Chemistry 122
Mines, Spring 2017
Answer Key, Problem Set 5 (With explanations)--Complete
1. WO1; 2. WO2; 3. MP; 4. MP (16.38); 5. MP (16.40); 6. MP (16.42); 7. WO3; 8. WO4; 9. MP; 10. WO5; 11. WO6;
12. MP; 13. WO7; 14. WO8; 15. WO9; 16. WO10; 17. MP; 18. MP; 19. WO11; 20. WO12; 21. WO13; 22. WO14;
23. WO15; 24. WO16; 25. MP; 26. WO17; 27. MP; 28. WO18; 29. MP; 30. WO19; 31. MP; 32. MP
-------------------------------Brønsted-Lowry Definition of Acids and Bases and Acid-Base Conjugates
1. WO1 16.36.
For each reaction [equation], identify the Brønsted-Lowry (BL) acid, the Brønsted-Lowry base, the
conjugate acid, and the conjugate base.
(a)
H3O+(aq) + I-(aq)
HI(aq) + H2O(l)
CH3NH3+(aq) + OH-(aq)
(b) CH3NH2(aq) + H2O(l)
(c)
CO32-(aq) + H2O(l)
HCO3-(aq) + OH-(aq)
H3O+(aq) + Br-(aq)
(d) HBr(aq) + H2O(l)
Answers:
BL Acid
HI
H2O
H2O
HBr
(a)
(b)
(c)
(d)
BL Base
H2O
CH3NH2
CO32H2O
Conjugate Base
IOHOHBr-
Conjugate Acid
H3O+
CH3NH3+
HCO3H3O+
Reasoning:
(1) A (Brønsted-Lowry) acid is a proton donor: a base is a proton acceptor. So you must look
at what a given species “ends up as” (i.e., what it “does” during the reaction) in order to
properly assess whether it is an acid or a base.
(2) Conjugates are pairs of species that differ in structure (and thus formula) by exactly one H+.
(Look closely at the relationship between the species in the two rectangles and the relationship between the species in
the two ovals.)
NOTE: The conjugate acid-base pair species are NOT the acid and base that react with one another! They
are on opposite sides of the equation (not the same side). A conjugate is formed when an acid or a
base “acts like” an acid or base (i.e., it gives away an H+ or accepts one).
2. WO2 16.37 (add (e) NH2 - ; (f) H2CO ; (g) C2H3O2- ) Write the formula for the conjugate base of each acid.
(a)
(b)
HCl
H2SO3
CB is
CB is
(c)
HCHO2
CB is
(d)
HF
CB is
ClHSO3CHO2F-
(e)
(f)
(g)
NH2H2CO
C2H3O2-
CB is NH2CB is HCOCB is C2H2O22-
Reasoning: To determine the formula of the conjugate base of an acid, you must remove exactly
one “H+”, which means to remove one “H” from the formula, and also subtract +1 from
whatever the charge on the acid formula was—that is, make the charge “one unit more
negative”.
3. MP
No answer in this key.
4. MP (16.38)
Write the formula for the conjugate acid of each base.
Mastering problem, so answers not shown. However, to determine the formula of the conjugate acid
of an base, you must do exactly the opposite of what was done in the prior problem. That is, you
must add exactly one “H+”, which means to add one “H” from the formula, and also add +1 to
whatever the charge on the acid formula was—that is, make the charge “one unit more positive”.
5. MP (16.40)
Both HCO3- and HS- are amphoteric. Write an equation to show how each substance [species!!] can act as
an acid, and another equation to show how it can act as a base.
**NOTE: The question 16.40 in Mastering is actually a bit different than the question asked in the
physical text 16.40. The answer below is therefore not the answer to the Mastering question,
PS5-1
Answer Key, Problem Set 5
although the concepts are obviously analogous. So I’ve left my answer here in case it might
help, but please make sure to realize these are not the answers to the Mastering problem, as it
is worded. Always read every question carefully! Don’t assume anything. 
NOTE: There are many possible answers to this problem because they did not specifically ask you to write out the “acid
ionization” and “base ionization” reaction equations for these species (even though those equations would represent one
set of correct answers to this question). As such, I will provide two correct answers for each, one of which is the acid (or
base) ionization equation and one of which is not. Please know that a species can “act as an acid” by giving its proton
away to any species—not just water. Acid ionization, however, is the (very specific) acid-base reaction in which an acid
gives its proton to H2O.
HCO3- acting as an acid:
 HCO3- needs to be a reactant
and CO32- must be a product.
HCO3- acting as a base:
 HCO3- needs to be a reactant
and H2CO3 must be a product.
HS- acting as an acid:
 HS needs to be a reactant
and S2- must be a product.
HCO3-(aq) + OH-(aq)
OR
HCO3-(aq)
 HS needs to be a reactant
and H2S must be a product.
