CHAPTER FOURTEEN
ACIDS AND BASES
For Review
1.
a. Arrhenius acid: produce H+ in water
b. Brnsted-Lowry acid: proton (H+) donor
c. Lewis acid: electron pair acceptor
The Lewis definition is most general. The Lewis definition can apply to all Arrhenius and
Brnsted-Lowry acids; H+ has an empty 1s orbital and forms bonds to all bases by accepting a
pair of electrons from the base. In addition, the Lewis definition incorporates other reactions not
typically considered acid-base reactions, e.g., BF3(g) + NH3(g) → F3BNH3(s). NH3 is
something we usually consider a base and it is a base in this reaction using the Lewis definition;
NH3 donates a pair of electrons to form the NB bond.
2.
a. The Ka reaction always refers to an acid reacting with water to produce the conjugate base of
the acid and the hydronium ion (H3O+). For a general weak acid HA, the Ka reaction is:
HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq) where A = conjugate base of the acid HA
This reaction is often abbreviated as: HA(aq) ⇌ H+(aq) + A(aq)
b. The Ka equilibrium constant is the equilibrium constant for the Ka reaction of some substance. For the general Ka reaction, the Ka expression is:
Ka =
[A  ][H 3O  ]
[H  ][A  ]
or K a 
[HA]
[HA]
(for the abbreviated Ka reaction)
c. The Kb reaction alwlays refers to a base reacting with water to produce the conjugate acid
of the base and the hydroxide ion (OH). For a general base, B, the Kb reaction is:
B(aq) + H2O(l) ⇌ BH+(aq) + OH(aq) where BH+ = conjugate acid of the base B
d. The Kb equilibrium constant for the general Kb reaction is: Kb =
[BH  ][OH  ]
[B]
e. A conjugate acid-base pair consists of two substances related to each other by the donating
and accepting of a single proton. The conjugate bases of the acids HCl, HNO2, HC2H3O2 and
H2SO4 are Cl, NO2, C2H3O2, and HSO4, respectively. The conjugate acids of the bases
NH3, C5H5N, and HONH2 are NH4+, C5H5NH+, and HONH3+, respectively. Conjugate acidbase pairs only differ by H+ in their respective formulas.
485
486
3.
CHAPTER 14
ACIDS AND BASES
a. Amphoteric: a substance that can behave either as an acid or as a base.
b. The Kw reaction is also called the autoionization of water reaction. The reaction always
occurs when water is present as the solvent. The reaction is:
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH(aq) or H2O(l) ⇌ H+(aq) + OH(aq)
c. The Kw equilibrium constant is also called the ion-product constant or the dissociation
constant of water. It is the equilibrium constant for the autoionization reaction of water:
Kw = [H3O+][OH] or Kw = [H+][OH]
At typical solution temperatures of 25C, Kw = 1.0 × 10 14.
d. pH is a mathematical term which is equal to the –log of the H+ concentration of a solution
(pH = –log[H+].
e. pOH is a mathematical terem which is equal to the –log of the OH concentration of a
solution (pOH = –log[OH]).
f.
The p of any quantity is the –log of that quantity. So: pKw = –logKw. At 25C, pKw = –log
(1.0 × 10 14 ) = 14.00.
Neutral solution at 25C: Kw = 1.0 × 10 14 = [H+][OH] and pH + pOH = 14.00
[H+] = [OH ] = 1.0 × 10 7 M; pH = pOH = –log(1.0 × 10 7 ) = 7.00
Acidic solution at 25C:
[H+] > [OH ]; [H+] > 1.0 × 10 7 M; [OH ] < 1.0 × 10 7 M; pH < 7.00; pOH > 7.00
Basic solution at 25C:
[OH] > [H+]; [OH] > 1.0 × 10 7 M; [H+] < 1.0 × 10 7 M; pOH < 7.00; pH > 7.00
As a solution becomes more acidic, [H+] increases, so [OH] decreases, pH decreases, and
pOH increases. As a solution becomes more basic, [OH] increases, so [H+] decreases, pH
increases, and pOH decreases.
4.
The Ka value refers to the reaction of an acid reacting with water to produce the conjugate base
and H3O+. The stronger the acid, the more conjugate base and H3O+ produced, and the larger the
Ka value.
Strong acids are basically 100% dissociated in water. Therefore, the strong acids have a Ka >> 1
because the equilibrium position lies far to the right. The conjugate bases of strong acids are terrible bases; much worse than water, so we can ignore their basic properties in water.
CHAPTER 14
ACIDS AND BASES
487
Weak acids are only partially dissociated in water. We say that the equilibrium lies far to the left,
thus giving values for Ka < 1. (We have mostly reactants at equilibrium and few products
present). The conjugate bases of weak acids are better bases than water. When we have a solution
composed of just the conjugate base of a weak acid in water, tbe resulting pH is indeed basic (pH
> 7.0). In general, as the acid strength increases, the conjugate base strength decreases, or as acid
strength decreases, the conjugate base strength increases. They are inversely related.
