13
Properties
of Solutions
Visualizing Concepts
13.1
+
+
The energy of the ion-solvent interaction is greater for Li t h a n N a . The smaller size
of the Li+ ion means that ion-dipole interactions with polar water molecules are
stronger.
13.5
Diagram (b) is the best representation of a saturated solution. There is some
undissolved solid with particles that are close together and ordered, in contact with a
solution containing mobile, separated solute particles. As much solute has dissolved as
can dissolve, leaving some undissolved solid in contact with the saturated solution.
13.9
(a)
Yes, the molarity changes with a change in temperature. Molarity is defined as
moles solute per unit volume of solution. If solution volume is different, molarity
is different.
(b)
No, molality does not change with change in temperature. Molality is defined as
moles solute per kilogram of solvent. Even though the volume of solution has
changed due to increased kinetic energy, the mass of solute and solvent have not
changed, and the molality stays the same.
The Solution Process
13.13
If the enthalpy released due to solute-solvent attractive forces (ΔH 3 ) is at least as large
as the enthalpy required to separate the solute particles (ΔH 1 ), the overall enthalpy of
solution (ΔH s oln) will be either slightly endothermic (owing to +ΔH 2 ) or exothermic.
Even if ΔH s oln is slightly endothermic, the increase in disorder due to mixing will cause
a significant amount of solute to dissolve. If the magnitude of ΔH 3 is small relative to
the magnitude of ΔH 1 , ΔH s oln will be large and endothermic (energetically unfavorable)
and not much solute will dissolve.
13.15
Analyze/Plan. Decide whether the solute and solvent in question are ionic, polar
covalent, or nonpolar covalent. Draw Lewis structures as needed. Then state the
appropriate type of solute-solvent interaction. Solve.
(a)
CCl 4 , nonpolar; benzene, nonpolar; dispersion forces
(b)
methanol, polar with hydrogen bonding; water, polar with hydrogen bonding;
hydrogen bonding
(c)
KBr, ionic; water, polar; ion-dipole forces
(d)
HCl, polar; CH 3 CN, polar; dipole-dipole forces
348
13 Properties of Solutions
13.17
13.19
Solutions to Exercises
(a)
Lattice energy is the amount of energy required to completely separate a mole of
solid ionic compound into its gaseous ions (Section 8.2). For ionic solutes, this
corresponds to ΔH 1 (solute-solute interactions) in Equation [13.1].
(b)
In Equation [13.1], ΔH 3 is always exothermic. Formation of attractive
interactions, no matter how weak, always lowers the energy of the system,
relative to the energy of the isolated particles.
(a)
ΔH s oln is determined by the relative magnitudes of the “old” solute-solute (ΔH 1 )
and solvent-solvent (ΔH 2 ) interactions and the new solute-solvent interactions
(ΔH 3 ); ΔH s oln = ΔH 1 + ΔH 2 + ΔH 3 . Since the solute and solvent in this case
experience very similar London dispersion forces, the energy required to separate
them individually and the energy released when they are mixed are
approximately equal.
(b)
ΔH 1 + ΔH 2 ≈ –ΔH 3 . Thus, ΔH s oln is nearly zero.
Mixing hexane and heptane produces a homogeneous solution from two pure
substances, and the randomness of the system increases. Since no strong
intermolecular forces prevent the molecules from mixing, they do so
spontaneously due to the increase in disorder.
Saturated Solutions; Factors Affecting Solubility
13.21
13.23
(a)
Supersaturated
(b)
Add a seed crystal. Supersaturated solutions exist because not enough solute
molecules are properly aligned for crystallization to occur. A seed crystal
provides a nucleus of already aligned molecules, so that ordering of the dissolved
particles is more facile.
Analyze/Plan. On Figure 13.17, find the solubility curve for the appropriate solute. Find
the intersection of 40°C and 40 g solute on the graph. If this point is below the solubility
curve, more solute can dissolve and the solution is unsaturated. If the intersection is on
or above the curve, the solution is saturated. Solve.
(a)
13.25
13.27
unsaturated
(b)
saturated
(c)
saturated
(d)
unsaturated
The liquids water and glycerol form homogenous mixtures (solutions), regardless of the
relative amounts of the two components. Glycerol has an –OH group on each C atom in
the molecule. This structure facilitates strong hydrogen bonding similar to that in
water. Like dissolves like and the two liquids are miscible in all proportions.
