vit C = (2.8
= 1.7
6.02 10 23 molec
10 mol vit C)
1 mol
–3
1021 molec vit C
Problem 2.60.
(a) To convert moles of aspartame to mass, we need the molar mass of aspartame, C14H18N2O5:
12.01 g
(14 mol C)
= 168.14 g
1 mol C
1.008 g
(18 mol H)
= 18.14 g
1 mol H
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14.01 g
1 mol N
16.00 g
(5 mol O)
1 mol O
(2 mol N)
= 28.02 g
= 80.00 g
molar mass aspartame = 294.30 g = 294 g (accurate enough for the rest of the data)
The mass of aspartame in 2.5 mol of aspartame is:
294 g
2.5 mol = (2.5 mol)
= 735 g = 7.4
1 mol aspartame
102 g = 0.74 kg
(b) To convert moles of aspirin to mass, we need the molar mass of aspirin, C9H8O4:
12.01 g
(9 mol C)
= 108.09 g
1 mol C
1.008 g
(8 mol H)
= 8.06 g
1 mol H
16.00 g
(4 mol O)
= 64.00 g
1 mol O
molar mass aspirin = 180.15 g = 180 g (accurate enough for the rest of the data)
The mass of aspirin in 0.040 mol of aspirin is:
0.040 mol = (0.040 mol)
180 g
= 7.2 g
1 mol aspirin
(c) To convert number of molecules of cholesterol, C27H46O, to mass, we first need to convert
number of molecules to number of moles (using Avogadro’s number) and then number of moles
to mass [using the molar mass, as in parts (a) and (b) of the problem]. The number of moles in
2.5 1023 molecules of cholesterol is:
2.5
1023 molec = (2.5
1023 molec)
1 mol
23
6.02 10 molec
= 0.42 mol
The molar mass of cholesterol is:
12.01 g
(27 mol C)
= 324.27 g
1 mol C
1.008 g
(46 mol H)
= 46.37 g
1 mol H
16.00 g
(1 mol O)
= 16.00 g
1 mol O
molar mass cholesterol = 386.64 g = 387 g (accurate enough for the rest of the data)
The mass of cholesterol in 0.42 mol cholesterol is:
387 g
= 163 g = 1.6
0.42 mol = (0.42 mol)
1 mol cholesterol
102 g = 0.16 kg
ACS Chemistry Chapter 2 suggested solutions
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Aqueous Solutions and Solubility
Chapter 2
(d) To convert number of molecules of caffeine, C8H10N4O2, to mass, we first need to convert
number of molecules to number of moles (using Avogadro’s number) and then number of moles
to mass [using the molar mass, as in parts (a) and (b) of the problem]. The number of moles in
1.2 1022 molecules of caffeine is:
1.2
1022 molec = (1.2
1022 molec)
1 mol
6.02 10 23 molec
= 0.020 mol
The molar mass of caffeine is:
12.01 g
(8 mol C)
= 96.08 g
1 mol C
1.008 g
(10 mol H)
= 10.08 g
1 mol H
14.01 g
(4 mol N)
= 56.04 g
1 mol N
16.00 g
(2 mol O)
= 32.00 g
1 mol O
molar mass caffeine = 194.20 g = 194 g (accurate enough for the rest of the data)
The mass of caffeine in 0.020 mol of caffeine is:
194 g
= 3.9 g
0.020 mol = (0.020 mol)
1 mol caffeine
Problem 2.61.
To find the number of atoms of carbon in 5 mg of niacin, we need a strategy to get from the
mass of a substance to the number of atoms of an element in that mass of the substance. This
problem involves understanding and applying the concept of the mole. One way to plan your
work is to reason backward from the desired answer, using the information given in the
problem, the mass of niacin, 5 mg, and its molecular structure:
H
H
H
C
C
C
N
O
C
C
C
H
N
H
H
The molecular formula for niacin, from its structure, is C6H6N2O, so each molecule of niacin
contains six atoms of carbon. (Note that you might write the formula a different way, ON2C6H6
for example, and that is OK. The format we used, C6H6N2O, is the conventional way chemists
write formulas for carbon-containing compounds: carbon first, hydrogen second, and then all
other elements in alphabetical order. Any format that shows the correct number of each atom in
the molecule is fine for stoichiometric problems like the one here.) If we know the number of
molecules of niacin in 5 mg of niacin, there are six times as many atoms of carbon in the
sample. If we know the number of moles of niacin in the sample, we use Avogadro’s number to
get the number of molecules in the sample. Finally, use the mass (in grams) of the sample of
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Aqueous Solutions and Solubility
niacin and its molar mass (from the formula) to get the number of moles of niacin. Going
forward through the solution, our strategy is summarized in this sequence:
mass moles molecules atoms
To find the number of moles of niacin in a 0.005 g [= (5 mg) 1 g 1000 mg ] sample, we need
the molar mass of niacin:
12.01 g
(6 mol C)
= 72.06 g
1 mol C
1.008 g
(6 mol H)
= 6.05 g
1 mol H
14.01 g
(2 mol N)
= 28.02 g
1 mol N
16.00 g
(1 mol O)
= 16.00 g
1 mol O
molar mass niacin = 122.13 g = 122 g (accurate enough for the rest of the data)
The number of moles of niacin is:
0.005 g niacin = (0.005 g niacin)
1 mol niacin
=4
122 g
10–5 mol niacin
The number of molecules of niacin is:
4
–5
10 mol niacin = (4
= 2.4
6.02 10 23 molec
10 mol niacin)
1 mol
–5
1019 molec niacin
The number on atoms of carbon in the sample is:
2.4
1019 molec niacin = (2.4
= 1.4
1019 molec niacin)
6 atoms carbon
1 molec niacin
1020 atoms carbon
Problem 2.62.
In order to determine the molar concentration, mol·L–1, of DNA in the given bacterium, we need
to know the number of moles of DNA and the volume of the bacterium in which it is found. We
are told that there is one molecule of DNA in the bacterium, so we can use Avogadro’s number
to find the number of moles of DNA:
1 mol
= 1.7 10–24 mol
1 molec DNA = (1 molec DNA)
23
6.02 10 molec
We are told that the bacterium is spherical and has a diameter, d, of 1 10–6 m. The formula for
the volume of a sphere is 4 3 r3 = 1 6 d3 (where radius, r, = d/2). We want the volume in
liters and one way to get it is to recall that 1 L = 1 dm3. Therefore, if we express the diameter of
the bacterium in decimeters, 1 dm = 10–1 m, the volume we get will be in dm3 (= liters).
ACS Chemistry Chapter 2 suggested solutions
31
Aqueous Solutions and Solubility
Chapter 2
(Another approach is to calculate the volume in cubic centimeters, which are equal to milliliters,
and thence to liters.) The required diameter of the bacterium is:
1 dm
d = 1 10–6 m = (1 10–6 m)
= 1 10–5 dm
10 –1 m
The volume, V, of the bacterium is:
V = 1 6 d3 = 1 6 (1 10–5 dm)3 = 5 10–16 dm3 = 5 10–16 L
The DNA concentration is:
mol DNA 1.7 10 –24 mol
[DNA] =
=
= 3 10–9 M
V
5 10 –16 L
Problem 2.63.
To find the number of sodium ions in 50 mL of blood serum, we need a strategy to get from the
volume of a solution to the number of ions of an element in that volume of the solution. One
way to plan your work is to reason backward from the desired answer, using the information
given in the problem, the volume of solution (serum), 50 mL, and its molarity, 0.14 M in NaCl.
If we know the number of moles of sodium ion in the sample, we use Avogadro’s number to get
the number of sodium ions in the sample. We know that every mole of NaCl dissolved in the
serum produces a mole of sodium ions, Na+(aq). Finally, use the molarity of the serum sample
and its volume (in liters) to get the number of moles of NaCl dissolved. Going forward through
the solution, our strategy is summarized in this sequence:
volume solution moles NaCl moles ions number ions
The volume of the serum sample is:
volume = 50 mL = (50 mL)
1L
1000 mL
= 0.050 L
The number of moles of NaCl in the serum sample is:
0.14 mol
0.050 L = ( 0.050 L)
= 7.0 10–3 mol
1L
The number of moles of sodium ions in the serum sample is:
1 mol Na +
–3
–3
7.0 10 mol NaCl = (7.0 10 mol NaCl)
= 7.0
1 mol NaCl
The number of molecules of sodium ions in the serum sample is:
6.02 10 23 molec
–3
+
–3
+
7.0 10 mol Na = (7.0 10 mol Na )
1 mol
10–3 mol Na+
= 4.2
1021 Na+
Problem 2.64.
For each solution in this problem we are asked to calculate the mass of solute present in a
known volume of solution of a known molar concentration. The route map shown in Worked
Example 2.50 (read from bottom to top) shows the sequence of steps required to get these
masses:
volume solution moles solute mass solute
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ACS Chemistry Chapter 2 suggested solutions
Chapter 2
Aqueous Solutions and Solubility
These steps can be combined:
mol solute
g solute
1 L solution 1 mol solute
In each case, we will need the molar mass of the solute and will calculate this first.
mass solute, g = (volume solution, L)
(a) The molar mass of K2Cr2O7 is:
39.10 g
(2 mol K)
= 78.20 g
1 mol K
52.00 g
= 104.00 g
(2 mol Cr)
1 mol Cr
16.00 g
= 112.00 g
(7 mol O)
1 mol O
molar mass K2Cr2O7 = 294.20 g = 294 g (accurate enough for the rest of the data)
The mass of K2Cr2O7 in 350 mL of 0.105 M K2Cr2O7 is:
mass K2Cr2O7 = (0.350 L)
0.105 mol K2 Cr2O 7
1 L solution
294 g
1 mol K2 Cr2O 7
= 10.8 g
(b) The molar mass of FeCl3·6H2O is:
55.85 g
(1 mol Fe)
= 55.85 g
1 mol Fe
35.45 g
= 106.35 g
(3 mol Cl)
1 mol Cl
1.008 g
(12 mol H)
= 12.10 g
1 mol H
16.00 g
= 112.00 g
(6 mol O)
1 mol O
molar mass FeCl3·6H2O = 286.30 g = 286 g (accurate enough for the rest of the data)
The mass of FeCl3·6H2O in 50 mL of 1.0 M FeCl3·6H2O is:
mass FeCl3·6H2O = (0.050 L)
1.0 mol FeCl3 ·6H2 O
1 L solution
286 g
1 mol FeCl3 ·6H2 O
= 14.3 g
(c) The molar mass of KCl is:
39.10 g
(1 mol K)
= 39.10 g
1 mol K
35.45 g
= 35.45 g
(1 mol Cl)
1 mol Cl
molar mass KCl = 74.55 g = 74.6 g (accurate enough for the rest of the data)
The mass of KCl in 0.3 L of 1.70 M KCl is:
1.70 mol KCl
mass KCl = (0.3 L)
1 L solution
74.6 g
1 mol KCl
= 38.0 g
ACS Chemistry Chapter 2 suggested solutions
33
Aqueous Solutions and Solubility
Or we get about 4
Chapter 2
102 g, if the volume is really only known to about 1 part in 3 ( 33%).
Problem 2.65.
For each solution in this problem we are asked to calculate the molar concentration, molarity =
moles per liter of solution, of a known mass of solute present in a known volume of solution. To
get the molarity, we need to convert the given mass to moles and divide by the solution volume
(in liters) to find the number of moles per liter:
mass solute moles solute molarity
These steps can be combined:
1 mol solute
1
molarity solute, mol·L–1 = (mass solute, g)
volume solution, L
g solute
In each case, we will need the molar mass of the solute and will calculate this first.