Base strength is directly related to the Kb value. The larger the Kb value, the more OH produced
from the Kb reaction, and the more basic the solution (the higher the pH). Weak bases have a Kb <
1 and their conjugate acids behave as weak acids in solution. As the strength of the base
increases, the strength of the conjugate acid gets weaker; the stronger the base, the weaker the
conjugate acid, or the weaker the base, the stronger the conjugate acid.
5.
Strong acids are assumed 100% dissociated in water, and we assume that the amount of H+
donated by water is negligible. Hence, the equilibrium [H+] of a strong acid is generally equal to
the initial acid concentration ([HA]0). Note that solutions of H2SO4 can be different from this as
H2SO4 is a diprotic acid. Also, when you have very dilute solutions of a strong acid, the H+
contribution from water by itself must be considered. The strong acids to memorize are HCl, HBr,
HI, HNO3, HClO4, and H2SO4.
Ka values for weak acids are listed in Table 14.2 and in Appendix 5 of the text. Because weak
acids only partially dissociate in water, we must solve an equilibrium problem to determine how
much H+ is added to water by the weak acid. We write down the Ka reaction, set-up the ICE table,
then solve the equilibrium problem. The two assumptions generally made are that acids are less
than 5% dissociated in water and that the H+ contribution from water is negligible.
The 5% rule comes from the assumptions that weak acids are less than 5% dissociated. When this
is true, the mathematics of the problem are made much easier. The equilibrium expression we get
for weak acids in water generally has the form (assuming an initial acid concentration of 0.10 M):
x2
x2

Ka =
0.10  x
0.10
The 5% rule refers to assuming 0.10 – x  0.10. The assumption is valid if x is less than 5% of the
number the assumption was made against ([HA]0). When the 5% rule is valid, solving for x is
very straight forward. When the 5% rule fails, we must solve the mathematical expression exactly
using the quadratic equation (or your graphing calculator). Even if you do have a graphing
calculator, reference Appendix A1.4 to review the quadratic equation. Appendix A1.4 also
discusses the method of successive approximations which can also be used to solve quadratic
(and cubic) equations.
6.
Strong bases are soluble ionic compounds containing the OH anion. Strong bases increase the
OH concentration in water by just dissolving. Thus, for strong bases like LiOH, NaOH, KOH,
RbOH, and CsOH, the initial concentration of the strong base equals the equilibrium [OH] of
water.
The other strong bases to memorize have +2 charged metal cations. The soluble ones to know are
Ca(OH)2, Sr(OH)2, and Ba(OH)2. These are slightly more difficult to solve because they donate 2
moles OH for every mole of salt dissolved. Here, the [OH] is equal to two times the initial
concentration of the soluble alkaline earth hydroxide salt dissolved.
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ACIDS AND BASES
Neutrally charged organic compounds containing at least one nitrogen atom generally behave as
weak bases. The nitrogen atom has an unshared pair of electrons around it. This lone pair of
electrons is used to form a bond to H+.
Weak bases only partially react with water to produce OH. To determine the amount of OH
produced by the weak acid (and, in turn, the pH of the solution), we set-up the ICE table using the
Kb reaction of the weak base. The typical weak base equilibrium expression is:
Kb =
x2
x2

0.25  x
0.25
(assuming [B]0 = 0.25 M)
Solving for x gives us the [OH] in solution. We generally assume that weak bases are only 5%
reacted with water and that the OH contribution from water is negligible. The 5% assumption
makes the math easier. By assuming an expression like 0.25 M – x  0.25 M, the calculation is
very straight forward. The 5% rule applied here is that if (x/0.25) × 100 is less than 5%, the
assumption is valid. When the assumption is not valid, then we solve the equilibrium expression
exactly using the quadratic equation (or by the method of successive approximations).
7.
Monoprotic acid: an acid with one acidic proton; the general formula for monoprotic acids is
HA.
Diprotic acid: an acid with two acidic protons (H2A)
Triprotic acid: an acid with three acidic protons (H3A)
H2SO4(aq) → HSO4(aq) + H+(aq)
HSO4(aq)
⇌
SO42(aq) + H+(aq)
K a1 >> 1; this is a strong acid.
K a 2 = 0.012; this is a weak acid.
When H2SO4 is dissolved in water, the first proton is assumed 100% dissociated because H2SO4 is
a strong acid. After H2SO4 dissociates, we have H+ and HSO4 present. HSO4 is a weak acid and
can donate some more protons to water. To determine the amount of H+ donated by HSO4, one
must solve an equilibrium problem using the K a 2 reaction for HSO4.
⇌
H+(aq) + H2PO4(aq)
K a1 = 7.5 × 10 3
H2PO4(aq) ⇌ H+(aq) + HPO42(aq)
K a 2 = 6.2 × 10 8
HPO42(aq) ⇌ H+(aq) + PO43(aq)
K a 3 = 4.8 × 10 13
H3PO4(aq)
When H3PO4 is added to water, the three acids that are present are H3PO4, H2PO4, and HPO42.
H3PO4, with the largest Ka value, is the strongest of these weak acids. The conjugate bases of the
three acids are H2PO4, HPO42, and PO43. Because HPO42 is the weakest acid (smallest Ka
value), its conjugate base (PO43) will have the largest Kb value and is the strongest base.