(a)
Dispersion interactions among nonpolar CH 3 (CH 2 ) 1 6 –chains dominate the
properties of stearic acid. It is more soluble in nonpolar CCl 4 than polar
(hydrogen bonding) water, despite the presence of the –COOH group.
349
13 Properties of Solutions
Solutions to Exercises
(b)
Dioxane can act as a hydrogen bond acceptor, so it will be more soluble than
cyclohexane in water.
13.29
13.31
13.33
Analyze/Plan. Hexane is a nonpolar hydrocarbon that experiences dispersion forces
with other C 6 H 1 4 molecules. Solutes that primarily experience dispersion forces will be
more soluble in hexane. Solve.
(a)
CCl 4 is more soluble because dispersion forces among nonpolar CCl 4 molecules
are similar to dispersion forces in hexane. Ionic bonds in CaCl 2 are unlikely to be
broken by weak solute-solvent interactions. For CaCl 2 , ΔH 1 is too large, relative
to ΔH 3 .
(b)
Benzene, C 6 H 6 , is also a nonpolar hydrocarbon and will be more soluble in
hexane. Glycerol experiences hydrogen bonding with itself; these solute-solute
interactions are less likely to be overcome by weak solute-solvent interactions.
(c)
Octanoic acid, CH 3 (CH 2 ) 6 COOH, will be more soluble than acetic acid
CH 3 COOH. Both solutes experience hydrogen bonding by –COOH groups, but
octanoic acid has a long, rod-like hydrocarbon chain with dispersion forces
similar to those in hexane, facilitating solubility in hexane.
(a)
Carbonated beverages are stored with a partial pressure of CO 2 (g) greater than 1
atm above the liquid. A sealed container is required to maintain this CO 2
pressure.
(b)
Since the solubility of gases increases with decreasing temperature, some CO 2 (g)
will remain dissolved in the beverage if it is kept cool.
Analyze/Plan. Follow the logic in Sample Exercise 13.3. Solve.
S H e = 3.7 × 10
–4
M/atm × 1.5 atm = 5.6 × 10
–4
M
Concentrations of Solutions
13.35
Analyze/Plan. Follow the logic in Sample Exercise 13.4.
(a)
(b)
350
Solve.
13 Properties of Solutions
13.37
Solutions to Exercises
Analyze/Plan. Given masses of CH 3 OH and H 2 O, calculate moles of each component.
(a)
Mole fraction CH 3 OH = (mol CH 3 OH)/(total mol)
(b)
mass % CH 3 OH = [(g CH 3 OH)/(total mass)] × 100
(c)
molality CH 3 OH = (mol CH 3 OH)/(kg H 2 O). Solve.
(a)
(b)
(c)
13.39
Analyze/Plan. Given mass solute and volume solution, calculate mol solute, then
molarity = mol solute/L solution. Or, for dilution, M c × L c = M d × L d . Solve.
(a)
(b)
(c)
13.41
M c × L c = M d × L d ; 3.50 M HNO 3 × 0.0250 L = ?M HNO 3 × 0.250 L
250 mL of 0.350 M HNO 3
Analyze/Plan. Follow the logic in Sample Exercise 13.5.
Solve.
(a)
(b)
13.43
The density of H 2 O = 0.997 g/mL = 0.997 kg/L.
Analyze/Plan. Assume 1 L of solution. Density gives the total mass of 1 L of solution.
The g H 2 SO 4 /L are also given in the problem. Mass % = (mass solute/total mass
solution)
100. Calculate mass solvent from mass solution and mass solute. Calculate
moles solute and solvent and use the appropriate definitions to calculate mole fraction,
molality, and molarity.
Solve.
(a)
mass percent is thus 0.4301 × 100 = 43.01% H 2 SO 4
(b)
In a liter of solution there are 1329 – 571.6 = 757.4 = 757 g H 2 O.
351
13 Properties of Solutions
Solutions to Exercises
(The result has 3 sig figs because 42.0 mol H 2 O limits the denominator to 3 sig
figs.)
(c)
(d)
13.45
Analyze/Plan. Given: 98.7 mL of CH 3 CN(l), 0.786 g/mL; 22.5 mL CH 3 OH, 0.791 g/mL.