(a) The molar mass of NaCl is:
22.99 g
(1 mol Na)
= 22.99 g
1 mol Na
35.45 g
(1 mol Cl)
= 35.45 g
1 mol Cl
molar mass NaCl = 58.44 g = 58.4 g (accurate enough for the rest of the data)
The molarity of NaCl in 0.120 L of solution containing 4.5 g NaCl is:
molarity NaCl = (4.5 g NaCl)
1 mol NaCl
58.4 g
1
0.120 L
= 0.64 M
(b) The molar mass of NH4Cl is:
14.01 g
(1 mol N)
= 14.01 g
1 mol N
1.008 g
= 4.03 g
(4 mol H)
1 mol H
35.45 g
= 35.45 g
(1 mol Cl)
1 mol Cl
molar mass NH4Cl = 53.49 g = 53.5 g (accurate enough for the rest of the data)
The molarity of NH4Cl in 0.25 L of solution containing 1.3 g NH4Cl is:
1 mol NH 4 Cl
1
= 0.097 M
53.5 g
0.25 L
Or we get about 0.10 M, if the mass is only known to a precision of about 1 part in 13.
molarity NH4Cl = (1.3 g NH4Cl)
(c) The molar mass of AgNO3 is:
107.9 g
= 107.9 g
(1 mol Ag)
1 mol Ag
14.01 g
= 14.01 g
(1 mol N)
1 mol N
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ACS Chemistry Chapter 2 suggested solutions
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Aqueous Solutions and Solubility
16.00 g
= 48.00 g
1 mol O
molar mass AgNO3 = 169.9 g = 170 g (accurate enough for the rest of the data)
(3 mol O)
The molarity of AgNO3 in 1.3 L of solution containing 1.85 g AgNO3 is:
1
1 mol AgNO3
= 0.0084 M
1.3 L
170 g
Or we get about 0.008 M, if the volume is only known to a precision of about 1 part in 13.
molarity AgNO3 = (1.85 g AgNO3)
Problem 2.66.
(a) We are asked to calculate the molar concentration, molarity = moles per liter of solution, of
a known mass of glucose present in a known volume of solution. To get the molarity, we need
to convert the given mass to moles and divide by the solution volume (in liters) to find the
number of moles per liter:
mass glucose moles glucose molarity
These steps can be combined:
1 mol glucose
1
molarity glucose, mol·L–1 = (mass glucose, g)
g glucose
volume solution, L
We will need the molar mass of glucose, C6H12O6, and will calculate this first.
12.01 g
(6 mol C)
= 72.06 g
1 mol C
1.008 g
= 12.10 g
(12 mol H)
1 mol H
16.00 g
= 96.00 g
(6 mol O)
1 mol O
molar mass glucose = 180.16 g
molarity glucose = (5.405 g glucose)
1 mol glucose
180.16 g
1
1.000 L
= 0.03000 M
(b) There are many ways to solve this part of the problem and any one that you can explain and
justify to get the correct answer is OK. The way presented here may be useful in other situations
involving moles/millimoles (mmol) and liters/milliliters (mL); these are especially common in
biochemistry. Since 1 L = 1000 mL, a solution that contains x moles of a solute in one liter of
solution will contain x millimoles of a solute in one milliliter of solution (one-thousandth of the
solute in one-thousandth of the solution). For example, the solution in part (a) contains 0.03000
mmol of solute in each 1.000 mL of solution. The “rule” is this: the numeric value of a solution
concentration in mmol·mL–1 equals the concentration in mol·L–1. In the problem we have here,
we wish to convert a desired number of millimoles to the equivalent volume (in mL) of solution
of a known concentration:
1.000 mL
0.950 mmol glucose = (0.950 mmol glucose)
= 31.67 mL
0.03000 mmol
ACS Chemistry Chapter 2 suggested solutions
35
Aqueous Solutions and Solubility
Chapter 2
You may have chosen a different route to calculate this volume, and, as we said at the
beginning, that’s OK. Whatever way you choose is perfectly acceptable — as long as you
understand the concepts and arrive at the correct answer.
Problem 2.67.
In order to prepare a 1.00 M solution of any solute, you have to dissolve one molar mass of the
solute in one liter of solution (or an equivalent fraction of the molar mass and the volume). For
this problem, we need to know the molar mass of the solutes each student used. The molar mass
of CuSO4 is:
63.55 g
(1 mol Cu)
= 63.55 g
1 mol Cu
32.07 g
(1 mol S)
= 32.07 g
1 mol S
16.00 g
= 64.00 g
(4 mol O)
1 mol O
molar mass CuSO4 = 159.62 g
The molar mass of CuSO4·5H2O adds the mass of the five moles of water, 90.08 g, to the mass
of the CuSO4 for a total of 249.70 g·mol–1. Both students used the correct procedure for making
a solution of known concentration, but the male student neglected to account for the water of
hydration and weighed too little of the solute: so prepared a solution of too low a concentration:
1 mol
159.60 g CuSO4·5H2O = (159.60 g CuSO4·5H2O)
249.70 g CuSO4 5H2 O
= 0.639 mol
He prepared a solution of too low a concentration: 0.639 mol·L–1 = 0.639 M. The female student
used a solid of the correct composition, CuSO4, and prepared a solution of the correct
composition: 1.000 mol·L–1 = 1.000 M.
Problem 2.68.
To make a solution with a fairly exact concentration of a solid solute, we need to weigh the
solute carefully, add the solid to a volumetric flask, dissolve the solid in some water in the flask,
and then make up the solution to the exact volume of the flask. Volumetric flasks are commonly
available in a variety of sizes: 10 , 25 , 50 , 100 , 250 , 500 , and 1000 mL. We need about 170
mL of 0.10 M NaOH. In order to waste as little as possible of the solid reagent, we should use
the smallest volumetric flask, 250-mL, that will give us enough of the solution. This is a
problem for which we know the concentration of the desired solution, 0.10 M, and its volume,
0.250 L, so we need to convert the volume to the equivalent number of moles of solute and then
moles to mass using the molar mass of NaOH (40.0 g·mol–1) to get the mass of NaOH that must
be dissolved. The calculation may be done in a single step by combining the two conversions:
0.10 mol NaOH
40.0 g
0.250 L soln = (0.250 L solution)
= 1.00 g NaOH
1 L solution
1 mol NaOH
You may not have known the standard sizes of volumetric flasks and have chosen a different
volume of solution to prepare. That’s OK. If you substitute your volume into this equation and
get the result you calculated, you solved the problem correctly.
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Problem 2.69.
We are first asked to determine the mass of sodium chloride, NaCl, required to make 250. mL
of a 0.90% (mass to volume %) sodium chloride (normal saline) solution and then to find the
molarity of this solution. The mass to volume % concentration is not discussed in the textbook,
but its meaning is clear: the mass of solute is a given percentage (parts in 100) of the volume of
the solution. The only ambiguities are the units to be used for mass and volume. In context, it
makes sense to use grams and milliliters, because they will give a reasonable result (where
“reasonable” means physically significant or possible). A 0.90% solution will have 0.90 g of
solute per 100. mL of solution. In this case, volume of solution is given, 250. mL, and we wish
to find the mass of NaCl solute:
0.90 g NaCl
250. mL solution = (250. mL solution)
= 2.25 g NaCl
100. mL solution
To find the molarity of this solution, we need to convert the mass of NaCl in a given volume (in
liters) to moles of NaCl, using the molar mass of NaCl (58.4 g·mol–1), and then divide by the
volume to get molarity (mol·L–1). We can either work with the mass of NaCl in 100. mL of
solution (0.90 g) or the mass in 250. mL (2.25 g). Let’s use the former:
0.90 g NaCl = (0.90 g NaCl)
1 mol
58.4 g NaCl
1
0.100 L
= 0.154 M
Check to see that using 2.25 grams of NaCl in 250. mL of aqueous solution to calculate the
molarity gives the same answer.
Problem 2.70.
We have a mass of solute, 25 g of urea, (NH2)2CO, in 2.5 L of solution (urine) and are asked to
calculate the molarity of the solution. We need to convert the mass of urea to moles, using the
molar mass of urea (60 g·mol–1), and then divide by the volume to get molarity (mol·L–1):
25 g urea = (25 g urea)
1 mol
60 g urea
1
2.5 L
= 0.17 M
Problem 2.71.
The label on a sports drink tells us that 240 mL of the solution contains 30 mg of potassium.
KH2PO4 is the only solution ingredient listed on the label that can provide this potassium. We
are asked to determine the mass of KH2PO4 required to provide 30 mg of potassium [as K+(aq)]
in the solution and to find the molarity of KH2PO4 in the solution. Note that each mole of
KH2PO4 that dissolves in the solution provides a mole of K+(aq) in the solution. There are
several ways to approach this problem, but they all require knowing the molar masses of K+
(39.1 g·mol–1) and of KH2PO4 (136 g·mol–1), because we need to convert masses to moles and
moles to masses in this problem. In one approach, we follow these pathways:
mass K+ moles K+ = moles KH2PO4
moles KH2PO4 mass KH2PO4
moles KH2PO4 molarity KH2PO4
ACS Chemistry Chapter 2 suggested solutions
37
Aqueous Solutions and Solubility
Chapter 2
Another approach takes these pathways:
mass K+ mass KH2PO4 (using relative masses in one mole KH2PO4)
mass K+ moles K+ molarity K+ = molarity KH2PO4
Note that the actual number of conversions in each approach is the same, so neither is more
efficient or preferred, although the second approach can be condensed into two steps. We will
go through both to show that the results are the same. In the first approach, we have (after
converting milligrams to grams and milliliters to liters):
0.030 g K+ = (0.030 g K+)
1 mol
39.1 g K+
= 7.7
10–4 mol K+ = 7.7
10–4 mol KH2PO4
7.7
10–4 mol KH2PO4 = (7.7
10–4 mol KH2PO4)
136 g KH2 PO 4
1 mol
= 0.10 g
7.7
10–4 mol KH2PO4 = (7.7
10–4 mol KH2PO4)
1
0.240 L
10–3 M KH2PO4
KH2PO4
= 3.2
In the second approach we have:
0.030 g K+ = (0.030 g K+)
136 (g KH 2 PO4 ) (mol KH2 PO4 )
+
–1
39.1 (g K ) (mol KH2 PO 4 )
0.030 g K+ = (0.030 g K+)
1 mol
39.1 g K+
1
0.240 L
= 3.2
–1
= 0.10 g KH2PO4
10–3 M K+
= 3.2 10–3 M KH2PO4
Thus, the results are the same, no matter which approach we use. If you used yet a different
approach, got the same results, and can explain the concepts behind what you did, that’s fine.
Problem 2.72.
We need to remember that the relationship of numbers of atoms of the elements to a given
number of molecules of a compound is the same as the relationship of numbers of moles of
atoms of the elements to the given number of moles of the compound. For example, in two
molecules of ammonium acetate, NH4C2H3O2, there are two atoms of nitrogen, N, and in two
moles of NH4C2H3O2 there are two moles of nitrogen atoms, N. Also, in two moles of
NH4C2H3O2, we have:
2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)
2 mol C
1 mol NH 4 C2 H 3O2
= 4 mol C
2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)
7 mol H
1 mol NH 4 C2 H 3O2
= 14 mol H
2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)
2 mol O
1 mol NH 4 C2 H 3O2
= 4 mol O
Summing the number of moles of all atoms (N, C, H, and O) in 2 mol NH4C2H3O2, we have 24
(= 2 + 4 + 14 + 4) mol of atoms.
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We are also asked for the total number of moles of ions in two moles of NH4C2H3O2. Although
no discrete molecules of NH4C2H3O2 are present in a crystal of the ionic compound, the
chemical formula indicates that the ratio of cations to anions is 1:1, so one formula unit
(equivalent to a molecule) contains one ammonium cation, NH4+, and one acetate anion,
C2H3O2–. Two formula units (molecules) of NH4C2H3O2 contain four ions (two cations and two
anions), so two moles of NH4C2H3O2 contain four moles of ions:
2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)
2 mol ions
1 mol NH 4 C2 H 3O2
= 4 mol ions
There is more than one way to solve a problem like this and, if you solved it differently, but got
the correct answers and can explain the concepts you used, that’s OK.
Problem 2.73.
