Download Chemistry 121 Mines PS13-1

Chemistry 121
Mines
Answer Key, Problem Set 13
1. WO1; 2. MP; 3. WO2; 4. MP; 5. WO3; 6. MP; 7. MP(13.34); 8. WO4; 9. WO5; 10. MP; 11. WO6; 12. MP;
13. WO7; 14. WO8; 15. MP
---------------------------------------Units of Concentration (and other physical quantities, like density, of solutions)
1. WO1.
A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting
volume was 104 mL. Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a
density of 1.00 g/mL.
The following is (with slight modification) the solution from the published solutions guide (Thomas Hummel,
Steven Zumdahl, and Susan Arena Zumdahl):
(= 0.981 M)
(= 1.02 m)
2. MP No answer to Mastering Problem in this key. However, see Solutions worksheet initial problems
(done in class) as well as Example 13.5 in Tro (which is not exactly the same kind of problem, but
is similar in the way you think about/approach it).
Molarity, Revisited (applied to mixtures of solutes or mixing of solutions)
3. WO2. Some K2Cr2O7, with a mass of 2.335 g, is dissolved in enough water to make 500. mL of
solution.
(a) What is the molarity of the potassium dichromate?
1 mole of K2Cr2O7 has a mass of: 2(39.1) + 2(52.0) + 7(16.0) = 294.2 g. So 2.335 g is
2.335/294.2 ths of a mole or 0.007937 mol:
PS13-1
Answer Key, Problem Set 13
2.335 g K 2 Cr2O7
 0.007937 mol K 2Cr2 O7
294.2 g/mol K 2 Cr2O7
Molarity =
0.007937 mol K 2Cr2O7
moles solute

 0.01587 = 0.0159 M K2Cr2O7
liter of solution
0.500 L solution
(b) What are the concentrations of the K+ and Cr2O72- ions?
[K+] = 2 x [K2Cr2O7] = 0.0317 M
(because there are two moles of K+ ions per mole of K2Cr2O7)
[Cr2O72-] = [K2Cr2O7] = 0.0159 M
(because there is one mole of Cr2O72- per mole of K2Cr2O7).
(c) If 300. mL of 0.054 M KNO3 were ADDED TO the 500. mL solution described in this
problem, what would the final concentration of K+ in the new solution be? (Note: assume
that the final volume of the solution is 800. mL).
Find TOTAL moles of K+ in the new solution, and divide by the TOTAL volume of solution in L.
Moles of K+ from the 500. mL solution: 0.007937 mol K2Cr2O7 x 2 = 0.01587 mol K+
Moles of K+ from the 300. mL solution: (0.054 moles KNO3/L)(0.300 L)(1 mole K+/mol KNO3) =
0.0162 mol K+
Total (new) volume = 800. mL = 0.800 L
So, [K + ]=
0.01587 + 0.0162 mol K + 0.03207 mol K +