CO32-(aq) + H3O+(aq)
+ H2O(l)
HCO3-(aq) + HSO4-(aq)
H2CO3(aq) + SO42-(aq)
OR
HCO3-(aq)
H2CO3(aq) + OH-(aq)
+ H2O(l)
HS-(aq) + F-(aq)
-
HS- acting as an base:
CO32-(aq) + H2O(l)
S2-(aq) + HF(aq)
OR
HS-(aq)
S2-(aq) + H3O+(aq)
+ H2O(l)
HS-(aq) + HCN(aq)
-
H2S(aq) + CN-(aq)
OR
HS-(aq)
+ H2O(l)
H2S(aq) + OH-(aq)
Acid Ionization Definition, Ka, and Relating Relative Acid Strength to Ka Values and to Nanoscopic
Pictures of Solutions of Acids
6. MP (16.42)
Classify each acid as strong or weak. If the acid is weak, write an expression for the acid ionization
constant (Ka)
No answer in key. However, learn the 6 common strong acids so that a) You won’t need to worry
about their Ka values (because they’re basically 100% ionized), and b) You will know that any acid
other than one of these is a weak acid. Separately, know that “acid ionization” means “an acid
reacting with water” (using the Bronsted-Lowry definition of acid) so that you can generate the acid
ionization reaction equation and Ka expression (Law of Mass Action) for any acid of interest.
7. WO3.
The following pictures represent aqueous solutions of three acids HA
(A = X, Y, or Z); water molecules have been omitted for clarity: NOTE:
The black circles represent oxygen atoms. The open circles represent
hydrogen atoms (charges are not indicated here).
Answer the following, with reasoning:
Answer: X-, Y-, and ZReasoning: A CB is a species with exactly one H+ removed from the (C)A
(a) What is the conjugate base of each acid?
(b) Arrange the three acids in the order of increasing acid strength.
Answer: HX < HZ < HY
Reasoning:
Ignoring the concentration dependence on % ionization of a weak acid (which is a very minor issue in the
scenario in this problem), the stronger the acid, the greater the fraction of dissolved formula units that will
exist as separated ions. HX has the smallest fraction of formula units existing as separated ions (1
out of 10), HZ has the next smallest (2 out of 10), and HY has the greatest (6 out of 6).
Answer: only HY
Reasoning: A strong acid has nearly 100% of its dissolved formula units existing as separated ions.
HY has 100% of its formula units existing as separated ions.
(c) Which acid, if any, is a strong acid?
PS5-2
Answer Key, Problem Set 5
(d) Which acid has the smallest value of Ka?
Answer: HX
Reasoning:
Ignoring the concentration dependence on % ionization of a weak acid (which is a very minor issue in the
scenario in this problem), the smaller the Ka value, the smaller the fraction of dissolved formula units that
will exist as separated ions. HX has the smallest fraction of formula units existing as separated ions
(see (b) above), so it has the smallest Ka.
8. WO4 16.44.
Rank the solutions in order of decreasing [H3O+]: 0.10 M HCl; 0.10 M HF; 0.10 M HClO; 0.10 M HC6H5O.
Answer: 0.10 M HCl >> 0.10 M HF > 0.10 M HClO > 0.10 M HC6H5O
Reasoning:
For solutions of acids with the same (prepared) concentration (same [HA]0), the percentage of
formula units that will ionize will be greater for the stronger the acid (i.e., the greater the Ka). This
is because a bigger Ka means the acid ionization reaction lies “farther to the right” (towards
products) at equilibrium, and H3O+ is always a product of acid ionization (by definition).
In this problem, HCl is the only strong acid, and so it is the strongest (it is essentially 100%
ionized) and thus has the greatest [H3O+]eq. From Table 16.5, the Ka’s for HF, HClO, and
HC6H5O at 25°C are 3.5 x 10-4, 2.9 x 10-8, and 1.3 x 10-10 respectively. Hence the order given.
Note: It is unfortunate that the correct order for this problem happens to be the exact order in which the acids appear in
the original question. Somebody wasn’t looking very carefully…
9. MP. No answer in key. This a visual (nanoscopic pictures) question about solutions of acids.
10. WO5.
Use Table 16.5 (and 16.3, which you should memorize) to order the following from the strongest to the weakest acid.
HClO2, H2O, HC7H5O2 (benzoic acid), HClO4
Answer: HClO4 (strong) > HClO2 (weak) > HC7H5O2 (weak) > H2O (“reference”)
Reasoning: HClO2 and HC7H5O2 are listed in the table with HClO2 having a larger Ka value than
HC7H5O2 (1.1 x 10-2 vs 6.5 x 10-5). So clearly HClO2 is the stronger acid. (“stronger” means
“better at giving away its H+; a larger K means “more product favored” [as long as stoichiometry is
identical between the two equations being compared], so in the case of acid ionization, a larger K means a
“greater tendency to give away its H+).