We are asked to determine the mass of calcium phosphate that can be made by mixing 125. mL
of 0.100 M calcium chloride with 125. mL of 0.100 M sodium phosphate. To solve this
problem, we will first need the formulas for calcium chloride (CaCl2), sodium phosphate
(Na3PO4), and calcium phosphate (Ca3(PO4)2). The formulas are based on balancing out the
known charges on the cations and anions, which we can get from Table 2.2, if we don’t
remember them. Combined with the solution concentration data, the first two formulas provide
us enough information to determine the molarity of the reactant ions, Ca2+(aq) and PO43–(aq), in
the solutions and, hence, the number of moles of each of these ions present in the mixed
solution. We see that there is one mole of Ca2+(aq) in solution for each mole of CaCl2 that is
dissolved. Thus, the concentration and number of moles of calcium cation is:
0.100 M CaCl2 =
0.100 mol CaCl2
1 L solution
2+
1 mol Ca
1 mol CaCl 2
= 0.100 M Ca2+(aq)
0.100 mol Ca 2+
0.100 M Ca (aq) =
(0.125 L solution) = 1.25
1 L solution
2+
10–2 mol Ca2+(aq)
The same chain of reasoning gives the concentration and number of moles of phosphate anion:
0.100 mol Na 2 PO4
0.100 M Na3PO4 =
1 L solution
3–
1 mol PO4
1 mol Na 2 PO4
= 0.100 M PO43–(aq)
0.100 mol PO3–
4
0.100 M PO4 (aq) =
(0.125 L solution) = 1.25
1 L solution
3–
10–2 mol PO43–
(aq)
Use the formula for the product of reaction, a precipitate of calcium phosphate (Ca3(PO4)2(s)),
to write a balanced reaction for its formation from the ions:
3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)
The equation tells us that the mole ratio of Ca2+(aq) to PO43–(aq) ions that react is 3:2. Thus, 1.5
(= 3 2 ) mol of Ca2+(aq) is required to react completely with 1 mol of PO43–(aq). Since our
mixed solution contains an equal number of moles of each ion, there is not enough Ca2+(aq) to
react with all the PO43–(aq), so Ca2+(aq) is the limiting reactant in the mixture. We use the ratio
ACS Chemistry Chapter 2 suggested solutions
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Aqueous Solutions and Solubility
Chapter 2
of moles of calcium ions in a mole of calcium phosphate to convert number of moles of calcium
ion that react to moles of solid calcium phosphate precipitated:
1 mol Ca 3 (PO 4 )2
1.25 10–2 mol Ca2+ = (1.25 10–2 mol Ca2+)
3 mol Ca 2+
= 4.17 10–3 mol Ca3(PO4)2(s)
The molar mass of Ca3(PO4)2 is 310.2 g, so the mass of Ca3(PO4)2(s) formed is:
4.17
10–3 mol Ca3(PO4)2 = (4.17
10–3 mol Ca3(PO4)2)
310.2 g Ca 3 (PO4 )2
1 mol Ca 3 (PO4 )2
= 1.29 g Ca3(PO4)2(s)
Problem 2.74.
The coefficients in the balanced ionic equation give the relative number of moles of each
reactant and product;
2Fe3+(aq) + SO32–(aq) + 3H2O(l) 2Fe2+(aq) + SO42–(aq) + 2H3O+(aq)
These ratios can also be expressed in millimoles (1 mol = 1000 mmol). See the solution for
Problem 2.66. The volume of 0.100 M SO32–(aq) needed to react exactly and completely with
24.0 mL of 0.200 M Fe3+(aq) is:
24.0 mL Fe3+ = (24.0 mL
2–
0.200 mmol Fe 3+ 1 mmol SO2–
1.000 mL SO3
3
Fe3+)
2–
1.000 mL Fe 3+
2 mmol Fe3+
0.100 mmol SO 3
= 24.0 mL SO32– solution
Problem 2.75.
(a) We are asked to determine the number of moles of each of the four ions, Na+, SO42–, Ba2+,
and Cl–, in a mixture prepared by mixing 50.0 mL of a 0.45 M Na2SO4 solution with 50.0 mL of
a 0.36 M BaCl2 solution. The number of moles of an ion in the mixture is the same as the
number moles of that ion in its original solution. We need to convert the volumes (in liters) to
moles of ionic compound solute, using the solution molarity, and then to moles of individual
ions, using the ratio of moles of ions to moles of solute:
0.45 mol Na 2 SO 4
0.050 L Na2SO4 = (0.050 L Na2SO4)
1L
+
2 mol Na
1 mol Na 2 SO4
= 0.045 mol Na+
0.050 L Na2SO4 = (0.050 L Na2SO4)
0.45 mol Na 2 SO 4
1L
2–
1 mol SO4
1 mol Na 2 SO4
= 0.023 mol SO42–
0.050 L BaCl2 = (0.050 L BaCl2)
0.36 mol BaCl2
1L
= 0.018 mol Ba2+
40
2+
1 mol Ba
1 mol BaCl 2
ACS Chemistry Chapter 2 suggested solutions
Chapter 2
Aqueous Solutions and Solubility
–
0.36 mol BaCl2
0.050 L BaCl2 = (0.050 L BaCl2)
1L
2 mol Cl
1 mol BaCl 2
= 0.036 mol Cl–
(b) We are asked to determine how many moles of Ba2+(aq) are required to react with all the
SO42–(aq) in the mixture, if the SO42–(aq) reacts with Ba2+(aq) to give BaSO4(s). We use the 1:1
reactant ratio to find the number of moles of Ba2+(aq) required:
2–
0.023 mol SO4 (aq) = (0.023 mol SO4
2–
2+
1 mol Ba
)
2–
1 mol SO 4
= 0.023 mol Ba2+(aq)
(c) We are asked to determine how many moles of SO42–(aq) are required to react with all the
Ba2+(aq) in the mixture, if the SO42–(aq) reacts with Ba2+(aq) to give BaSO4(s). We use the 1:1
reactant ratio to find the number of moles of SO42–(aq) required:
1 mol SO2–
4
0.018 mol Ba (aq) = (0.018 mol Ba )
1 mol Ba 2+
2+
2+
= 0.018 mol SO42–(aq)
(d) To determine whether Ba2+(aq) or SO42–(aq) is the limiting reactant in this mixture, we
compare the number of moles of each ion in the solution, from part (a), to the number of moles
of that ion, from parts (b) and (c), required to react completely with the other ion. If the solution
does not contain the requisite number of moles of one of the ions, that ion is the limiting
reactant, since it will be used up before all the other ion has reacted. In this case, we see that
0.023 mol Ba2+(aq) are required to react with all the SO42–(aq), but the solution contains only
0.018 mol Ba2+(aq). Therefore, Ba2+(aq) is the limiting reactant in this mixture and it will react
with 0.018 mol SO42–(aq), leaving about 0.005 mol SO42–(aq) behind in solution.
Problem 2.76.
The objective of this problem is to use our solubility rules, including the few exceptions we
have noted, to predict the products of mixing four pairs of aqueous solutions and to determine
the limiting reagent for each reaction and the mass of the precipitate (assuming that all
precipitation reactions go to completion). To find the limiting reagent in each case we will also
need to write a net ionic equation for the reaction that occurs
(a) 125 mL of 0.15 M BaBr2 are mixed with 125 mL of 0.15 M Na3PO4.
The mixed solution contains a multiply-charged cation, Ba2+(aq) and a multiply-charged anion,
PO43–(aq), which react to form an insoluble precipitate:
3Ba2+(aq) + 2PO43–(aq) Ba3(PO4)2(s)
The number of moles of each of these ions in the mixture is:
0.15 mol BaBr2
0.125 L BaBr2 = (0.125 L BaBr2(aq))
1 L BaBr2 (aq)
2+
1 mol Ba
1 mol BaBr2
= 0.019 mol Ba2+(aq)
0.125 L Na3PO4 = (0.125 L Na3PO4(aq))
0.15 mol Na 3 PO 4
1 L Na 3 PO 4 ( aq)
3–
1 mol PO4
1 mol Na 3PO4
= 0.019 mol PO43–(aq)
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Chapter 2
The equation tells us that the mole ratio of Ba2+(aq) to PO43–(aq) ions that react is 3:2. Thus, 1.5
(= 3 2 ) mol of Ba2+(aq) is required to react completely with 1 mol of PO43–(aq). Since our
mixed solution contains an equal number of moles of each ion, there is not enough Ba2+(aq) to
react with all the PO43–(aq), so Ba2+(aq) is the limiting reactant in the mixture. We use the ratio
of moles of barium ions in a mole of barium phosphate to convert number of moles of barium
ion that react to moles of solid barium phosphate precipitated:
1 mol Ba 3 (PO 4 )2
0.019 mol Ba2+ = (0.019 mol Ba2+)
= 6.3 10–3 mol Ba3(PO4)2(s)
2+
3 mol Ba
The molar mass of Ba3(PO4)2 is 602 g, so the mass of Ba3(PO4)2(s) formed is:
6.3
10–3 mol Ba3(PO4)2 = (6.3
10–3 mol Ba3(PO4)2)
602 g Ba 3 (PO4 )2
1 mol Ba 3 (PO4 )2
= 3.8 g Ba3(PO4)2(s)
(b) 85 mL of 0.40 M NH4Cl are mixed with 65 mL of 0.50 M KNO3.
The mixed solution contains only singly-charged cations and anions which form soluble ionic
compounds, so there is NO APPARENT REACTION in this case.
(c) 85 mL of 0.40 M (NH4 )2S are mixed with 65 mL of 0.50 M ZnCl2.
The mixed solution contains a multiply-charged cation, Zn2+(aq) and a multiply-charged anion,
S2–(aq), which react to form an insoluble precipitate:
Zn2+(aq) + S2–(aq) ZnS(s)
The number of moles of each of these ions in the mixture is:
0.065 L ZnCl2 = (0.065 L ZnCl2(aq))
0.50 mol ZnCl2
1 L ZnCl2 (aq )
2+
1 mol Zn
1 mol ZnCl 2
= 0.033 mol Zn2+(aq)
0.085 L (NH4 )2S = (0.085 L (NH4 )2S(aq))
0.40 mol (NH4 )2 S
1 L (NH4 )2 S(aq)
2–
1 mol S
1 mol (NH 4 )2 S
= 0.034 mol S2–(aq)
These ions react in a 1:1 ratio, so the ion whose concentration is lower, the Zn2+(aq), will be
completely used up in the reaction. Thus, Zn2+(aq) is the limiting reactant and the mass of
ZnS(s), molar mass = 97.4 g, that will be formed is:
1 mol ZnS
97.4 g ZnS
0.033 mol Zn2+ = (0.033 mol Zn2+)
= 3.2 g ZnS(s)
2+
1 mol Zn
1 mol ZnS
(d) 15.0 mL of 0.20 M AgNO3 are mixed with 15.0 mL of 0.40 M NaBr.
The mixed solution contains only singly-charged cations and anions, but these include Ag+(aq)
and Br–(aq), which form a precipitate, AgBr(s), that is an exception to our rules:
Ag+(aq) + Br–(aq) AgBr(s)
The number of moles of each of these ions in the mixture is:
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ACS Chemistry Chapter 2 suggested solutions
Chapter 2
Aqueous Solutions and Solubility
0.20 mol AgNO3
0.0150 L AgNO3 = (0.0150 L AgNO3(aq))
1 L AgNO 3 (aq )
+
1 mol Ag
1 mol AgNO3
= 0.0030 mol Ag+(aq)
0.40 mol NaBr
1 mol Br –
0.0150 L NaBr = (0.0150 L NaBr(aq))
1 mol NaBr
1 L NaBr(aq)
–
= 0.0060 mol Br (aq)
These ions react in a 1:1 ratio, so the ion whose concentration is lower, the Ag+(aq), will be
completely used up in the reaction. Thus, Ag+(aq) is the limiting reactant and the mass of
AgBr(s), molar mass = 188 g, that will be formed is:
0.0030 mol Ag+ = (0.030 mol Ag+)
1 mol AgBr
1 mol Ag+
188 g AgBr
1 mol AgBr
= 5.6 g AgBr(s)
Problem 2.77.
(a) What is the white precipitate formed when 50. mL of an aqueous 0.1 M SrCl2 solution are
mixed with 50. mL of an aqueous 0.1 M Na3PO4 solution? The mixed solution contains a
multiply-charged cation, Sr2+(aq) and a multiply-charged anion, PO43–(aq), which react to form
an insoluble precipitate, Sr3(PO4)2(s). The complete ionic equation representing the reaction that
occurs in the mixed solution is:
3Sr2+(aq) + 6Cl–(aq) + 6Na+(aq) + 2PO43–(aq) Sr3(PO4)2(s) + 6Cl–(aq) + 6Na+(aq)
(b) The complete ionic reaction shows that all the chloride ion from its original solution is still
present in the mixture when the precipitation is complete. The number of moles of chloride and
its mass may be calculated from the number of moles in its original solution volume:
0.050 L SrCl2(aq) = (0.050 L SrCl2(aq))
0.1 mol SrCl 2
1 L SrCl2 (aq )
–
2 mol Cl
1 mol SrCl 2
= 0.01 mol Cl–(aq)
0.01 mol Cl– = (0.01 mol Cl–)
35.45 g
1 mol Cl –
= 0.35 g Cl–
The concentrations of the solutions are only given with one significant digit, so we really should
not report this result with two figures, but we will need the greater precision for the later
calculation to show the mixed solution’s electrical neutrality.