 0.04009  0.040 M K +
0.800 L
0.800 L
Meaning of Solubility, Dynamic Equilibrium, Saturated vs. Unsaturated, Concentrated vs. Dilute
4. MP(video only). This is not actually a “problem”. It was put here for you to watch the video (first half,
mainly) in order to help you visualize what is going on in parts of the problem below (WO3). Also see
Section 13.4 in Tro (especially Figure 13.9 on p. 581).
5. WO3.
Consider the following snapshots of several different beakers, all at the same temperature, that have been
sitting around for a while. Assume that if you waited awhile and looked again, the contents of the beakers would look
essentially the same as they do now. Each contains water, which is not explicitly shown; each also contains a salt
(see the key at the right), but not necessarily the SAME salt as in the others (i.e., one might have AgCl, one might have
NaBr, etc.)
(i)
(ii)
(iii)
(iv)
dissolved
anion
dissolved
cation
solid salt
(undissolved)
[NOTE: For all of the following, you should NOT consider the undissolved solid to be a part of the solution; it
is a separate “phase” (a solid). Only ions that are dissolved are a part of a solution.]
Answers: 1. (iii) and (iv); 2. (ii); 3. (iv); 4. (iv)
PS13-2
Answer Key, Problem Set 13
Explanations:
1. Which solution(s) are saturated? (iii) and (iv)
These are the only ones that (you know) are saturated because there is some solid left
undissolved (the solution is “full” or “saturated” with that solid).
2. Which solution is the most concentrated? (ii)
Concentration refers to HOW MANY PARTICLES are dissolved in a given volume of solution.
The more particles there are in a given volume the higher the concentration. Since all volumes
here are essentially the same, the one with the most particles DISSOLVED is the one with the
highest concentration.
3. Which solution is the most dilute? (iv)
Is the most dilute since it has the lowest concentration (fewest number of particles dissolved in a
given volume).
4. Which beaker contains the salt that is the least soluble? (iv).
The least soluble salt is the one that has the lowest concentration WHEN SATURATED.
Beaker (iv) is the most dilute AND it is saturated, so it must contain the salt that is the least
soluble.
Factors that affect Solubility
6. MP. No answer to Mastering Problem in this key. Use the info in the problem (interactive diagram), as
well as in the slides in Ppt27 or in the text.
7. MP(13.34). No answer to Mastering Problem in this key. Use the idea that “like dissolves like”, where
“like” means “similar intermolecular forces”. See explanation to WO4 below.
8. WO4. Which substance is most soluble in water? Give your reasoning.
(i) C6H6
(ii) C2H5OH
(iii) CH3Cl
(iv) CO2
Comments/Explanation: To best predict solubility of a molecular substance in a molecular
solvent, identify the intermolecular forces of attraction of the solute and solvent and then compare.
The likelihood is that the solutes that have the greatest similarity of intermolecular forces
with the solvent will be the most soluble in it. Why? At this point in your chemical education,
the best explanation is that energy plays a partial role in dissolution. If the energy cost of
separating the solute molecules from one another and the solvent molecules from one another are
not at least partially compensated for by an energy release that comes from the solute molecules
interacting with the solvent molecules, then the dissolution process tends to be less favorable. If
the solute is polar or can hydrogen bond, the energy “compensation” will not likely come close to
the energy “cost” unless the solvent is polar or can hydrogen bond as well (and the same logic
applies if the solvent is polar). This is the basis for the simple phrase “Like dissolves like”. Polar
solutes tend to dissolve in polar solvents, nonpolar solutes tend to dissolve in nonpolar solvents,
and polar and nonpolar things do not dissolve in one another. And solutes that can participate in
hydrogen bonding will be most soluble in solvents that can as well.
So in this case, the solvent is water. Water is polar and its molecules can hydrogen bond to one
another. C6H6 and CO2 are nonpolar molecules and so they will not be very soluble in water.
CH3Cl is polar but cannot participate in hydrogen bonding. CH3OH is polar and can participate in
hydrogen bonding, and so is predicted to be the most soluble in water (and it is).
PS13-3
Answer Key, Problem Set 13
9. WO5.
Which solvent, water or carbon tetrachloride (CCl4) would you choose to dissolve each of the following?
(a) KrF2
(b) SF2 (c) SO2
(d) CO2
(e) MgF2
(f) CH2O
(C is in the middle)
(g) CH2=CH2
The following is essentially the solution from the published solutions guide (Thomas Hummel, Steven
Zumdahl, and Susan Arena Zumdahl), although I’ve tweaked it a bit by adding parenthetical comments.
Polar (only one polar bond!),
so more soluble in H2O
Nonpolar (like all hydrocarbons),
so more soluble in CCl4
PS13-4
Answer Key, Problem Set 13
10. MP. No answer to Mastering Problem in this key. Watch the video and think about how the solubility
of solids and gases each depend on temperature.
11. WO6. Under what conditions of temperature and pressure is air (or most any other gas) most
soluble in water? Give some reasoning.
Answer: low temperature and high (partial) pressure
(Note that these are the same two “conditions”
under which a gas behaves most NON-ideally!)
Reasoning/Comments: The solubility of gases depends on both temperature and pressure (unlike
solids and liquids, which are only significantly dependent on temperature). It is most clear to see the
effects of each if you start out with a system at dynamic equilibrium at a given T and P and then
imagine how the “opposing” rates are affected by either a T change or a P change.
Temperature effect: The solubility of gases decreases with increasing temperature, and so it
increases with decreasing temperature. This makes sense if you think of the molecules of a gas
already being separated from each other, but when in they are dissolved, they are surrounded by