Use the density and volume of each component to calculate mass and then moles of
each component. Use the definitions to calculate mole fraction, molality, and molarity.
Solve.
(a)
13.47
(b)
Assuming CH 3 OH is the solute and CH 3 CN is the solvent,
(c)
The total volume of the solution is 121.2 mL, assuming volumes are additive.
Analyze/Plan. Given concentration and volume of solution use definitions of the
appropriate concentration units to calculate amount of solute; change amount to moles
if needed. Solve.
(a)
(b)
Assume that for dilute aqueous solutions, the mass of the solvent is the mass of
solution. Use proportions to get mol KCl.
(c)
Use proportions to get mass of glucose, then change to mol glucose.
352
13 Properties of Solutions
13.49
Solutions to Exercises
Analyze/Plan. When preparing solution, we must know amount of solute and solvent.
Use the appropriate concentration definition to calculate amount of solute. If this
amount is in moles, use molar mass to get grams; use mass in grams directly. Amount
of solvent can be expressed as total volume or mass of solution. Combine mass solute
and solvent to produce the required amount (mass or volume) of solution. Solve.
(a)
Weigh out 1.5 g KBr, dissolve in water, dilute with stirring to 0.75 L (750 mL).
(b)
Mass of solution is required, but density is not specified. Use molality to calculate
mass fraction, and then the masses of solute and solvent needed for 125 g of
solution.
Thus,
In 125 g of the 0.180 m solution, there are
Weigh out 2.62 g KBr, dissolve it in 125 – 2.62 = 122.38 = 122 g H 2 O to make
exactly 125 g of 0.180 m solution.
(c)
Using solution density, calculate the total mass of 1.85 L of solution, and from the
mass % of KBr, the mass of KBr required.
0.120 (2035 g soln) = 244.2 = 244 g KBr
Dissolve 244 g KBr in water, dilute with stirring to 1.85 L.
(d)
Calculate moles KBr needed to precipitate 16.0 g AgBr. AgNO 3 is present in
excess.
Weigh out 0.0852 mol KBr (10.1 g KBr), dissolve it in a small amount of water,
353
13 Properties of Solutions
Solutions to Exercises
and dilute to 0.568 L.
13.51
Analyze/Plan. Assume a solution volume of 1.00 L. Calculate the mass of 1.00 L of
solution and the mass of HNO 3 in 1.00 L of solution. Mass % = (mass solute/mass
solution) × 100. Solve.
13.53
Analyze. Given: 80.0% Cu, 20.0% Zn by mass; density = 8750 kg/m . Find: (a) m of Zn
(b) M of Zn
3
(a)
Plan. In the brass alloy, Zn is the solute (lesser component) and Cu is the solvent
3
(greater component). m = mol Zn/kg Cu. Assume 1 m → 8750 kg of brass. 80.0%
is Cu, 20.0% is Zn. Change g Zn → mol Zn and solve for m.
Solve.
8750 kg brass – 7000 kg Cu = 1750 kg Zn
(b)
13.55
3
Plan. M = mol Zn/L brass. Use mol Zn from part (a). Change 1 m → L brass and
calculate M.
Solve.
Analyze. Given: 4.6% CO 2 by volume (in air), 1 atm total pressure. Find: partial
pressure and molarity of CO 2 in air.
Plan. 4.6% CO 2 by volume means 4.6 mL of CO 2 could be isolated from 100 mL of air,
at the same temperature and pressure. According to Avogadro’s Law, equal volumes of
gases at the same temperature and pressure contain equal numbers of moles. By
inference, the volume ratio of CO 2 to air, 4.6/100 or 0.046, is also the mole ratio.
Solve.
M = mol CO 2 /L air = n/V.
PV = nRT, M = n/V = P/RT
354
13 Properties of Solutions
Solutions to Exercises
Colligative Properties
13.57
freezing point depression, ΔT f = K f (m); boiling point elevation, ΔT b = K b (m);
osmotic pressure, π = M RT; vapor pressure lowering, P A = χ A P A °
13.59
The vapor pressure over the sucrose solution is higher than the vapor pressure over the
glucose solution. Since sucrose has a greater molar mass, 10 g of sucrose contains fewer
particles than 10 g of glucose. The solution that contains fewer particles, the sucrose
solution, will have the higher vapor pressure.