(c) The number of moles of each of the other ions added to the mixed solution is:
0.1 mol SrCl 2
0.050 L SrCl2(aq) = (0.050 L SrCl2(aq))
1 L SrCl2 (aq )
2+
1 mol Sr
1 mol SrCl 2
= 0.005 mol Sr2+(aq)
0.1 mol Na 3 PO4
0.050 L Na3PO4(aq) = (0.050 L Na3PO4(aq))
1 L Na 3 PO 4 (aq)
+
3 mol Na
1 mol Na 3PO4
= 0.015 mol Na+(aq)
ACS Chemistry Chapter 2 suggested solutions
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Aqueous Solutions and Solubility
0.1 mol Na 3 PO4
0.050 L Na3PO4(aq) = (0.050 L Na3PO4(aq))
1 L Na 3 PO 4 (aq)
Chapter 2
3–
1 mol PO4
1 mol Na 3PO4
= 0.005 mol PO43–(aq)
The equation in part (a) tells us that all of the Na+(aq) added to the mixture is still present after
the precipitation reaction is complete. Now we need to find out whether any Sr2+(aq) or PO43–
(aq) ions remain unreacted. Without doing any calculations, we can predict which ion will still
be present after the precipitation is complete. Note that we found in part (b) that 0.01 mol of Cl–
(aq) is present in the mixture and now we have 0.015 mol of Na+(aq) also left in the mixture.
But the solution must be electrically neutral, so there must be more negative ions present. The
only possibility is that some PO43–(aq) ions remain unreacted. Thus, it must be the case that
Sr2+(aq) is the limiting reactant. Let’s see if that is the case and find out how much PO43–(aq)
remains unreacted.
The equation in part (a) tells us that the mole ratio of Sr2+(aq) to PO43–(aq) ions that react is 3:2.
Thus, 1.5 (= 3 2 ) mol of Sr2+(aq) is required to react completely with 1 mol of PO43–(aq). Since
our mixed solution contains an equal number of moles of each ion, there is not enough Sr2+(aq)
to react with all the PO43–(aq), so Sr2+(aq) is the limiting reactant in the mixture, as we reasoned
in the previous paragraph. We use the ratio of moles of strontium ions in a mole of strontium
phosphate to convert number of moles of strontium ion that react to moles of solid strontium
phosphate precipitated and then to find the number of moles of PO43–(aq) left unreacted:
1 mol Sr3 (PO4 )2
= 0.0017 mol Sr3(PO4)2(s)
0.005 mol Sr2+ = (0.005 mol Sr2+)
3 mol Sr 2+
Since each mole of Sr3(PO4)2 contains two moles of PO43–, we have 0.0034 mol of PO43–
removed from solution as Sr3(PO4)2(s). The amount of PO43–(aq) remaining in solution is
0.0016 mol [= (0.005 mol) – (0.0034 mol)].
(d) The total number of moles of negative charge on the remaining PO43–(aq) anions, from part
(c), is 0.005 mol [= (3 charges·mol–1)(0.0016 mol)]. This amount of negative charge added to
that from Cl–(aq), 0.01 mol, is equivalent to the positive charge from the Na+(aq), 0.015 mol, so
the solution is electrically neutral, as we said in the reasoning in part (c).
Problem 2.78.
We are asked whether gases are very soluble in water. There is no simple answer to this
question, because the answer depends upon the interactions of the gas with water. Non-polar gas
molecules like nitrogen and oxygen have only limited solubility because they are so volatile and
interact only weakly by dispersion forces with the water. At the other end of the solubility scale
are gases like ammonia and hydrogen chloride that hydrogen bond strongly with and/or react
with water to form quite concentrated solutions. Intermediate solubilities are shown by gases
like carbon dioxide that have polar bonds (dipole-dipole attractions with water) and react
somewhat with water (as CO2(aq) does to form small amounts of H2CO3(aq)). In all cases, the
solubility of gases is temperature dependent and decreases as temperature increases.
Problem 2.79.
We are asked to predict whether the noble gases (He, Ne, Ar, Kr, and Xe) have a low solubility
in water (less than 1 g·L–1) or a high solubility in water (greater than 10 g·L–1). Recall from
Section 2.11 that only when water reacts with a gas will the gas have a very high solubility.
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Aqueous Solutions and Solubility
Since the noble gases are all rather unreactive (that is why they are called noble -- aloof and not
mingling well with the more “common” elements), they show no tendency to react with or bond
with water. The electron distribution around the noble gas nuclei is spherically symmetric,
which makes them non-polar and further suggests that they are insoluble in water.
Problem 2.80.
(a) We are asked to determine how many moles of nitrogen gas, N2(g), dissolve in 10.0 L of
water at a temperature of 25 C and gas pressure of 101 kPa (one atmosphere). The data in
Table 2.6 give the solubility of N2(g) under these conditions as 0.018 g·kg–1 (grams of gas per
kilogram of water). The density of water at 25 C and one atmosphere pressure is 1.00 kg·L–1
[Chapter 1, Figure 1.29(a)], so 10.0 L of water is 10.0 kg of water and the mass of N2 (g) that
dissolves in this much water is 0.18 g [= (0.018 g·kg–1)(10.0 kg)]. Thus, the number of moles of
nitrogen dissolved in 10.0 L of water is:
0.18 g N2 = (0.18 g N2)
1 mol N2
28 g N2
= 6.4
10–3 mol N2
(a) We are asked to determine how many moles of oxygen gas, O2(g), dissolve in 0.100 L of
water at a temperature of 25 C and gas pressure of 101 kPa (one atmosphere). The data in
Table 2.6 give the solubility of O2(g) under these conditions as 0.039 g·kg–1. As in part (a),
0.100 L of water is 0.100 kg of water and the mass of O2 (g) that dissolves in this much water is
0.0039 g [= (0.039 g·kg–1)(0.100 kg)]. Thus, the number of moles of oxygen dissolved in 0.100
L of water is:
0.0039 g O2 = (0.0039 g O2)
1 mol O2
32 g O2
= 1.2
10–4 mol O2
Problem 2.81.
The solubility of H2(g) in water is temperature dependent. The solubility at 25 C is 7.68 10–4
mol L–1 and, at 0 C, the solubility is 9.61 10–4 mol L–1. Note that the solubility of this nonpolar molecule is quite low at both temperatures. There is evidently very little attraction of
water molecules for hydrogen molecules. There are only dispersion interactions and with only
two electrons, it is difficult to induce dipoles in hydrogen molecules. Since there is very little
attraction between H2 and H2O, perhaps the explanation for the temperature dependence of the
solubility lies in the motion of H2 at different temperatures. At 25 C, H2 are moving faster than
0 C. Thus, more molecules will have enough energy to escape into the gas phase at 25 C than
at 0 C. This means that the solubility of H2 should be greater at 0 C than at 25 C, as observed.
Problem 2.82.
The solution that is formed when hydrogen bromide gas, HBr(g), dissolves in water to form an
acidic solution is called hydrobromic acid, just as the analogous solution of HCl(g) is called
hydrochloric acid,
Problem 2.83.
A saturated solution of HBr(g) in water (hydrobromic acid) is approximately 8.9 M in HBr(aq)
and the density of the solution is about 1.5 kg·L–1. We are asked to express the solubility of
ACS Chemistry Chapter 2 suggested solutions
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Aqueous Solutions and Solubility
Chapter 2
HBr(g) in units of g·kg–1. The density of the solution tells us that a liter of the solution has a
mass of 1.5 kg and we know, from the molarity, that this liter contains 8.9 mol HBr. The molar
mass of HBr is 80.9 g, so the mass of HBr dissolved in the liter of solution is:
80.9 g HBr
8.9 mol HBr = (8.9 mol HBr)
= 7.2 102 g HBr = 0.7 kg HBr
1 mol HBr
The mass of water in the solution is 0.8 kg [= (1.5 kg solution) – (0.7 kg solute)], so the
concentration of HBr(aq) (in units of g·kg–1)is:
7.2 10 2 g HBr
= 9 102 g·kg–1
0.8 kg water
Note that the mass of HBr(g) that dissolves in a kilogram of water is greater than the mass of
HCl(g) that dissolves in water, 695 g·kg–1, but the number of moles of HBr(g) that dissolve,
about 11, is only about half as much as the moles of HCl(g) that dissolve, about 19. On a molar
basis, HCl(g) is much more soluble.
Problem 2.84.
(a) We are asked to illustrate each of the potential interactions that might explain the high
solubility of HCl(g) in diethyl ether. For simplicity, diethyl ether, C2H5O C2H5, is written as R–
O–R in these sketches showing (i) dipole-dipole, (ii) hydrogen bonding, and (iii) ionization of
HCl in ether solution:
(b) An experiment that might be done to eliminate or confirm one or more of these possibilities
is testing the electrical conductivity of the solution, If it conducts, then we know that at least
some HCl ionization, interaction (iii), must occur. This does not rule out contributions to the
solubility from the other interactions. If the solution does not conduct, ionization is ruled out as
a contributor to the solubility. The other interactions are hard to distinguish experimentally.
NOTE: Stephen Branz, San Jose State University, has provided an interesting discussion of the
possible reasons for the high solubility of HCl(g) in diethyl ether. If ionization reactions are
primarily responsible for the solubility of HCl(g), we would expect the dielectric strength of the
solvent to greatly affect the solubility. Solubility data for HCl(g) in water, MeOH, EtOH, and
Et2O are from the Merck Index, 9th ed., and all other data are from standard reference tables.
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Solvent
Solubility
g·(100 mL)–1 at 30 °C
Dielectric Constant
water
67.3
78.4
methanol
43.0
33.6
ethanol
38.1
25.0
diethyl ether
19.5
4.3
benzene
“soluble” *
2.3
pentane
n/a #
1.8
chloroform
n/a #
4.8
* no quantitative data available; several sources report “soluble.”
# presumably very low
Although there is indeed a reduction in the solubility of HCl going from water to the alcohols
and ether, the differences are (qualitatively) not reflective of the differences in dielectric
constants. Another way to look at the data is to calculate the solubility relative to the number of
oxygen atoms in a 100 mL sample of the solvent.
Solvent
g HCl / mol “O” (in 100 mL solvent)
water
methanol
ethanol
diethyl ether
12.1
17.4
22.2
20.3
Assuming that there is a specific interaction between HCl and the oxygen of the solvent (and
this might be anywhere from formation of the oxonium ion, R2OH+, to a strong hydrogen bond),
it is justifiable to consider the solubility relative to the number of oxygen atoms in the solvent.
By this measure, of the four oxygenated solvents, water is the worst solvent for HCl! A (perhaps
naive) qualitative explanation is that of the four solvents in the table, only water has two H’s
and two lone pairs for hydrogen bonding (or oxonium ion formation). Water must sacrifice H
bonds (among water molecules) in order to interact with/ionize HCl. The other three solvents
each have an “excess of lone pairs” and can more effectively solvate the HCl.
The foregoing discussion is based on correlations with no physical data (such as conductivity,
for which we have found no data) to back it up, but the correlation is quite good. Other support
for a possible specific interaction (or interactions) between HCl and ether molecules should be
noted:
(a) Although we don’t have quantitative data for benzene, strong hydrogen bonding to the pi
system is quite well documented and undoubtedly accounts for the high solubility of HCl in
benzene.
(b) The boiling points of ether and pentane (35-36 °C) are nearly identical, reflective of their
very similar sizes and shapes. If the dipole moment of ether were of great importance in
interactions with itself or other molecules, we would predict that its boiling point would be
higher than that of pentane.
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Chapter 2
(c) There are many well-known organic reactions that support the formation of low
concentrations of ionic species in non-polar solvents (for example, electrophilic aromatic
substitution).
(d) Some non-polar solvents can be surprisingly good solvents for ionic species (for example,
chloroform dissolves very high concentrations of simple organic ammonium chlorides even
though those same salts are quite insoluble in ether -- these solvents have very similar dielectric
constants).