Both acids are considered “weak acids” because they are not 100% ionized [K values are not >>
1], but are better acids than water [Ka values are larger than Kw, or 1.0 x 10-14 at 25°C]. Thus,
HClO4, one of the six common strong acids, must be stronger than both of these, and H2O, which
is the reference to whom everyone is compared (and so it defines the border between weak and
negligible), must be weaker than both of these. (Note: All weak acids are stronger than water,
by definition, and all negligible acids are weaker than water, by definition.)
Base Ionization Definition, Kb
11. WO6 16.88
Write equations showing how each weak base ionizes water to form OH-. Also write the corresponding
expression for Kb.
NOTE: This question is basically asking you to write out the equation for “base ionization” of a given
base, along with the expression for Kb (i.e., the Law of Mass Action)
HCO3-(aq) + OH-(aq)
(a) CO32-:
CO32-(aq) + H2O(l)
(b) C6H5NH2
C6H5NH2 (aq) + H2O(l)
C6H5NH3+(aq) + OH-(aq)
(c) C2H5NH2:
C2H5NH2(aq) + H2O(l)
C2H5NH3+(aq) + OH-(aq)
PS5-3
Answer Key, Problem Set 5
Strategy: Remember that “base ionization” is “a base reacting with water” (using the Bronsted-Lowry
definition of a base). Namely, just write the equation in which the base accepts (takes) a proton (H+)
from a water molecule. The products will therefore be the conjugate acid of the base, and OH- (the
conjugate base of water, which acts as the acid in this reaction).
Relating Relative Base Strength to Strengths of Conjugate Acids (Applying “The stronger the acid, the
weaker its conjugate base”, etc.)
12. MP. No answer in key. The first few parts of this Item are related to conjugate acid-base pairs, but
the last part involves the concept that “The stronger the acid, the weaker its conjugate base”, so
I put the problem here.
13. WO7 16.46.
Pick the stronger base from each pair.
(a) ClO4- or ClO2-
ClO2- HClO2 is a weak acid whereas HClO4 is a strong acid. The conjugate base of the weaker
acid will be the stronger base. (In this case, since HClO4 is a strong acid, its conjugate,
ClO4- is a negligible base, which is obviously weaker than any weak base (like ClO2-).
(b) Cl- or H2O
H2O
HCl is a strong acid, so its conjugate (Cl-) is a negligible base. A negligible base, by
definition, is a poorer base than H2O. (Water is the reference to which all acids and bases
are compared. If you are worse at accepting protons than water is, you are, by definition,
a “negligible” base (in water).)
(c) CN- or ClO-
CN-
From Table 16.5, it is clear that HCN (Ka = 4.9 x 10-10 at 25°C) is a weaker acid than HClO
(Ka = 2.9 x 10-8 at 25°C). Thus its conjugate (CN-) must be a stronger base than HClO’s
conjugate (ClO-).
NOTE: I used the conceptual reasoning here that “the stronger an acid is, the weaker will be its
conjugate base”. I describe the same kind of reasoning below (in WO3). The idea is that
being really good at “giving away an H+”, means that what’s left behind after the giving (i.e.,
the conjugate) must not be very good at “taking” an H+. This is important and solid reasoning.
However, one could also answer certain kinds of these questions (i.e., ones in which Ka’s are
known) by utilizing the relationship: KaKb = Kw. That is, one can either solve directly for each
 K 
Kb   w  and then compare the Kb’s (whomever’s Kb is higher is the stronger base), or just
 Ka 
use that equation to show that “the larger an acid’s Ka, the smaller will be its Kb”. This is
essentially “proof” of the qualitative statement made at the beginning of this paragraph.
14. WO8.
Use Table 16.5 (and appropriate reasoning!) to order the following from the strongest to weakest base.
ClO2-, H2O, C7H5O2-, ClO4-, OH-
Answer: OH- (strong) > C7H5O2- (weak) > ClO2- (weak) > H2O (“reference”) > ClO4- (negligible)
Reasoning: (very short): Applying the idea that “the stronger the acid, the weaker is its conjugate
base” will yield the order shown above for the four bases other than H2O based on the ordering of
the acids in the prior problem. Then H2O is placed between the weakest weak base (ClO2-) and
the negligible base (ClO4-). (One must be able to identify ClO4- as negligible [see below]).
Reasoning (more complete):
From the prior problem, we know that HClO2 and HC7H5O2 are both weak acids, but HClO2 is
stronger. That means HClO2 gives away its H+ better (i.e., with a greater tendency) than does
HC7H5O2, which means that ClO2- (the conjugate base of HClO2) must be poorer at taking an H+ than
C7H5O2- (the conjugate base of HC7H5O2).
PS5-4
Answer Key, Problem Set 5
Since both HClO2 and HC7H5O2 are weak acids, their conjugates must also be weak bases
(“weak gives weak—strong gives negligible”! Strong and negligible are the “extremes”; weak is in
the middle.) Thus both are stronger bases than water, the reference.