13.61
(a)
Analyze/Plan. H 2 O vapor pressure will be determined by the mole fraction of
H 2 O in the solution. The vapor pressure of pure H 2 O at 338 K (65°C) = 187.5 torr.
Solve.
(b)
Analyze/Plan. For this problem, it will be convenient to express Raoult’s law in
terms of the lowering of the vapor pressure of the solvent, ΔP A .
ΔP A = P A ° – χ A P A ° = P A ° (1 – χ A ). 1 – χ A = χ B , the mole fraction of the solute
particles
ΔP A = χ B P A °; the vapor pressure of the solvent (A) is lowered according to the
mole fraction of solute (B) particles present.
Solve.
y = 1.0366 = 1.04 mol C 3 H 8 O 2
This result has 3 sig figs because (18.9 × 0.0521 = 0.983) has 3 sig figs.
13.63
Analyze/Plan. At 63.5°C,
Let G = the mass of H 2 O
and/or C 2 H 5 OH. Solve.
(a)
Multiplying top and bottom of the right side of the equation by 1/G gives:
355
13 Properties of Solutions
Solutions to Exercises
(b)
χ E th = 0.2812, P E th = 0.2812 (400 torr) = 112.48 = 112 torr
= 112.5 torr + 125.8 torr = 238.3 = 238 torr
(c)
13.65
(a)
Because NaCl is a soluble ionic compound and a strong electrolyte, there are 2
mol dissolved particles for every 1 mol of NaCl solute. C 6 H 1 2O 6 is a molecular
solute, so there is 1 mol of dissolved particles per mol solute. Boiling point
elevation is directly related to total moles of dissolved particles; 0.10 m NaCl has
more dissolved particles so its boiling point is higher than 0.10 m C 6 H 1 2O 6 .
(b)
Analyze/Plan. ΔT = K b m; K b for H 2 O is 0.51 °C/m (Table 13.4)
Solve.
Check. Because K b for H 2 O is so small, there is little real difference in the boiling
points of the two solutions.
(c)
13.67
In solutions of strong electrolytes like NaCl, electrostatic attractions between ions
lead to ion pairing. Ion pairing reduces the effective number of particles in
solution, decreasing the change in boiling point. The actual boiling point is then
lower than the calculated boiling point for a 0.1 M solution.
Analyze/Plan. Follow the logic in Sample Exercise 13.10. Solve.
The more nonvolatile solute particles, the higher the boiling point of the solution. Since
LiBr and Zn(NO 3 ) 2 are electrolytes, the particle concentrations in these solutions are
0.10 m and 0.15 m, respectively (although ion-ion attractive forces may decrease the
effective concentrations some-what). Thus, the order of increasing boiling points is:
0.050 m LiBr < 0.120 m glucose < 0.050 m Zn(NO 3 ) 2
13.69
Analyze/Plan. ΔT = K(m); first, calculate the molality of each solution. Solve.
(a)
0.22 m
(b)
(c)
356
13 Properties of Solutions
Solutions to Exercises
Solve. Then, f.p. = T f – K f (m); b.p. = T b + K b (m); T in °C
m
13.71
Tf
-Kf(m)
f.p.
Tb
+Kb(m)
b.p.
(a)
0.22
−114.6
−1.99(0.22) = −0.44
−115.0
78.4
1.22(0.22) = 0.27
78.7
(b)
0.820
−63.5
−4.68(0.820) = −3.84
−67.3
61.2
3.63(0.820) = 2.98
64.2
(c)
0.325
0.0
−1.86(0.325) = −0.604
−0.6
100.0
0.51(0.325) = 0.17
100.2
Analyze. Given freezing point of solution and mass of solvent, calculate mass of solute.
Plan. Reverse the logic in Sample Exercise 13.9. Use ΔT f = K f (m) to calculate the
required molality, and then apply the definition of molality to calculate moles and
grams of C 2 H 6 O 2 .
Solve. f.p. of solution = 5.00°C; f.p. of solvent (H 2 O ) = 0.0°C
Δ T f = 5.00
m=
= K f (m); 5.00
= 1.86
/m(m)
= 2.688 = 2.69 m C 2 H 6 O 2
m=
= C 2 H 6 O 2 = m × kg H 2 O
2.688 m C 2 H 6 O 2 × 1.00 kg H 2 O = 2.688 = 2.69 mol C 2 H 6 O 2
2.688 m C 2 H 6 O 2 ×
= 166.84 = 167 g C 2 H 6 O 2
13.73
Analyze/Plan. π = M RT; T = 25°C + 273 = 298 K; M = mol C 9 H 8 O 4 /L soln
Solve.