Problem 2.85.
We would expect hydrogen chloride gas, HCl(g), to be less soluble in hexane than in water.
Hydrogen chloride can react with water to form hydronium cations and chloride anions,
equation (2.16). This reaction with the solvent increases the solubility of HCl(g) in water. There
is no parallel reaction of HCl with hexane, CH3CH2CH2CH2CH2CH3, a non-polar solvent, so its
hexane solubility is much lower.
Problem 2.86.
(a) The data in Table 2.6 are for the solubility of the gases at 101 kPa (one atmosphere)
pressure. Assume that the amount of a gas that dissolves in water is directly proportional to its
pressure over the solution; the lower the pressure, the less gas dissolved. (This is Henry’s Law
for the solubility of non-polar gases in liquids.) In air, the nitrogen pressure is 80% of one
atmosphere, so only 80% as much nitrogen will dissolve: (0.80)[0.018 g·(kg water)–1] =
0.014 g·(kg water)–1. Oxygen pressure is 20% of one atmosphere: (0.20)[0.039 g·(kg water)–1] =
0.008 g·(kg water)–1.
(b) In one kilogram of water saturated with gases from the air, there are 0.014 g of nitrogen and
0.008 g of oxygen, or 0.022 g total of these gases. The mass percent of oxygen is:
0.008 g O2
100% = 36%
0.22 g air
(c) In a mole of air (Avogadro’s number of nitrogen and oxygen molecules), 80% of the
molecules are nitrogen and 20% are oxygen. The mass of nitrogen in a mole of air is
(0.80 mol)(28 g·mol–1) = 22.4 g. The mass of oxygen is (0.20 mol)(32 g·mol–1) = 6.4 g. The
total mass of a mole of air is 28.4 g and the mass percent of oxygen in air is:
6.4 g O2
100% = 22%
28.8 g air
The mass percent of oxygen in the gases dissolved in water is greater than the mass percent of
oxygen in the (major) gases in the atmosphere.
Problem 2.87.
An acid is a compound whose aqueous solution has a pH below 7.
Problem 2.88.
A base is a compound whose aqueous solution has a pH above 7.
Problem 2.89.
An acid solution that has a pH of 1 contains a higher concentration (100 times higher) of
hydronium ions than an acid solution that has a pH of 3. From the data given, this is the only
48
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Chapter 2
Aqueous Solutions and Solubility
conclusion we can draw. We have no way to characterize acid A relative to acid B, because we
do not know the concentrations of acid A and acid B in the solutions.
Problem 2.90.
Aqueous solutions with pH < 7, that is [H3O+(aq)] > 1.0 10-7 M and [OH–(aq)] < 1.0 10-7 M,
are acidic. Aqueous solutions with pH > 7, that is [H3O+(aq)] < 1.0 10-7 M and [OH–(aq)] >
1.0 10-7 M, are basic. Aqueous solutions with pH = 7, that is [H3O+(aq)] = 1.0 10-7 M and
[OH–(aq)] = 1.0 10-7 M, are neither acidic nor basic. Each of these solutions is characterized
as acidic, basic, or neither on the basis of the definitions just stated.
(a) pH < 7
acidic
+
-7
(b) [H3O (aq)] = 1.0 10 M
neither
–
-7
(c) [OH (aq)] > 1.0 10 M
basic
(d) pH > 7
basic
(e) [H3O+(aq)] > 1.0 10-7 M
acidic
(f) [OH–(aq)] < 1.0 10-7 M
acidic
+
-7
(g) [H3O (aq)] < 1.0 10 M
basic
–
-7
(h) [OH (aq)] = 1.0 10 M
neither
Problem 2.91.
We calculate the pH of each of these solutions using the definition of pH [equation (2.18)]:
pH = –log10[H3O+(aq)]. This is very easy to do on any scientific calculator, because they all
have a “log” key that returns the logarithm to the base 10, log10, of whatever value is displayed.
That is how these results were calculated. The number of significant digits following the
decimal point in a pH value is determined by the number in the pre-exponential factor in the
concentration value. In all the cases here, the pre-exponential has two digits and the pH has two
digits after the decimal point.
(a) [H3O+(aq)] = 1.0 10-2 M
pH = 2.00
+
-10
(b) [H3O (aq)] = 1.0 10 M
pH = 10.00
+
-4
(c) [H3O (aq)] = 5.0 10 M
pH = 3.30
+
-8
(d) [H3O (aq)] = 5.0 10 M
pH = 7.30
Problem 2.92.
If the reaction of HCl(g) and water goes to completion to form H3O+(aq) and Cl–(aq), the molar
concentration of HCl(aq) in the solution is the same as the molar concentration of H3O+(aq). In
this problem, we are asked to find the [HCl(aq)], which is equal to [H3O+(aq)], that will result
in a solution of a given pH. Use Figure 2.24 to find the [H3O+(aq)] that corresponds to a given
pH and that will also be the appropriate [HCl(aq)].
(a) For a pH = 4, Figure 2.24 shows that [H3O+(aq)] = 10–4 M, so we need [HCl(aq)] = 10–4 M.
(b) For a pH = 2, Figure 2.24 shows that [H3O+(aq)] = 10–2 M, so we need [HCl(aq)] = 10–2 M.
Problem 2.93.
By analogy with HCl(g) being named hydrogen chloride and HCl(aq) named hydrochloric acid,
we can name other gaseous hydrides that dissolve to form acidic solutions:
HI(g) is hydrogen iodide.
HI(aq) is hydroiodic acid.
ACS Chemistry Chapter 2 suggested solutions
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Chapter 2
H2S(g) is hydrogen sulfide.
H2S(aq) is hydrosulfuric acid.
Problem 2.94.
The two movies in the Web Companion, Chapter 2, Section 2.12, page 3, are an attempt to show
in animation what the figure at the bottom of that page and Figure 2.26 in the textbook try to
illustrate. What the figures and movies try to show is how hydronium cations and hydroxide
anions “migrate” in aqueous solution by transfer of a hydrogen ion, H+ (a proton), from one
molecule or ion to another ion or molecule. Four frames from the hydronium ion movie are
shown here:
In the first frame (on the left), one of the hydrogens from the hydronium at the far left of the
frame is hydrogen bonded to the adjacent water molecule and by a small movement of this
hydrogen atomic core, the hydronium ion has transferred a proton to form a new hydronium ion
(center of second frame). This new hydronium ion goes on in frames three and four to transfer a
proton (not the one it got in the previous transfer) to a third water molecule. Although the water
molecules and hydronium are moving about, they don’t move far during these transfers, but a
hydronium ion has moved almost all the way across the frame and, in the movie, is shown
finally bonded to the oxygen atom on the far right of the frame. A similar sequence of proton
transfers is shown in the hydroxide ion movie where transfers of a proton from water to
hydroxide result in the movement of a hydroxide ion. [Theoretical calculations and
experimental reaction rates are consistent with this mechanism for the migration of hydronium
and hydroxide ions through aqueous solutions. It is only a matter of time until very fast
spectroscopic methods will be able to observe events like these “directly” to see if they actually
occur as we imagine.]
Problem 2.95.
(a) The balanced chemical reaction equation for the reaction of phosphorus pentoxide, P2O5(s),
with water to form a solution of phosphoric acid, (HO)3PO(aq) (or H3PO4(aq)) is:
P2O5(s) + 3H2O(l) 2H3PO4(aq)
(b) To find the molarity of the phosphoric acid solution formed when 1.42 g of phosphorus
pentoxide is mixed with 250. mL of water, we need to convert moles of phosphorus pentoxide
(molar mass = 142 g) to moles of phosphoric acid, using the reaction equation in part (a), and
then divide by the volume of solution, in liters, to get the molarity. We assume that the volume
of liquid does not change significantly when this small amount of phosphorus pentoxide is
added and reacted.
1.42 g P2O5 = (1.42 g P2O5)
= 8.0
50
1 mol P2 O5
142 g P2 O5
2 mol H3 PO 4 (aq)
1 mol P2 O5
10-2 M H3PO4(aq)
ACS Chemistry Chapter 2 suggested solutions
1
0.250 L
Chapter 2
Aqueous Solutions and Solubility
Problem 2.96.
The name is given for each of these ionic compounds with oxyanions. The arsenate oxyanion,
AsO43–, is named by analogy with its phosphorus analog, the phosphate oxyanion, PO43–.
(a) Ca(HSO4)2
calcium hydrogen sulfate
(b) Na2CO3
sodium carbonate
(c) Al2(HPO4)3
aluminum hydrogen phosphate
(d) Ce2SO4
cesium sulfate
(e) KHCO3
potassium hydrogen carbonate
(f) Na3AsO4
sodium arsenate
Problem 2.97.
The Lewis structures (showing all nonbonding electron pairs as a pair of dots and all covalent
bonds as lines) for the nitrate, ethanoate (acetate), and hydrogen sulfate oxyanions are:
O
O
O
H
O
C
C
N
H
O
O
H
O
S
O
H
O
nitrate
ethanoate
hydrogen sulfate
The hydrogen sulfate oxyanion is shown with six bonding electron pairs around the central
sulfur atomic core, as in Table 2.7. The marginal note on page 126 of the textbook indicates
that, in Chapters 5 and 6, we will find that third (and higher) period elements often
accommodate more than four bonding electron pairs and that the structure with six bonding
electron pairs and several double bonds seems to fit the properties of the oxyanion better than a
structure with four bonding pairs (an octet).
Problem 2.98.
Each Brønsted-Lowry acid and base is identified with an A or B, respectively, below the species
in these reactions. When necessary, the reaction in the problem is written out in is complete
ionic form to make identification of the acid and base possible.
(a)
H2S(g) + H2O(l)
A
B
HS–(aq) + H3O+(aq)
B
A
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
+
–
Na (aq) + OH (aq) + H3O+(aq) + Cl–(aq)
Na+(aq) + Cl–(aq) + H2O(l) + H2O(l)
B
A
B
A
The Na+(aq) and Cl–(aq) ions are spectator ions in this reaction, which, in aqueous solution, is
just a reaction between hydronium cation and hydroxide anion to produce two molecules of
water. Water can act as either a Brønsted-Lowry acid or base, depending upon whether it is a
proton donor or proton acceptor in the reaction.
(b)
NH3(g) + HCl(g)
NH4+Cl–(s)
B
A
A B
In this case, the molecule, HCl(g) is the acid (proton donor). This is unlike part (b), where
HCl(aq) does not exist as a molecule in aqueous solution, but as the hydronium cation and
chloride anion.
(c)
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Chapter 2
Problem 2.99.
Each Brønsted-Lowry acid and base is identified with an A or B, respectively, below the species
in these reactions. When necessary, the reaction in the problem is written out in is complete
ionic form to make identification of the acid and base possible.
(a)
NO2-(aq) + H3O+(aq)
B
A
HNO2(aq) + H2O(l)
A
B
(b) 2H3O+(aq) + 2ClO4–(aq) + Mg2+(OH–)2(s)
Mg2+(aq) + 2ClO4–(aq) + 4H2O(l)
2H3O+(aq) + 2ClO4–(aq) + Mg2+(OH–)2(s)
Mg2+(aq) + 2ClO4–(aq) + 2H2O(l) + 2H2O(l)
A
B
B
A
In this reaction, the hydronium ion from a perchloric acid, HOClO3(aq), solution reacts with the
hydroxide ion in solid magnesium hydroxide, Mg2+(OH–)2(s). The magnesium hydroxide
formula is written out this way to make explicit the presence of the ions in the solid. Like the
perchlorate anion, the magnesium cation is essentially a spectator ion in this reaction. See the
solution for Problem 2.98(b) for a discussion of the role of water molecules.
HNO3(aq) + Al3+(OH–)3(s)
Al3+(aq) + 3NO–4(aq) + 3H2O(l) + 3H2O(l)
3H3O+(aq) + 3NO3–(aq) + Al3+(OH–)3(s)
A
B
B
A
This reaction is modeled after the one in part (b), which was written out in full as a model.