ClO4-, being the conjugate to a strong acid (HClO4), must be even poorer at accepting an H+ than
water, so it is a negligible base (in water). (At equilibrium, there is essentially no HClO4 in a
solution of HClO4, which means that ClO4- cannot take H+ away from water to essentially any
extent!).
OH- does not appear in any table. However, since OH- is the conjugate base to H2O, and H2O is,
by definition, a weaker acid than any “weak acid”, its conjugate base must be stronger than the
(weak) conjugates of any weak acid.
NOTE: I have labeled OH- a “strong” base above for two reasons: 1) Soluble hydroxides are considered
“strong bases” (e.g., NaOH, KOH), and 2) The equilibrium constant for the reaction OH- + H2O
H2O + OH- is effectively “1” (since the products [numerator] are the same as the reactants
[denominator]. Technically, it is “neither reactant favored nor product favored”, but since all weak
bases have Kb values less than 1, it is not unreasonable to lump OH- in with the strong bases.
Autoionization of Water, Kw, and the Definition of Acidic, Basic, and Neutral
15. WO9. 16.13
Answer:
Write an equation for the autoionization of water and an expression for the ion product constant for
water (Kw). what is the value of Kw at 25°C?
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq);
Kw  [H3O+][OH-]
You should not have to memorize anything other than “water autoionization is “water reacting with
water” (using the Bronsted-Lowry definitions of acid and base). In a sense, this reaction is like the
acid ionization reaction for water and the base ionization reaction for water.
This is where the relationship 1.0 x 10-14  [H3O+][OH-] (at 25°C only)
“comes from”!
16. WO10. What is the definition of a neutral solution (in an acid-base sense, not “electrically”)? Relate this to the
chemical equation for the autoionization of water (see prior problem). Is a sample of pure
water at any temperature neutral, even though the concentration of [H3O+] will vary with
temperature? Explain.
Answer: A neutral solution is one in which the concentration of hydronium ion equals the
concentration of hydroxide ion: [H3O+]  [OH-]
It’s relation to the equation for autoionization is this: If all you have is water (i.e., a sample of pure
water), then the only reaction that occurs in solution is autoionization. In autoionization, for every
mole of H3O+ that forms, one mole of OH- will form also, because the coefficients are both one.
Thus, in a sample of pure water, no matter what the value of Kw is, the concentration of
hydronium ion will equal the concentration of hydroxide, and thus the solution is neutral. When
the temperature varies, changing the value of Kw, the amount of water that undergoes this
reaction (net) will change, thus changing the concentration of hydronium ion at equilibrium, but
whether is a greater or lesser amount of reaction that occurs, the concentration of hydronium ion
will still equal that of hydroxide ion (i.e., if more autoionization occurs, [H3O+] will increase, but the
[OH-] will increase just as much!). Thus a sample of pure water will always necessarily be
neutral. **NOTE: When more autoionization takes place and [H3O+] increases, the pH will
decrease. But that does not mean the solution will be more acidic, because the solution will still
be neutral (with [H3O+] still equalling [OH-])!**
17. MP. No answer in key.
NOTES: 1) Please note that the authors of this question treat H+ as the equivalent of H3O+. As
mentioned in class, I don’t care for this approach because it makes it look like the acid
molecule is “dissociating” rather than “giving a proton to a water molecule”. Physically
PS5-5
Answer Key, Problem Set 5
speaking, the H+ ion does not exist to any appreciable extent in aqueous solution—it attaches
to a water molecule to create a hydronium ion.
2) I’m not a big fan of pOH, but I was not about to skip this problem just because it has pOH in
it. Just remember that pOH = -log[OH-]. I would not try to learn any special relationships
between pH and pOH, although if you do use such a relationship correctly, I will not take off
points, obviously.
Definition of pH, and Interconversions Between pH, [H3O+], and [OH-] (and pOH); Revisit “Acidic, Basic,
Neutral” at T’s other than 25°C.
18. MP. No answer in key. This problem is similar to the following problem.
19. WO11. 16.52. Complete the table. (All solutions are at 25°C.)
NOTE: Underlined values were given in the problem. (See below for strategy / reasoning.)
[H3O+] (M)
[OH-] (M)
pH
Acidic or Basic?
-14
(a)
3.5 x 10-3 M
1.0 x 10 -14
 2.63.. 
(b) 3.8 x 10 - 7
1.0 x 10
 2.857..
3.5 x 10 - 3
 2.9 x 10-12 M
3.8 x 10-7 M
-log(3.5 x 10-3)  2.455.. 
Acidic
2.46
([H3O+] > [OH-])
-log(2.63 x 10-8) = 7.579.. 
Basic
7.58
([OH-] > [H3O+])
2.6 x 10-8 M
(c)
1.8 x 10-9 M
1.0 x 10 -14
 5.55.. 
1.8 x 10 - 9
-log(1.8 x 10-9)  8.744.. 
Basic
8.74
([OH-] > [H3O+])
5.6 x 10-6 M
(d)
10-7.15 7.07..