13.75
Analyze/Plan. Follow the logic in Sample Exercise 13.12 to calculate the molar mass of
adrenaline based on the boiling point data. Use the structure to obtain the molecular
formula and molar mass. Compare the two values.
Solve.
The molecular formula is C 9 H 1 3NO 3 , MM = 183 g/mol. The values agree to 2 sig figs,
357
13 Properties of Solutions
Solutions to Exercises
the precision of the experimental value.
13.77
Anayze/Plan. Follow the logic in Sample Exercise 13.13.
mol = M × L = 5.128 × 10
13.79
(a)
–5
× 0.210 L = 1.077 × 10
–5
Solve.
= 1.08 × 10
–5
mol lysozyme
Analyze/Plan. i = Π (measured) / Π (calculated for a nonelectrolyte);
Π (calculated) = M RT.
Solve.
i = 0.674 atm/0.2445 atm = 2.756 = 2.8
(b)
The van’t Hoff factor is the effective number of particles per mole of solute. The
closer the measured i value is to a theoretical integer value, the more ideal the
solution. Ion-pairing and other interparticle attractive forces reduce the effective
number of particles in solution and reduce the measured value of i. The more
concentrated the solution, the greater the ion-pairing and the smaller the
measured value of i.
Colloids
13.81
13.83
13.85
(a)
In the gaseous state, the particles are far apart and intermolecular attractive
forces are small. When two gases combine, all terms in Equation [13.1] are
essentially zero and the mixture is always homogeneous.
(b)
The outline of a light beam passing through a colloid is visible, whereas light
passing through a true solution is invisible unless collected on a screen. This is
the Tyndall effect. To determine whether Faraday’s (or anyone’s) apparently
homogeneous dispersion is a true solution or a colloid, shine a beam of light on it
and see if the light is scattered.
(a)
hydrophobic
(d)
hydrophobic (but stabilized by adsorbed charges)
(b)
hydrophilic
(c)
hydrophobic
Proteins form hydrophilic colloids because they carry charges on their surface (Figure
13.28). When electrolytes are added to a suspension of proteins, the dissolved ions form
ion pairs with the protein surface charges, effectively neutralizing them. The protein’s
capacity for ion-dipole interactions with water is diminished and the colloid separates
into a protein layer and a water layer.
358
13 Properties of Solutions
Solutions to Exercises
Additional Exercises
13.87
The outer periphery of the BHT molecule is mostly hydrocarbon-like groups, such as
–CH 3 . The one –OH group is rather buried inside, and probably does little to enhance
solubility in water. Thus, BHT is more likely to be soluble in the nonpolar hydrocarbon
hexane, C 6 H 1 4, than in polar water.
13.90
(a)
(b)
13.93
P R n = χ R nP t otal; P R n = 3.5 × 10
–6
(32 atm) = 1.12 × 10
–4
= 1.1 × 10
–4
atm
(a)
1 L soln = 826 g soln; g CH 3 CN = 826 – 156.3 = 669.7 = 670 g CH 3 CN
(b)
(c)
13.96
Analyze. Given vapor pressure of both pure water and the aqueous solution and moles
H2O find moles of solute in the solution.
Plan. Use vapor pressure lowering,
to calculate χA, mole fraction solvent,
and then use the definition of mole fraction to calculate moles solute particles. Because
NaCl is a strong electrolyte, there is one mole NaCl for every two moles solute particles.
Solve.
0.80818 (0.115 + mol ions) = 0.115; 0.80818 (mol ions) = 0.115 – 0.092940
mol ions = 0.02206/0.80818 = 0.02730 = 0.0273
mol NaCl = mol ions/2 = 0.02730/2 = 0.01365 = 0.0137 mol NaCl
13.99
(a)
0.100 m K 2 SO 4 is 0.300 m in particles. H 2 O is the solvent.