(c)
(d)
HCN(aq) + NaOH(aq)
Assuming that hydrocyanic acid, HCN(aq), like hydrochloric acid, transfers a proton to a water
molecule to form H3O+(aq) and CN–(aq), we can write the reaction equation as:
H3O+(aq) + CN–(aq) + Na+(aq) + OH–(aq)
Na+(aq) + CN–(aq) + H2O(l) + H2O(l)
A
B
B
A
[At this point in this chapter, weak acids have not been introduced, so we are not forced to
consider whether the reaction could also be written with the aquated hydrogen cyanide
molecule, HCN(aq), as the acid:
HCN(aq) + Na+(aq) + OH–(aq)
Na+(aq) + CN–(aq) + H2O(l)
A
B
B
A
This, of course, changes the interpretation of the reaction somewhat and makes it clearer, as will
be discussed in Chapter 6, that there is a competition between OH–(aq) a CN–(aq) for protons
in this solution.]
Problem 2.100.
The Lewis structures (showing all nonbonding electron pairs as a pair of dots and all covalent
bonds as lines) for HOCO2–, (HO)2PO2–, and HOPO32– (modeled on the structures for the
corresponding oxyacids in Table 2.7) are:
O
O
C
H
O
H
hydrogen carbonate
52
O
P
2
O
O
O
H
O
dihydrogen phosphate
O
P
O
H
O
monohydrogen phosphate
ACS Chemistry Chapter 2 suggested solutions
Chapter 2
Aqueous Solutions and Solubility
Problem 2.101.
(a) The Brønsted-Lowry acids and bases in two possible acid-base reactions of ammonia and
water are identified by placing an A below each acid and a B below each base.
H2O(l) + NH3(g)
H3O+(aq) + NH2–(aq)
B
A
A
B
OH–(aq) + NH4+(aq)
H2O(l) + NH3(g)
A
B
B
A
The ammonia and water molecules switch roles in these two possible reactions.
(b) We predict that the second equation represents the correct acid-base reaction. Because
oxygen is more electronegative than nitrogen, the O–H bonding electrons in water are held more
closely by the oxygen than the N–H bonding electrons in ammonia. Thus, in comparison with
the H+ in the N–H bond, the H+ in the O–H bond can interact more effectively with non-bonding
electron pairs on other molecules. The result is that H+ from a water molecule is transferred to a
Brønsted-Lowry base more readily than the H+ from an ammonia molecule. If the second
reaction is the correct one, then solutions of NH3 in water will have extra OH–(aq) and will be
basic. The fact that the pH > 7 in aqueous ammonia solutions indicates that there is hydroxide
present in concentrations greater than found in neutral water (pH = 7) and confirms our
prediction.
Problem 2.102.
To explain the electrical conductivity observed in a solution of methylamine, CH3NH2(g),
dissolved in water, we need to consider how ions are formed in the interaction of the
methylamine and water molecules. By analogy with the reaction of ammonia with water, we can
write:
CH3NH2(aq) + H2O(l)
CH3NH3+(aq) + OH–(aq)
Since the electrical conduction by the solution is weak, we conclude that only a few ions are
produced, that is, the reaction does not proceed far toward products. We indicate this with an
equilibrium arrow in the balanced equation to show that the system comes to equilibrium before
all the methylamine has reacted.
Problem 2.103.
To predict/decide whether an aqueous solution of ethylene glycol, HOCH2CH2OH will be basic,
acidic, or neutral and whether it will conduct an electric current, we need to consider how the
molecules will interact with water molecules. Other compounds we know about with –OH
groups bonded to carbon are alcohols, like methanol (CH3OH) and ethanol (CH3CH2OH), and
compounds with multiple –OH groups, like glucose (se Figure 2.6 in the textbook). These
compounds are miscible with water (the alcohols) or very soluble (glucose), so it is not
surprising that ethylene glycol, a small molecule with two –OH groups, should be miscible with
water. The reason all these compounds are so soluble is that they can form several hydrogen
bonds with water and the non-polar part of the molecule does not disrupt the water structure
enough to be a major factor in their solubility. For all these molecules, including ethylene
glycol, the interactions with water do not involve any proton transfers that would form ionic
products, so the solutions do not conduct an electric current and are not acidic or basic.
ACS Chemistry Chapter 2 suggested solutions
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Chapter 2
Problem 2.104.
Our task in this problem is to use Le Chatelier’s principle to predict and clearly explain the
outcome of changes in the reaction conditions for this equilibrium reaction system:
CH3C(O)OH + HOCH2CH3
CH3C(O)OCH2CH3 + H2O
acetic acid
ethanol
ethyl acetate
(a) Starting with 0.1 mole of acetic acid and 0.1 mole of ethanol, would more, less, or the same
amount of ethyl acetate be formed, if water is added to the reaction mixture? Addition of water
is a disturbance to the equilibrium system, which will respond by reacting to use up some of the
added water. The reaction will proceed from the right to the left, using up both water and ethyl
acetate. The amount of ethyl acetate will be less when the extra water is added.
(b) Would a mixture of 0.2 mole of acetic acid and 0.1 mole of ethanol form more, less, or the
same amount of ethyl acetate as a mixture of 0.1 mole of acetic acid and 0.1 mole of ethanol?
Addition of acetic acid (0.2 mol instead of 0.1 mol) is a disturbance to the equilibrium system,
which will respond by reacting to use up some of the added acetic acid. The reaction will
proceed from the left to the right, using up both acetic acid and ethanol and forming more ethyl
acetate. The amount of ethyl acetate will be more when the extra water is added.
Problem 2.105.
Our task in this problem is to use Le Chatelier’s principle to predict and/or clearly explain the
outcome of the changes in the reaction conditions for two equilibrium reaction systems:
(a) To explain why the solubility of carbon dioxide is greater in an aqueous sodium hydroxide
solution, Na+(aq) + OH–(aq), than in water itself, we need to examine the reactions of dissolved
carbon dioxide and see how increasing the hydroxide anion concentration would affect them.
The dissolution equilibrium reactions are:
CO2(g) + H2O(l)
(HO)2CO(aq)
H3O+(aq) + HOCO2–(aq)
(HO)2CO(aq) + H2O(l)
Both of the products of the second reaction, H3O+(aq) and HOCO2–(aq), can react with OH–(aq)
to “use them up”:
H3O+(aq) + OH–(aq) 2H2O(l)
HOCO2–(aq) + OH–(aq) H2O(l) + CO32–(aq)
These two equations are also reversible and could be shown with equilibrium arrows, but the
directional arrow is used to show how the H3O+(aq) and HOCO2–(aq) are used up by added OH–
(aq) anion. Removing the two products of the dissolution reactions is a disturbance to the
system, which reacts by forming more of these products (which are, in turn, used up until all the
hydroxide anion has reacted). The direction of the response to the disturbance is shown by this
series of reactions with directional arrows:
CO2(g) + H2O(l) (HO)2CO(aq)
(HO)2CO(aq) + H2O(l)
H3O+(aq) + HOCO2–(aq)
H3O+(aq) + OH–(aq) 2H2O(l)
HOCO2–(aq) + OH–(aq) H2O(l) + CO32–(aq)
The overall reaction, in the presence hydroxide anion is the sum of these four reactions:
CO2(g) + 2OH–(aq) H2O(l) + CO32–(aq)
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(b) To predict what will happen to the concentration of calcium cation, [Ca2+(aq)], if sodium
sulfate (solid), Na2SO4(s), is added to a saturated aqueous solution of sparingly soluble calcium
sulfate, we need know what equilibrium reaction might be disturbed and what happens to the
added sodium sulfate. Sodium sulfate is a soluble ionic compound, so it will dissolve in the
solution to give Na+(aq) and SO42–(aq) ions. The equilibrium reaction in a saturated solution of
calcium sulfate is the equilibrium dissolution of the solid ionic compound:
CaSO4(s)
Ca2+(aq) + SO42–(aq)
Sodium sulfate is a soluble ionic compound, so it will dissolve in the solution to give Na+(aq)
and SO42–(aq) ions. The added SO42–(aq) anions are a disturbance to the equilibrium system and
the system will respond by trying to use up some of the ions by reacting to form more CaSO4(s).
This requires using up some of the Ca2+(aq) cation, so we predict that [Ca2+(aq)] will decrease
when Na2SO4(s) is added to a saturated aqueous solution of CaSO4(s).
Problem 2.106.
We are to represent each of these statements as a complete balanced chemical equation:
(a) Carbonic acid is formed when carbon dioxide reacts with water.
CO2(g) + H2O(l) (HO)2CO(aq)
(b) Calcium carbonate (limestone) reacts with carbonic acid to form an aqueous solution of
calcium hydrogen carbonate.
CaCO3(s) + (HO) 2CO(aq) Ca2+(aq) + 2HOCO2–(aq)
(c) Calcium hydrogen carbonate reacts with calcium hydroxide to form calcium carbonate
precipitate. [The implication of this statement is that one or both of the reactants is in aqueous
solution. Calcium hydroxide is a sparingly soluble ionic compound, so we might write the
reaction of solid calcium hydroxide with a solution of calcium hydrogen carbonate. We’ll do
this as well as write the reaction for both ionic compounds in solution.]
Ca2+(aq) + 2HOCO2–(aq) + Ca(OH)2(s) 2CaCO3(s) + 2H2O(l)
Ca2+(aq) + 2HOCO2–(aq) + Ca2+(aq) + 2OH–(aq)
2CaCO3(s) + 2H2O(l)
(d) Calcium hydrogen carbonate reacts with sodium hydroxide to form calcium carbonate
precipitate and the water-soluble salt sodium carbonate. [Assume both ionic compounds are in
aqueous solution.]
Ca2+(aq) + 2HOCO2–(aq) + 2Na+(aq) + 2OH–(aq)
2CaCO3(s) + 2Na+(aq) + CO32–(aq) + 2H2O(l)
(e) The mixing of aqueous solutions of sodium hydrogen carbonate and sodium hydroxide is
exothermic. A reaction has occurred but there is no precipitate.
Na+(aq) + HOCO2–(aq) + Na+(aq) + OH–(aq) 2Na+(aq) + CO32–(aq) + H2O(l)
Since the reaction is exothermic, you may show “energy” as a product, if you wish. In this case,
we are given this information, because there is no visible evidence (precipitate, gas formation,
color change, etc.) that a reaction has occurred.
Problem 2.107.
(a) As a droplet of solution containing calcium cations and hydrogen carbonate anions
evaporates from the tip of a stalactite, the concentration of these solute ions increases. The
ACS Chemistry Chapter 2 suggested solutions
55
Aqueous Solutions and Solubility
Chapter 2
volume of solution is less, so the number of ions per unit volume is greater, that is, a higher
concentration.
(b) The ions present in the droplet of solution in part (a) were a result of reaction (2.40):
Ca2+(aq) + 2HOCO22–(aq)
CaCO3(s) + H2O(l) + CO2(g)
The reaction is reversible and is shown here as an equilibrium. The increased concentrations of
the products of reaction (2.40), which were noted in part (a), are a disturbance to the system. Le
Chatelier’s Principle states that the system will respond by minimizing the disturbance, that is,
by reacting in a way that decreases the concentrations. The reaction that does this is the reverse
of reaction (2.40), which will proceed to precipitate solid calcium carbonate and release carbon
dioxide gas into the surrounding air in the cave.
(c) The result of the reaction in part (b) is to precipitate some solid calcium carbonate, which
causes the stalactite to grow downward (very slowly). If the drop of liquid should drop off the
tip of the stalactite, it will end up directly below and the same evaporation and precipitation
process on the floor of the cave will build stalagmites upward toward the stalactites. If they
meet, they form a column, which is another common structure in limestone caves.
Problem 2.108.
(a) We are asked what gas is evolved when calcium carbonate, a solid, is placed in an aqueous
solution of HCl(g). We know that HCl(g) dissolves and reacts completely with water to produce
hydronium cations and chloride anions (hydrochloric acid solution):
HCl(g) + H2O(l) H3O+(aq) + Cl–(aq)
Calcium carbonate is a sparingly soluble solid whose dissolution in water is represented by this
equilibrium reaction:
(i)
CaCO3(s)
Ca2+(aq) + CO32–(aq)
The carbonate ion product of this dissolution is a Brønsted-Lowry acid that can react with the
hydronium cation in the solution [from the dissolved HCl(g)] to form hydrogen carbonate anion
and this anion can react further with another hydronium ion to form carbonic acid:
(ii)
CO32–(aq) + 2H3O+(aq)
(HO)2CO(aq) + 2H2O(l)
We know that carbonic acid is formed in water when carbon dioxide gas dissolves and we can
write this equilibrium reaction in reverse to show how carbon dioxide gas is released by the
decomposition of carbonic acid:
(iii) (HO)2CO(aq)
CO2(g) + H2O(l)
This series of reactions shows that carbon dioxide is likely to be the gas evolved when calcium
carbonate is placed in hydrochloric acid solution.