7.1 x 10-8 M
1.0 x 10 -14
1.41..
7.07.. x 10 - 8
7.15
Basic
([OH-] > [H3O+])
1.4 x 10-7 M
Strategy / Reasoning:
Utilize the two relationships below in order to get between pH and [H3O+] and between [H3O+] and
[OH-]. Then apply the definition of “acidic” ([H3O+] > [OH-]) or “basic” ([OH-] > [H3O+]) to make that
conclusion.
pH = -log[H3O+]
and
[H3O+][OH-]  Kw ( 1.0 x 10-14 at 25°C)
NOTE: Although not independent of the first equation above, I will include the following (3rd)
[H3O  ]  10 pH
relationship for convenience:
(which helps you get pH from [H3O+]. as in part (d) of this problem)
If T = 25 C, what is the pH of a solution having [OH-] = 1.3 x 10-15 M? State the concentration of H3O+ in this
solution explicitly somewhere as well.
20. WO12.
Answers: [H3O+] = 7.7 M; pH = 0.89
Reasoning:
At 25 C, [H3O+][OH-] = 1.0 x 10-14
 [H3 O+ ] 
1.0 x 10-14 1.0 x 10-14

 7.69 M
[OH- ]
1.3 x 10-15
(NOTE: The above result is 7.69 M, a concentration of H+. This is NOT a pH value!!)
PS5-6
Answer Key, Problem Set 5
Apply the definition of pH:
pH =  log[H3O+ ]   log(7.69)  0.89
NOTE: If the concentration of H3O+ is greater than 1 M, the pH will be less than zero! This is no big deal (just quite
acidic!), but since many people have learned that the pH scale goes from 0 to 14, they find this troubling or
problematic.
21. WO13. 16.53.
Like all equilibrium constants, the value of Kw depends on temperature. At body temperature (37°C),
Kw = 2.4 x 10-14. What is the [H3O+] and pH of pure water at body temperature? Is the sample in this problem
acidic, basic, or neutral?
Answers: [H3O+]  1.5 x 10-7 M; pH  6.81; sample is neutral
Strategy / Work:
1) Recognize that in a sample of “pure water”, the only reaction occurring (for all practical
purposes) is autoionization:
H2O + H2O
H3O+ + OH-
Kw
To get concentrations at equilibrium, “allow some water to autoionize” (like an ICE problem in
which you let x  [H3O+] that forms as equilibrium is established).
2) Because of the stoichiometry, the system at equilibrium must have [H3O+]  [OH-] (If you were
to create an ICE table, you’d have initial concentrations of H3O+ and OH- equal to zero, and then
the change row would have + x for both species since the stoichiometry is 1 : 1.)
Thus, for pure water: x2  Kw
(where x  [H3O+]eq  [OH-]eq)
3) In this problem, since Kw  2.4 x 10-14, x 
2.4 x 10 -14  1.549... x 10 7  [H3O+]
4) pH  -log[H3O+]  -log(1.55 x 10-7)  6.809 6.81
5) Recognize that despite the pH not being 7.0, the solution MUST be neutral (since [H3O+] = [OH-],
right?)! Do not forget that the definition of “neutral” is [H3O+]  [OH-] and that neutral only
ends up “being” pH of 7.00 if the temperature is 25°C!
22. WO14.
A solution has a pH of 7.0 but is basic. Explain clearly what this means and how it can be so! (Hint: Apply
the definition of “basic” and the definition of pH!) ADD: What can you say about the value of [H3O+][OH-] in
this solution? And what about the value of Kw? Finally, what can you conclude about the T of the solution?
Answers: [H3O+]  1.0 x 10-7 M and [OH-] > [H3O+]  Kw  [H3O+][OH-] > 1.0 x 10-14
 T  25C (i.e., this can be so as long as T is not 25C; pH of 7.0 means “neutral” only at 25C!)
Reasoning:
The pH is 7.0. That must mean (based on our definition of pH) that [H3O+]  1.0 x 10-7 M.
(It turns out that this definition is only approximately true; it applies for extremely dilute solutions only. But I will not hold
you accountable for anything but this definition. Ask me if you want to know more.)
The solution is basic. That must mean that [OH-] > [H3O+], period. Why? Because THAT is the
definition of “basic”!
THUS:
If [H3O+]  1.0 x 10-7 M and [OH-] > [H3O+], [OH-] MUST be greater than 1.0 x 10-7 M!
This means that [H3O+] x [OH-] {= (1.0 x 10-7) x (a number bigger than 1.0 x 10-7)} is NOT equal to
1.0 x 10-14 (it must be greater than that).
PS5-7
Answer Key, Problem Set 5
If [H3O+] x [OH-] is bigger than 1.0 x 10-14, then Kw must be bigger than 1.0 x 10-14. Since Kw = 1.0
x 10-14 at 25 C, that must mean the temperature of the given solution is NOT 25C!