solvent molecules. That is, it takes energy for a dissolved gas molecule to escape the liquid, but it
doesn’t take energy for a gas molecule to be “captured” by the solvent. So an increase in
temperature should increase the rate of escape more than the rate of “capture”, and so there will be
less gas dissolved at dynamic equilibrium (which means a decreased solubility).
Pressure effect. Consider gas dissolved in equilibrium with undissolved gas above it. The rates in
and out of the solvent are the same. If you increase the partial pressure of the gas above the liquid,
that means there will be MORE particles in a given volume, and so the rate of collisions into the
solvent will become greater. More particles will end up in the (given) amount of solvent, and so the
concentration will go up as dynamic equilibrium is reestablished (which means an increased
solubility).
Colligative Properties (Including strong electrolyte considerations)
12. MP. No answer to Mastering Problem in this key. This is about colligative properties (specifically
vapor pressure lowering), a review of which is in the video. One can also look at my Ppt28
lecture/notes and the text.
13. WO7. Which aqueous solution (these are NOT pure substances!!!) has a) The smallest
freezing point? b) The smallest boiling point? Explain.
(i) 0.2 M Ca(NO3)2
(ii) 0.2 M CH3OH
(iii) 0.2 M MgSO4
(iv) 0.2 M K3PO3
Answers: (a) Smallest freezing point: 0.2 M K3PO3 (highest [particles] = 0.8 M)
(b) Smallest boiling point: 0.2 M CH3OH (lowest [particles] = 0.2 M)
Explanation: The presence of particles dissolved in a solvent decreases the “escaping tendency” of
solvent molecules, and thus inhibits both boiling and freezing. Thus the freezing point will decrease and
the boiling point will increase with an increase in the concentration of PARTICLES dissolved (where
“particle” means either an ion or a molecule).
So, in a solution with the highest concentration of particles, the following will be true: the freezing
point will be decreased the most (freezing point will be lowest) and the boiling point will be raised the
most (boiling point will be greatest).
It thus follows that the solution with the smallest concentration of particles will be the one that has
the smallest decrease in freezing point (and thus the highest freezing point) as well as the smallest
increase in boiling point (and thus the smallest boiling point).
In order to determine the concentration of particles, one must first identify whether the solute is an
electrolyte or a nonelectrolyte. If it is a nonelectrolyte, then the concentration of particles simply equals
PS13-5
Answer Key, Problem Set 13
the concentration of solute dissolved. But if the solute is an electrolyte, one must then determine how
many ions there are per formula unit AND whether or not the electrolyte is strong or weak (not a part of
THIS particular problem, but it could be [if a weak acid is given, for example]). Then one can determine
the total concentration of particles from the concentration of solute and the number of ions per formula
unit.
Ca(NO3)2 is ionic, so it is a strong electrolyte and it will separate into its ions in aqueous solution. In
this case, that means 3 ions per formula unit (one Ca2+ and two NO3- ions). Thus the concentration
of particles in a 0.2 M Ca(NO3)2 solution is actually 0.6 M particles.
CH3OH is molecular (and not an electrolye) so a 0.2 M solution of CH3OH will have a concentration
of particles of just 0.2 M. (lowest of all four). ***The “OH” here does not indicate that this is ionic! If
it were indicating OH-, then the cation would have to be CH3+ and we have never seen this as a
cation before [because it isn’t]). -OH attached to a carbon is called an “alcohol” group; CH3OH is
methanol, remember?!***
MgSO4 is ionic, but only has two ions per formula unit (Mg2+ and SO42-) so the concentration of
particles in a 0.2 M MgSO4 solution is 0.4 M particles.
K3PO3 is ionic and has four ions per formula unit (three K+ ions and one PO33- ion), so a 0.2 M
solution of K3PO3 will have a concentration of 0.8 M particles. (highest of all four).
14. WO8
(a) What is the mass percent concentration of the following solution: 0.135 g of KBr is dissolved in 5.00 mL of water.
(b) What is the molality of the same solution?
(c) What would be the freezing point of the solution, assuming a kf of 1.86 C/m for water and a normal freezing point of 0.00 C?
Answers: (a) 2.63 %; (b) 0.227 m KBr; (c) -0.84 C
Reasoning:
(a) mass percent =
mass of solute
x 100 , so you need to find two things:
(total) mass of solution
mass of solute and mass of the solution. Mass of the solute here is given, but the volume of
solvent (water, here) is given. So we need to use the density of water (assume it to be 1.00
g/mL) to calculate the mass of water from its volume: 5.00 mL x 1.00 g/mL = 5.00 g.
Mass of solute: Given as 0.135 g
Mass of solution = mass of solute plus mass of solvent = 0.135 g + 5.00 g
Mass percent 
0.135 g
x 100  2.629..  2.63%
0.135 g  5.00 g
(b)
Molality =
moles of solute
, so we need to find two things:
kg of solvent
the number of moles of KBr and the mass (in kg) of water.
Moles KBr: Since the molar mass of KBr is 39.1 + 79.9 = 119.0 g/mol, 0.135 g is a small fraction
of a mole indeed! Specifically it is:
PS13-6
Answer Key, Problem Set 13
0.135 g
 0.001134.. mol KBr
119.0 g/mol

1 kg
Mass of water: 5.00 g   5.00 g x
1
000 g

m
0.001134.. mol KBr
 0.2268.. 
0.00500 kg H 2 O
7
2
2
.
0
So the molality 

  0.00500 kg

(c) Tf = kf x mparticles = 1.86 C/m x (0.227 m) x 2 moles ions / mol KBr = 0.844 C
Thus, the freezing point is depressed by 0.844 C. If the fp of (pure) water is 0.00 C, then the fp
of this particular aqueous solution is 0.844 C below 0.00 C:
fp(sol’n) = 0.00 – 0.844 C = -0.844 C = -0.84 C
15. MP. No answer to Mastering Problem in this key. This is about colligative properties (specifically
osmosis, which isn’t itself a colligative property, but osmotic pressure is…), a review of which is
in the video. One can also look at my Ppt28 lecture/notes and the text.
PS13-7