(b) The net ionic equation for the reaction that produces the gas when solid calcium carbonate is
placed in hydrochloric acid solution is the sum of reactions (i), (ii), and (iii) in part (a):
CaCO3(s) + 2H3O+(aq)
Ca2+(aq) + CO2(g) + 3H2O(l)
(c) Only a tiny amount of calcium carbonate (marble, chalk) dissolves in water, which is
represented by equation (i) in part (a). The addition of hydronium ion [from the dissolved
HCl(g)] is a disturbance to this solubility equilibrium because the hydronium ion reacts with
one of the products, carbonate anion and reduces its concentration. In response, LeChatelier’s
principle, more of the solid calcium carbonate dissolves to try to increase the carbonate anion
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Aqueous Solutions and Solubility
concentration. If there is a good deal of hydronium present, it continues to react (and the
calcium carbonate continues to dissolve) forming more and more carbonic acid, because the
presence of a lot of hydronium ion displaces equilibrium equation (ii) toward products. The
increase in concentration of carbonic acid is a disturbance to equilibrium reaction (iii), which
responds, LeChatelier’s principle, by trying to use it up to form products, one of which, carbon
dioxide gas, we observe as it leaves the reaction solution.
Problem 2.109.
The molecular level representation of sucrose in the Web Companion Chapter 2, Section 2.2,
page 3 shows that sugar molecules have several regions of positive and negative charge, which
are due to the presence of a number of –OH groups that are shown in the molecular structure on
this page. Thus, in its solid crystals, the sugar molecules are oriented in specific directions, as
shown in the movie, that maximize their positive-negative attractions and hydrogen bonding
between molecules. When you put pressure on the crystal (by trying to crush it, for example),
these many oriented interactions strongly resist being reoriented and the crystal breaks into
smaller pieces as some of the interactions are finally broken. The molecular level representation
of grease on page 5, shows that the molecules are long chains of hydrocarbons that have
essentially no polarity. Attractions of these molecules for one another are through nondirectional dispersion (induced dipole) interactions and their packing in the solid (or semisolid)
form is rather random, as shown in the movie. Thus, when pressure is applied, the hydrocarbon
molecules can slide over each other because their attractions are non-directional and any
orientation of the molecules with respect to one another is as good as any other.
Problem 2.110.
(a) The dissolution of calcium sulfate, CaSO4(s), in water can be represented as:
Ca2+(aq) + SO42–(aq)
CaSO4(s)
About 2 g of calcium sulfate dissolve in a liter of water, so we can calculate the number of
moles that dissolve in a liter, using the molar mass of CaSO4(s) (= 136 g·mol–1), and from the
stoichiometry of the dissolution reaction find the molarities of the ions in the saturated solution:
1 mol CaSO 4
2 (g CaSO4)·L = [2 (g CaSO4)·L ]
136 g CaSO4
–1
–1
2+
1 mol Ca
1 mol CaSO4
= [Ca2+(aq)]
= 0.015 mol·L–1
The stoichiometry of the dissolution shows that [Ca2+(aq)] = [SO42–(aq)] = 0.015 mol·L–1.
Thus, a saturated solution of calcium sulfate is 0.015 M in both ions. From the table in Consider
This 2.54, we find that seawater contains 0.010 M Ca2+(aq) and 0.028 M SO42–(aq). Seawater
contain less calcium ion than in a saturated solution of calcium sulfate, but almost twice as
much sulfate ion as in the saturated calcium sulfate solution. It’s hard to tell from these data
whether seawater is saturated with calcium sulfate, because, at this point in the textbook, our
analysis of solubility has not provided a way to handle a situation like this. [The remainder of
the problem is designed to help get a qualitative feeling for what is going on and Chapter 9 will
provide a quantitative way to analyze solubilities in a solution like seawater.]
(b) Solubility equilibria, like the one represented in the equation in part (a), should respond (Le
Chatelier’s principle) to the disturbance caused by adding more of the cation or anion (by
adding a soluble ionic compound containing the cation or anion, for example). Since the ions
are on the product side of the solubility equilibrium, the system will react by trying to remove
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some of the added ion by formation of more solid. Removing some of the added ion also, of
course, removes some of the other ion, the counterion. In the case of the calcium sulfate in this
problem, addition of SO42–(aq) (by adding some sodium sulfate, Na2SO4, for example) to a
saturated solution of calcium sulfate will result in formation of a little more solid calcium
sulfate and a decreased concentration of Ca2+(aq). Looked at from another point of view, the
amount of calcium sulfate that will dissolve in a solution that already contains SO42–(aq) will be
lower, the solid will be less soluble, than in pure water. A saturated solution under these
conditions will contain a lower concentration of Ca2+(aq) than a saturated solution in pure
water. The ionic compound will be less soluble than it would be in pure water because Le
Chatelier’s principle favors the precipitation (or decreased solubility).
(c) Seawater contains more sulfate ion than is present in the saturated solution of calcium sulfate
we calculated in part (a). Since seawater contains more of the anion than could have been
present from CaSO4(s) alone, we can imagine that the solution is formed when CaSO4(s)
dissolves in a solution that already contains some of the anion. From part (b), we find that the
solubility of the CaSO4(s) will be lower in such a solution. This is the direction of the effect we
observe; there is less calcium cation in seawater than would be present in a saturated solution in
pure water. Although we still cannot be certain, this analysis makes it more likely that we can
conclude that seawater is saturated with CaSO4(s), in the presence of extra sulfate anion from
some other source. [In Chapter 9, we will find that the product of the concentrations of the
cation and anion characterizes the solubility equilibrium for a saturated solution of a sparingly
soluble salt like CaSO4. For the saturated solution in water, we have
[Ca2+(aq)][SO42–(aq)] = (0.015 M)(0.015 M) = 2.2 10–4 M2
For seawater, we have
[Ca2+(aq)][SO42–(aq)] = (0.010 M)(0.028 M) = 2.8 10–4 M2
These two products are almost the same and reinforce the conclusion that seawater is saturated
with CaSO4, in the presence of extra sulfate anion.]
(d) Since seawater already contains a substantial concentration of calcium anion, the solubility
of calcium carbonate will be lower than it would be in pure water. The solubility of calcium
carbonate is very low in pure water, so lowering it even further helps explain why seashells do
not redissolve (at least not rapidly) in the sea.
Problem 2.111.
We are asked to find the molarity of nitrate, NO3–(aq), anion in a sample of salt water with a
density of 1.02 g·mL–1 that contains 17.8 ppm (parts per million, by mass) of NO3–(aq). If we
can find the mass of NO3–(aq) in a known volume of the solution, we can use the molar mass of
NO3–(aq) (= 62.01 g) to get the number of moles and then the molarity. The concentration, 17.8
ppm (by mass) means that exactly one million grams of solution contain 17.8 g of NO3–(aq).
We can use this information to convert the density (mass per milliliter) to grams and then to
moles and molarity:
1.02 g·mL–1 sol’n = (1.02 g·mL–1 sol’n)
= 2.93
58
17.8 g NO–3
1 10 6 g sol'n
–
1 mol NO3
–
62.01 g NO3
10–4 mol·L–1
ACS Chemistry Chapter 2 suggested solutions
1000 mL
1L
Chapter 2
Aqueous Solutions and Solubility
Problem 2.112.
(a) We are asked to find the molarity of sodium chloride, NaCl, in a solution of salt water
(representing seawater) that has a density of 1.025 g·mL–1 and contains 3.50% by weight (mass)
of NaCl. If we can find the mass of NaCl in a known volume of the solution, we can use the
molar mass of NaCl (= 58.44 g) to get the number of moles and then the molarity. The
concentration, 3.50% by mass means that exactly one hundred grams of solution contain 3.50 g
of NaCl. We can use this information to convert the density (mass per milliliter) to grams and
then to moles and molarity:
1.025 g·mL–1 sol’n = (1.025 g·mL–1 sol’n)
3.50 g NaCl
1 10 2 g sol'n
1 mol NaCl
58.44 g NaCl
1000 mL
1L
= 0.614 mol·L–1
(b) In Consider This 2.54(b) we found that the sum of the concentrations of positive charges in
seawater is 0.606 M (and for negative charges 0.603 M). The result in part (a) is consistent
(within about 1.3%) with the answer to Consider This 2.54(b), because the solution in part (a)
has a 0.614 M concentration of positive charge from the Na+ cation (and the same concentration
of negative charge from the Cl– anion).
Problem 2.113.
We are given a good deal of information about a solution prepared by a student. The solution
contains 5.15 g of a compound dissolved in 10.0 g of water, has a density of 1.34 g·mL–1, and is
2.7 M. We are asked to determine which of these ionic compounds is the solute: (NH4)2SO4, KI,
CsCl, or Na2S2O3. If we can find out how many moles of solute are dissolved, we can determine
its molar mass and compare the result with the molar masses of the possible solutes. The known
molarity of the solution, 2.7 M, is equal to the number of moles dissolved divided by the
volume (liters) of the solution. To get the volume of solution, we divide the total mass of the
solution 15.15 g (= 5.15 g + 10.0 g) by its density:
15.15 g sol’n = (15.15 g sol’n)
1 mL sol'n
1.34 g sol'n
1L
1000 mL
Therefore, using the molarity of the solution, we have:
2.7 mol solute
0.0113 L sol’n = (0.0113 L sol’n)
1 L sol'n
= 0.0113 L sol’n
= 0.0305 mol solute
It is not really legitimate to carry the third significant figure in this result, since the molarity has
an uncertainly of about 3% (1 part in 27), but we’ll carry it along and check whether it makes a
difference later. Now we know that 5.15 g of the solute is 0.0305 mol of the solute, so we can
find its molar mass:
5.15 g solute
molar mass =
= 169 g·mol–1
0.0305 mol solute
The molar masses of the four possible compounds are: (NH4)2SO4 , 132 g·mol–1; KI, 166 g·mol–
1
; CsCl, 168 g·mol–1; and Na2S2O3, 158 g·mol–1. The molar mass we calculated has about a 3%
uncertainty from the uncertainty in the second significant figure of molarity, so the range of
possibilities is about 164 to 174 g·mol–1. The possibilities, therefore, are KI or CsCl for the
compound the student dissolved. The calculation fits CsCl best, but the uncertainty makes this
identification unsure. We would need a better value for the molarity to be sure.
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Problem 2.114.
We are asked to find the approximate percentage of the water molecules that is protonated in
aqueous solutions with different pHs. We can use Figure 2.24, or the relationship, [H3O+(aq)] =
10–pH, to find the molarity of hydronium ion (protonated water molecules) in each solution and
compare this molarity to the molarity of the water, 55.5 M, to get the fraction/percentage of the
water molecules that are protonated.
(a) At pH 7, we have [H3O+(aq)] = 10–7 M, so the percent of molecules protonated is:
% H2O protonated =
10
–7
+
mol [H3 O (aq)]
100% = 2
55.5 mol H2 O(l )
10–7 %
(b) At pH 6, we have [H3O+(aq)] = 10–6 M, so the percent of molecules protonated is:
% H2O protonated =
10
–6
+
mol [H3 O (aq)]
100% = 2
55.5 mol H2 O(l )
10–6 %
(a) At pH 4, we have [H3O+(aq)] = 10–4 M, so the percent of molecules protonated is:
% H2O protonated =
10
–4
+
mol [H 3O (aq )]
100% = 2
55.5 mol H2 O(l )
10–4 %
Problem 2.115.
(a) Figure 2.24 shows that [OH–(aq)] = 10–11 M and [H3O+(aq)] = 10–3 M in aqueous solution at
pH 3. At this pH, the numeric value of the mathematical product [H3O+(aq)]·[OH–(aq)] = (10–
3
M)(10–11 M) = 10–14 M2.