It turns out that the T must be greater than 25C,
because the autoionization of water is an endothermic
reaction. From our study of equilibrium properties of
systems, if a reaction is endothermic, then a T increase
will lead to an increase in the value of K. Remember?
From another text*, I found the following data (see right)
describing how Kw depends on T. Note that Kw
increases as T increases.
T (C)
*Quantitative Chemical Analysis, 2nd Ed. (1987); Harris,
D. C.; W. H. Freeman and Company.
Kw
pKw
-14
0
0.114 x 10
20
0.681 x 10-14
14.9435
14.167
24
1.00 x 10-14
14.000
25
1.01 x 10-14
13.9965
30
1.47 x 10-14
13.833
40
2.92 x 10-14
13.535
60
9.61 x 10-14
13.017
Definition of Ksp and its Associated Chemical Equation; Relation to Molar Solubility (in Pure Water)
23. WO15; 17.85
Write balanced equations and expressions for Ksp for the dissociation of each ionic compound:
17.85
Ba2+(aq) + SO42-(aq); Ksp  [Ba2+][SO42-]
(a) BaSO4(s)
Pb2+(aq) + 2 Br-(aq); Ksp = [Pb2+][Br-]2
(b) PbBr2(s)
2 Ag+(aq) + CrO42-(aq); Ksp  [Ag+]2[CrO42-]
(c) Ag2CrO4(s)
Added by me:
3 Sr2+(aq) + 2 PO43-(aq); Ksp  [Sr2+]3[CO32-]2
(d) Sr3(PO4)2 (s)
(e) Fe(OH)3 (s)
Fe3+(aq) + 3 OH-(aq); Ksp = [Fe3+][OH-]3
(f) Ag3PO4 (s)
3 Ag+(aq) + PO43- (aq); Ksp  [Ag+]3[PO43-]
24. WO16. 17.88(b).
(b) Ag2CrO4
NOTE:
I have left off the “eq” subscript
designations from these K
expressions merely for
convenience. Please be aware
that these are equilibrium
concentrations nonetheless!
Use the Ksp values in Table 17.2 to calculate the molar solubility of each compound in pure water.
Answer: 6.54 x 10-5 M
Strategy
1) Recognize that this is a “Find the concentrations at equilibrium” kind of problem in which you are
given (or can figure out) the initial situation as well as K. In such problems, it is useful to create
an ICE table and define an x to represent a species’ “change”. **One difference here is that you
are not given the balanced chemical equation—you must create that yourself (as in the prior
problem). Be very careful to write the proper charges of the ions made as well as the correct
coefficients.
2) In a Ksp context, x is essentially always defined as the “moles of solid that dissolve in exactly 1 L
of solution” and as such, x is the “molar solubility” of the solid (what is asked for in this problem)
3) Procees as in any ICE / x type of problem. That is, use stoichiometry to figure out expressions
for the equilibrium concentrations (i.e., fill out the ICE table correctly), substitute into the K
expression, and solve for x. Unlike in other problem types, if the only thing asked for is the molar
solubility, then you are essentially done when you “find x”. However, it is always a good idea to
solve for all concentrations and check your answer by plugging back into the K expression to see
if you get a value of K close to the given value.
Execution of Strategy:
(a) Ag2CrO4(s)
2 Ag+(aq) + CrO42-(aq); Ksp  [Ag+]2[CrO42-]  1.12 x 10-12
PS5-8
Answer Key, Problem Set 5
I
C
E
"Ag2CrO4(s)"
[Ag+] (M)
[CrO42-] (M)
excess
- “x”
leftover
0
+ 2x
2x
0
+x
x
Ksp  [Ag+]2[CrO42-]  (2x)2(x)  4x3  1.12 x 10-12 
3
1.12 x 10 -12
 6.542 x 10 - 5
4
 The (molar) solubility of Ag2CrO4(s) (in pure water) is 6.54 x 10-5 M
Check: [Ag+]eq = 2(6.54 x 10-5) = 1.31 x 10-4 M
Ksp = (1.31 x 10-4)2(6.54 x 10-5) = 1.118..x 10-12 (v. close)
[CrO42-]eq = 6.54 x 10-5 M
25. MP. No answer in key. NOTE: This problem asks you to calculate the molar solubility of Fe(OH)3 in
pure water at a temperature at which the Ksp is given/known. I assigned this before WO17
because this has an interactive “worked example” video that I thought would be useful.
However, since this problem has no hints, if you are struggling with it, I invite you to look at my
“Strategy” for this kind of problem shown in WO17 below.
26. WO17. 17.90(b,c).Use the given molar solubilities in pure water to calculate Ksp for each compound.
(b) Ag2SO3; molar solubility = 1.55 x 10-5 M
(c) Pd(SCN)2; molar solubility = 2.22 x 10-8 M
Answers: (b) 1.49 x 10-14; (c) 4.38 x 10-23
Strategy:
1) Recognize that writing out the balanced equation for dissolution is needed (so that the Ksp
expression can be properly written, and so that the proper stoichiometry can be clearly “seen”.)