ΔT f = K f m = –1.86(0.300) = –0.558; T f = 0.0 – 0.558 = –0.558°C = –0.6°C
(b)
ΔT f (nonelectrolyte) = –1.86(0.100) = –0.186; T f = 0.0 – 0.186 = –0.186°C = –0.2°C
T f (measured) = i × T f (nonelectrolyte)
From Table 13.5, i for 0.100 m K 2 SO 4 = 2.32
T f (measured) = 2.32(–0.186°C) = –0.432°C = –0.4°C
359
13 Properties of Solutions
13.102
Solutions to Exercises
(a)
Assume 1000 g of solution.
(b)
If there are 4.12 mol of KSCN, there are 8.24 moles of ions. There are then 33.3
mol H 2 O/8.24 mol ions ≈ 4 mol H 2 O for each mol of ions, or 4 water molecules
for each ion. This is too few water molecules to completely hydrate the anions
and cations in the solution.
For a solution that is this concentrated, one would expect significant ion-pairing,
because the ions are not completely surrounded and separated by H 2 O
molecules. Because of ion-pairing, the effective number of particles will be less
than that indicated by m and M, so the observed colligative properties will be
significantly different from those predicted by formulas for ideal solutions. The
observed freezing point will be higher, the boiling point lower, and the osmotic
pressure lower than predicted.
Integrative Exercises
13.104
Since these are very dilute solutions, assume that the density of the solution ≈ the
density of H 2 O ≈ 1.0 g/mL at 25°C. Then, 100 g solution = 100 g H 2 O = 0.100 kg H 2 O.
(a)
(b)
Molality and molarity are numerically similar when kilograms solvent and liters
solution are nearly equal. This is true when solutions are dilute, so that the
density of the solution is essentially the density of the solvent, and when the
density of the solvent is nearly 1 g/mL. That is, for dilute aqueous solutions such
as the ones in this problem, M ≈ m.
(c)
Water is a polar solvent; the solubility of solutes increases as their polarity
increases. All the fluorocarbons listed have tetrahedral molecular structures. CF 4 ,
a symmetrical tetrahedron, is nonpolar and has the lowest solubility. As more
360
13 Properties of Solutions
Solutions to Exercises
different atoms are bound to the central carbon, the electron density distribution
in the molecule becomes less symmetrical and the molecular polarity increases.
The most polar fluorocarbon, CHClF 2 , has the greatest solubility in H 2 O. It may
act as a weak hydrogen bond acceptor for water.
(d)
S g = k P g . Assume M = m for CHCIF 2 . P g = 1 atm
This value is greater than the Henry’s law constant for N 2 (g), because N 2 (g) is
nonpolar and of lower molecular mass than CHCIF 2 . In fact, the Henry’s law
constant for nonpolar CF 4 ,
is similar to the value for N 2 ,
13.108
(a)
(b)
If the lattice energy (U) of the ionic solid (ion-ion forces) is too large relative to
the solvation energy of the gaseous ions (ion-dipole forces), ΔH s oln will be too
large and positive (endothermic) for solution to occur. This is the case for solutes
like NaBr. Lattice energy is inversely related to the distance between ions, so salts
+
with large cations like (CH 3 ) 4 N have smaller lattice energies than salts with
+
simple cations like Na . The smaller lattice energy of (CH 4 ) 3 NBr causes it to be
more soluble in nonaqueous polar solvents. Also, the –CH 3 groups in the large
cation are capable of dispersion interactions with the –CH 3 (or other nonpolar
groups) of the solvent molecules. This produces a more negative solvation energy
for the salts with large cations.
Overall, for salts with larger cations, U is smaller (less positive), the solvation
energy of the gaseous ions is more negative, and ΔH s oln is less endothermic.
These salts are more soluble in polar nonaqueous solvents.
13.111
The process is spontaneous with no significant change in enthalpy, so we suspect that
there is an increase in entropy. In general, dilute solutions of the same solute have
greater entropy than concentrated ones, because the solute particles are more free to
move about the solution. There are a greater number of equivalent environments
available to the solute. In the limiting case that the more dilute solution is pure solvent,
there is a definite increase in entropy as the concentrated solution is diluted. In Figure
13.23, there may be some entropy decrease as the dilute solution loses solvent, but this
is more than offset by the entropy increase that accompanies dilution. There is a net
361
13 Properties of Solutions
Solutions to Exercises
increase in entropy of the system, going from the left to center panels in Figure 13.23.
362