(b) For any pH you choose, the mathematical product [H3O+(aq)]·[OH–(aq)] = 10–14 M2. This
observation certainly suggests that there must be some underlying molecular phenomenon that
is responsible for the constancy of this product of concentrations. In pure water, the only source
of hydronium and hydroxide ions is the reaction of water with itself:
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
Recall that in pure water, [H3O+(aq)] = [OH–(aq)] = 10–7 M and, therefore, as in any aqueous
solution, [H3O+(aq)]·[OH–(aq)] = 10–14 M2. If the concentration of hydroxide anion is increased
(by adding a tiny bit of sodium hydroxide, for example) in this system, the equilibrium is
disturbed and the system will respond (Le Chatelier’s principle) by trying to use up some of the
added hydroxide. This will, of necessity require reaction of some of the hydronium cation to
form some water, that is, the equilibrium reaction, as written, will go toward reactants. The
result will be a solution with [H3O+(aq)] < 10–7 M and [OH–(aq)] > 10–7 M. Apparently, given
our observation about the product of concentrations, [H3O+(aq)]·[OH–(aq)], the quantitative
decrease in [H3O+(aq)] and increase in [OH–(aq)] exactly compensate one another.
Problem 2.116.
(a) If the sulfur dioxide, SO2(g), molecule has a permanent dipole moment, the S–O bond
dipoles must not cancel out. Sulfur dioxide must have a bent structure, similar to water. See
Chapter 1, Section 1.6, for a discussion of bond dipoles and molecular dipole moments.
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Aqueous Solutions and Solubility
(b) If a solution of SO2 in water conducts an electric current, the dissolved gas must have
formed ions in the solution. From Section 2.13, we know that the oxides of nonmetals often
dissolve and react with water to give a solution of an oxyacid. For sulfur dioxide, the reaction is:
SO2(g) + H2O(l)
(HO)2SO(aq)
The product, sulfurous acid, can transfer one or both protons to water to form ions:
H3O+(aq) + HOSO2–(aq)
(HO)2SO(aq) + H2O(l)
H3O+(aq) + SO32–(aq)
HOSO2–(aq) + H2O(l)
Although sulfurous acid transfers only about 10% of its protons to water, enough ions will be
present in the solution to conduct electricity.
Problem 2.117.
(a) The name “bicarbonate” differentiates the HCO3– ion from the carbonate ion, CO32–. It could
be that the “bi” refers to the fact that it takes two HCO3– ions to be “equivalent” to one CO32–
ion when forming ionic compounds and in their reactions. For example, when calcium
carbonate is reacted with an acid, carbon dioxide gas is formed and one mole of gas is formed
for every two moles of hydronium ion that react:
CaCO3(s) + 2H3O+(aq) Ca2+(aq) + CO2(g) + 2H2O(l)
However, when calcium bicarbonate is reacted with an acid two moles of carbon dioxide gas are
formed for every two moles of hydronium ion that react:
Ca(HOCO2)2(s) + 2H3O+(aq) Ca2+(aq) + 2CO2(g) + 4H2O(l)
Ratios like these, amount of gas formed for a given amount of acid used, for example, were the
only kind of data scientists had when molecular formulas and atomic masses were not known.
Thus, the names for the constituents of compounds had names reflecting such observations and
results.
(b) Sodium ions are monopositive, Na+, and phosphate anions have a 3– charge, PO43–, so the
ionic compound they form must be Na3PO4, trisodium phosphate = TSP.
Problem 2.118.
(a) The equivalence point in a reaction occurs when the moles of one reactant are exactly
stoichiometrically equivalent to the moles of the other. In this problem, we are reacting an acid,
(HO)2SO2(aq), with a base dissolved in water, KOH(aq). The reaction that occurs is between
hydronium cations from the acid and hydroxide anions from the base, but we can also write the
reaction in terms of the two compounds:
(HO)2SO2(aq) + 2KOH(aq) 2H2O(l) + 2K+(aq) + SO42–(aq)
We want to know the volume of 0.075 M sulfuric acid solution, (HO)2SO2(aq), required to
reach the equivalence point of the reaction with 1.00 g of KOH(s) dissolved in 75 mL of water.
Use the molar mass of KOH (= 56.1 g) to find the number of moles of KOH we have. Then
determine the number of moles of (HO)2SO2 required to react with this amount of KOH and,
finally, convert moles of (HO)2SO2 to volume of sulfuric acid solution, using the molarity of the
sulfuric acid solution. A single equation that accomplishes these three steps is:
ACS Chemistry Chapter 2 suggested solutions
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Aqueous Solutions and Solubility
Chapter 2
1 mol KOH
56.1 g KOH
1.00 g KOH = (1.00 g KOH)
1 mol (HO)2 SO2
2 mol KOH
1 L sol'n
0.075 mol (HO)2 SO2
= 0.119 L = 119 mL sulfuric acid solution
(b) In the second part of this problem, we are to find the volume of 0.075 M sulfuric acid
solution required to reach the equivalence point of the reaction with 1.00 g of KOH(s) dissolved
in 150 mL of water. The amount of KOH here is exactly the same as in part (a). Note that the
amount of water the KOH is dissolved in was never considered in part (a), because this volume
is irrelevant. All that matters for the calculation is the mass/moles of KOH. Therefore, the
volume of 0.075 M sulfuric acid solution required here is the same as in part (a), 119 mL.
Problem 2.119.
(a) The reaction between drain cleaner, NaOH(s), and vinegar, a 0.9M aqueous solution of
ethanoic acid, CH3C(O)OH(aq), actually occurs between hydroxide anions as they dissolve in
the aqueous vinegar and the hydronium cations from the acid, but we can write a balanced
chemical reaction equation for the reaction between the drain cleaner and vinegar in terms of the
compounds:
CH3C(O)OH(aq) + NaOH(s) H2O(l) +Na+(aq) + CH3C(O)O–(aq)
(b) To find the minimum volume of vinegar required to react completely with one pound
(454 g) of drain cleaner [that is, for the reactants to be equivalent by the stoichiometric equation
in part (a)], use the molar mass of NaOH (= 40 g) to find the number of moles of NaOH we
have. Then determine the number of moles of CH3C(O)OH(aq) required to react with this
amount of NaOH and, finally, convert moles of CH3C(O)OH(aq) to volume of vinegar, ethanoic
acid solution, using the molarity of the ethanoic acid solution. A single equation that
accomplishes these three steps is:
454 g NaOH = (454 g NaOH)
1 mol NaOH
40 g NaOH
1 mol CH2 C(O)OH
1 mol NaOH
1 L sol'n
0.9 mol CH2 C(O)OH
= 13 L vinegar (ethanoic acid solution)
Problem 2.120.
(a) We are asked to find the number of moles of hydronium ion, H3O+(aq), in an acid-rain
acidified pond with a volume of 4.5 104 m3 (1 m3 = 1000 L) and [H3O+(aq)] = 5.0 10–5 M
(pH = 4.30). Use the volume and molarity to calculate number of moles:
4.5
4
3
10 m = (4.5
1000 L
10 m )
1 m3
4
3
5.0 10 –5 mol H3 O+ (aq)
1L
= 2.25 103 mol H3O+(aq) 2.3 103 mol H3O+(aq)
The experimental data (volume of the pond and its pH) do not justify three significant figures
(better than 1% uncertainty) in the moles of hydronium, but we will carry the three in the next
calculation and round off at the end.
(b) You wish to explain how much lime, CaO(s), will be required to react with 90% of the
hydronium ion in the pond by this reaction:
2H3O+(aq) + CaO(s) 3H2O(l) + Ca2+(aq)
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This equation shows us that we need one mole of lime for every two moles of hydronium ion
that we want to react. We want to react 90% of the H3O+(aq): [2.25 103 mol H3O+(aq)](.90) =
2.03 103 mol H3O+(aq). Thus, the number of moles of CaO(s) needed is:
103 mol H3O+)
(2.03
1 mol CaO
+
2 mol H3 O
= 1.0
103 mol CaO(s) [about 1000 mol CaO]
One mole of CaO has a mass of 56 g, so 1.0 103 mol CaO has a mass of 56 103 g = 56 kg. A
mass of one kilogram is 2.2 pounds, so we need (56 kg)(2.2 pound·kg–1) = 123 pounds of
CaO(s) to react with 90% of the hydronium ion in the pond. Since the lime is purchased in
40-pound bags, we will need to purchase three bags and use all of two bags and about half of
the third.
(c) After the lime has been added to react with 90% of the hydronium ion, 10% of the original
number of moles of H3O+(aq) will remain, that is, [2.3 103 mol H3O+(aq)](0.10) =
2.3 102 mol H3O+(aq) are still present in the pond. The molar concentration, [H3O+(aq)], is:
2.3 10 2 mol H3O+
= 5.1
[H3O (aq)] =
4.5 10 7 L
+
10–6 M
The pH of the pond is now about 5.3 [–log(5.1 10–6 M) = 5.29], that is, about one pH unit
higher (less acidic) than before the addition of the lime. Note that reduction of the [H3O+(aq)]
by a factor of 10 increases the pH by one unit. You can also see this relationship in Figure 2.24.
Problem 2.121.
(a) When chickens pant in warm weather, the shells of their eggs are less sturdy than usual
because they contain too little calcium carbonate, which is what make an eggshell hard. The
precipitation of calcium carbonate is being affected because the concentration of carbonate in
the chicken’s bloodstream goes down when more than the normal amount of carbon dioxide is
excreted. The directional flow of the reactions (by La Chatelier’s principle) is:
CaCO3(s) Ca2+(aq) + CO32–(aq)
CO32–(aq) + H2O(l)
HOCO2–(aq) + H2O(l)
HOCO2–(aq) + OH–(aq)
(HO)2CO(aq) + OH–(aq)
(HO)2CO(aq) CO2(g) + H2O(l)
As the chicken pants and more CO2(g) leaves, CaCO3(s) tends to dissolve, as shown, rather than
precipitate to form strong eggshells.
(b) To counteract the loss of CO2(g) in the chickens’ breath, the farmers give them carbonated
water, seltzer water, to drink. Chickens apparently like the seltzer water better than plain water
and drink even more than usual, so the method is even more effective than might have been
expected.
Problem 2.122.
The animated movie in the Web Companion Chapter 2, Section 2.6, page 3, showing the
interaction of chloride and silver ions provides one way to visualize the precipitation process
that we usually write like this:
Ag+(aq) + Cl–(aq)
AgCl(s)
ACS Chemistry Chapter 2 suggested solutions
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Aqueous Solutions and Solubility
Chapter 2
The movie tries to suggest some of the complexity of the molecular/ionic interactions that result
in the formation of a precipitate when appropriate cations and anions are mixed. The ions are
surrounded by water molecules in their hydration layers and these interactions with the water
molecules compete with the interactions of the cations with the anions. However, the cations
and anions sometimes come together to form an ion pair, AgCl(aq), in which some of the waters
of hydration from the individual ions are lost. When these ion pairs meet one another in a
favorable orientation, larger groups of ions (aggregates) can form, Ag2Cl2(aq), with the likely
loss of more waters of hydration. This process continues with the aggregates growing as more
ion pairs and individual ions join, until the aggregate is no longer solvated enough to hold the
mass in solution and it falls out of solution as a solid, AgxClx(s). Ion pairs and individual ions
continue to join this solid and ultimately form the visible crystal that we represent as AgCl(s),
with no indication of the unknown (very large) number of each ion present. All these processes
are reversible, but go in the direction of precipitation until the solution reaches saturation and
the concentrations of the ions are in equilibrium with the solid crystals. It is not easy to translate
the molecular level animation to a symbolic representation, particularly because the number of
waters of hydration on all the species is unknown. Assuming that each ion has six waters of
hydration and that about half are lost by the aggregates at each step, we can write these
molecular level reactions:
Ag+(H2O)6 + Cl–(H2O)6
AgCl(H2O)6 + 6H2O(l)
AgCl(H2O)6 + AgCl(H2O)6
Ag2Cl2(H2O)6 + 6H2O(l)
Ag2Cl2(H2O)6 + AgCl(H2O)6
Ag3Cl3(H2O)6 + 6H2O(l)
And so on and on the process continues until the aggregates are too large to be held in solution
by their waters of hydration and, finally, we might write:
AgnCln(aq)
AgCl(s) + some H2O(l) molecules leaving the solid surface
Remember that these symbolic representations still simplify what is going on, since many of
each of these kinds of reactions are happening simultaneously in the solution. A “simple”
reaction like precipitation is really not so simple.
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ACS Chemistry Chapter 2 suggested solutions