2) Although this could be done later, I would just go ahead and write out the Ksp expression right
next to the balanced equation.
3) Recognize that the molar solubility represents the “moles of solid that dissolve per liter of
solution”. This means that the concentration of each dissolved ion at equilibrium can be
determined using the balanced equation (i.e., stoichiometry).
4) Plug the equilibrium concentration values into the Ksp expression and calculate Ksp.
Execution of Strategy (for Ag2SO3):
Ag2SO3(s)
2 Ag+(aq) + SO32-(aq); Ksp  [Ag+]2[SO32-]
If 1.55 x 10-5 moles of solid dissolve in a liter, then 2 x (1.55 x 10-5)  3.10 x 10-5 moles of Ag+
must form (per liter)  [Ag+]eq = 3.10 x 10-5 M ;
[SO32-]eq = 1.55 x 10-5 M
(because one SO32- forms per formula unit of Ag2SO3 that dissolves)
Thus, Ksp  (3.10 x 10-5)2(1.55 x 10-5)  1.489.. 1.49 x 10-14
Execution of Strategy (for Pd(SCN)2):
Pd(SCN)2(s)
Pd2+(aq) + 2 SCN-(aq); Ksp = [Pd2+][SCN-]2
[Pd2+]eq = 2.22 x 10-8 M (because one Pd2+ forms per FU of Pd(SCN)2 dissolved)
[SCN-]eq = 2(2.22 x 10-8) M = 4.44 x 10-8 M (because two SCN- ions form per FU of Pd(SCN)2 dissolved)
Thus, Ksp  (2.22 x 10-8)(4.44 x 10-8)2  4.376.... 4.38 x 10-23
PS5-9
Answer Key, Problem Set 5
27. MP. No answer in key. More practice with “molar solubility from Ksp” and “Ksp from molar solubility” kinds of
problem. Please note that the “solubility product constant” is the same thing as “Ksp”.
Molar Solubilities In Solutions with Some of a “Product Ion” Present
28. WO18. 17.95.
Calculate the molar solubility of barium fluoride in each liquid or solution [at 25°C]. [Use Ksp from
Table 17.2.]
(a) pure water
(b) 0.10 M Ba(NO3)2
Answers: (a) 1.83 x 10-2 M; (b) 7.5 x 10-3 M; (c) 1.1 x 10-3 M
(c) 0.15 M NaF
Strategy:
Same as in Problem 13, except that in (b) and (c), the initial concentration of both ions is not
zero. Be careful to take this into account properly in the ICE table, and then use the “small x”
approximation to simplify the algebra once the Ksp expression is substituted into and you are
“solving for x”. NOTE: The molar solubility is still “x” in (b) and (c) (It is not, for example, the
equilibrium concentration of [Ba2+] in (b)).
Execution of Strategy (a):
BaF2(s)
I
C
E
Ba2+(aq) + 2 F-(aq); Ksp  [Ba2+][F-]2  2.45 x 10-5
" BaF2(s)"
[Ba2+] (M)
[F-] (M)
excess
- “x”
leftover
0
+x
x
0
+ 2x
2x
Ksp  [Ba2+][F-]2  (x)(2x)2  4x3  2.45 x 10-5 
3
2.45 x 10 -5
 1.829 x 10 - 2
4
 The (molar) solubility of BaF2(s) (in pure water) is 1.83 x 10-2 M
Check: [Ba2+]eq = 1.83 x 10-2 M
[F-]eq = 2(1.83 x 10-2) M = 3.66 x 10-2 M
Ksp = (1.83 x 10-2)(3.66 x 10-2)2 = 2.451..x 10-5 (v. close)
Execution of Strategy (b):
BaF2(s)
I
C
E
Ba2+(aq) + 2 F-(aq); Ksp  [Ba2+][F-]2  2.45 x 10-5
" BaF2(s)"
[Ba2+] (M)
[F-] (M)
excess
- “x”
leftover
0.10
+x
0.10 + x
0
+ 2x
2x
small x approx; 0.10  x  0.10
Ksp  [Ba2+][F-]2  (0.10 + x)(2x)2  
      (0.10)(2x)2  2.45 x 10-5
 4x 2 
2.45 x 10 -5
 x
0.10
2.45 x 10 -5
 0.007826
4(0.10)
Check approx:
0.0078
x 100  7.8% Oops!
0.10
The approximation isn’t quite “good” based on
the 5% “rule”. See below for how to address.
It’s very rare for the small x approximation not to work for situations such as this (a common
ion situation—where one of the products is present to begin with). It only occurred here
because Ksp was not really all that small. If that happens, one can do one more “iteration” as
follows to get a more accurate value for x:
PS5-10
Answer Key, Problem Set 5
Use the approximate value for x just determined in place of the “small x” approximation.
That is, use 0.10 + x  0.10 + 0.0078 = 0.1078 (Note: only do this for the “0.10 + x” term, and
don’t round to the proper # of SFs [yet]). Now we have:
(0.1078)(2x)2  2.45 x 10-5  4 x 2 
2.45 x 10 -5
 x
0.1078
2.45 x 10 -5
 0.007537..
4(0.1078)
In this case, I’d do the check first and then make the final conclusion:
Check: [Ba2+]eq = 0.10 + 0.007537 M = 0.107 M
Ksp = (0.107)(1.507 x 10-2)2 = 2.43..x 10-5 (v. close)
[F-]eq = 2(0.007537) M = 1.507 x 10-2 M
 The (molar) solubility of BaF2(s) in 0.10 M Ba(NO3)2 is 7.5 x 10-3 M
**The molar solubility of BaF2 here is not equal to the final (equilibrium) concentration of
Ba2+ ion because most of that Ba2+ ion was present before the dissolution reaction
occurred. If this confuses you, recall that x is defined to be “the moles of solid that
dissolve” (to reach equilibrium), not the moles of the cation present at equilibrium.**
Execution of Strategy (c):
BaF2(s)
I
C
E
Ba2+(aq) + 2 F-(aq); Ksp  [Ba2+][F-]2  2.45 x 10-5
" BaF2(s)"
[Ba2+] (M)
[F-] (M)
excess
- “x”
leftover
0
+x
x
0.15
+ 2x
0.15 + 2x
small x approx; 0.15  2 x  0.15
Ksp  [Ba2+][F-]2  (x)(0.15 + 2x)2  

 (x)(0.15)2  2.45 x 10-5
 x
2.45 x 10 -5
 x  0.001088
0.15 2
Check approx: 2(0.001088) x 100  1.5%
0.15
 The (molar) solubility of BaF2(s) in 0.15 M NaF is 1.1 x 10-3 M
Check: [Ba2+]eq = 1.09 x 10-3 M
-
Ksp = (1.09 x 10-3)(0.152)2 = 2.52..x 10-5 (close)
-3
[F ]eq = 0.15 + 2(1.09 x 10 ) M = 0.152 M
29. MP. No answer in key. Another “Find the molar solubility from Ksp” kind of problem (one in pure water and
another in a solution of a “common ion”).
“Will a Precipitate Form?” Kinds of Problems
30. WO19. 17.104.
Predict whether or not a precipitate will form if you mix 175.0 mL of a 0.0055 M KCl solution with
145.0 mL of a 0.0015 M AgNO3 solution [at 25°C]. Identify the precipitate, if any. [Use Ksp from Table 17.2.].
Answer: A precipitate of AgCl(s) should form
Strategy:
1) Recognize that if a precipitate would form, it would have to be AgCl(s), because all nitrate salts
are soluble, and all compounds with K+ as the cation are soluble (1st semester).
2) Recognize that since Ksp is what is given (in 17.2), you should create the balanced chemical
equation (and Ksp expression) for AgCl dissolution.
PS5-11
Answer Key, Problem Set 5
3) Once this equation is set (see #2), realize that “precipitate forms” means “reverse reaction
occurs”. Thus, you need to determine Q (initially, but after mixing) and see if it is greater than
K. If so, a precipitate should form. If not, the solution is unsaturated (and no precipitate will
form.)
4) Use the dilution factor approach (or M1V1 = M2V2) to determine the initial concentrations of Ag+
and Cl- (also recognizing that KCl ionizes 100% into K+ and Cl- and that AgNO3 ionizes into
Ag+ and NO3-) and then substitute into the Qsp expression to find Q. Then compare to Ksp as
noted in 3) to make the final conclusion.
Execution of Strategy:
AgCl(s)
Ag+(aq) + Cl-(aq); Ksp  [Ag+][Cl-]  1.77 x 10-10 (from Table 17.2)
“Reverse reaction occurs”  a precipitate will form  check for Q > K
175.0 mL of 0.0055 M KCl(aq) + 145.0 mL of 0.0015 M AgNO3(aq) 
Vtotal = 175.0 + 145.0 = 320.0 mL


 175.0 mL 
[Cl-]0  0.0055 M Cl- 
  0.00301M Cl
 320.0 mL 


Qsp  (3.01 x 10-3)(6.80 x 10-4)  2.05..x 10-6
 145.0 mL 
[Ag+]0  0.0015 M Ag 
  0.000680 M Ag
 320.0 mL 
Since 2.0 x 10-6 > 1.77 x 10-10, Qsp > Ksp and thus a precipitate (of AgCl) will form
31. MP. No answer in key. Another precipitation kind of problem (multiple parts).
32. MP. No answer in key. This is a nice LeChatelier’s type of equilibrium problem related to solubility
equilibria. Be very careful with these kinds of problems. For some reason, some student tend
to get the thinking backwards on this!
PS5-12