CHAPTER 14
ACIDS AND BASES
489
See Sample Exercises 14.15-14.17 on the strategies used to solve for the pH of polyprotic acids.
The strategy to solve most polyprotic acid solutions is covered in Sample Exercise 14.15. For
typical polyprotic acids, K a1 >> K a 2 (and K a 3 if a triprotic acid). Because of this, the dominant
producer of H+ in solution is just the K a1 reaction. We set-up the equilibrium problem using the
K a1 reaction and solve for H+. We then assume that the H+ donated by the K a 2 (and K a 3 if
triprotic) reaction is negligible that is, the H+ donated by the K a1 reaction is assumed to be the H+
donated by the entire acid system. This assumption is great when K a1 >> K a 2 (roughly a 1000 fold
difference in magnitude).
Sample Exercises 14.16 and 14.17 cover strategies for the other type of polyprotic acid problems.
This other type is solutions of H2SO4. As discussed above, H2SO4 problems are both a strong acid
and a weak acid problem in one. To solve for the [H+], we sometimes must worry about the H+
contribution from HSO4. Sample Exercise 13.16 is an example of an H2SO4 solution where the
HSO4 contribution of H+ can be ignored. Sample Exercise 14.17 illustrates an H2SO4 problem
where we can’t ignore the H+ contribution from HSO4.
8.
a. H2O and CH3CO2
b. An acid-base reaction can be thought of as a competiton between two opposing bases. Since
this equilibrium lies far to the left (Ka < 1), CH3CO2 is a stronger base than H2O.
c. The acetate ion is a better base than water and produces basic solutions in water. When we
put acetate ion into solution as the only major basic species, the reaction is:
CH3CO2 + H2O ⇌ CH3CO2H + OH
Now the competition is between CH3CO2 and OH for the proton. Hydroxide ion is the
strongest base possible in water. The equilibrium above lies far to the left, resulting in a K b
value less than one. Those species we specifically call weak bases ( 10 14 < Kb < 1) lie
between H2O and OH in base strength. Weak bases are stronger bases than water but are
weaker bases than OH.
The NH4+ ion is a weak acid because it lies between H2O and H3O+ (H+) in terms of acid
strength. Weak acids are better acids than water, thus their aqueous solutions are acidic. They
are weak acids because they are not as strong as H3O+ (H+). Weak acids only partially
dissociate in water and have Ka values between 10 14 and 1.
For a strong acid HX having Ka = 1 × 106, the conjugate base, X, has Kb = Kw/Ka
= 1.0 × 10 14 / 1 × 106 = 1 × 10 20 .
The conjugate bases of strong acids have extremely small values for K b; so small that they are
worse bases than water (Kb << Kw). Therefore, conjugate bases of strong acids have no basic
properties in water. They are present, but they only balance charge in solution and nothing else.
The conjugate bases of the six strong acids are Cl, Br, I, NO3, ClO4, and HSO4.
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ACIDS AND BASES
Summarizing the acid-base properlties of conjugates:
a. The conjugate base of a weak acid is a weak base ( 10 14 < Kb < 1)
b. The conjugate acid of a weak base is a weak acid ( 10 14 < Ka < 1)
c. The conjugate base of a strong acid is a worthless base (Kb << 10 14 )
d. The conjugate acid of a strong base is a worthless acid (Ka << 10 14 )
Identifying/recognizing the acid-base properties of conjugates is crucial in order to understand the
acid-base properties of salts. The salts we will give you will be salts containing the conjugates
discussed above. Your job is to recognize the type of conjugate present, and then use that
information to solve an equilibrium problem.
9.
A salt is an ionic compound composed of a cation and an anion.
Weak base anions: these are the conjugate bases of the weak acids having the HA general
formula. Table 14.2 lists several HA type acids. Some weak base anions derived from the acids in
Table 14.2 are ClO2, F, NO2, C2H3O2, OCl, and CN.
Garbage anions (those anions with no basic or acidic properties): these are the conjugate bases of
the strong acids having the HA general formula. Some neutral anions are Cl , NO3, Br, I, and
ClO4.
Weak acid cations: these are the conjugate acids of the weak bases which contain nitrogen. Table
14.3 lists several nitrogen-containing bases. Some weak acid cations derived from the weak bases
in Table 14.3 are NH4+, CH3NH3+, C2H5NH3+, C6H5NH3+, and C5H5NH+.
Garbage cations (those cations with no acidic properties or basic properties): the most common
ones used are the cations in the strong bases. These are Li+, Na+, K+, Rb+, Cs+, Ca2+, Sr2+, and
Ba2+.
We mix and match the cations and anions to get what type of salt we want. For a weak base salt,
we combine a weak base anion with a garbage cation. Some weak base salts are NaF, KNO2,
Ca(CN)2, and RbC2H3O2. To determine the pH of a weak base salt, we write out the K b reaction
for the weak base anion and determine Kb (= Kw/Ka). We set-up the ICE table under the Kb
reaction, and then solve the equilibrium problem to calculate [OH] and, in turn, pH.
For a weak acid salt, we combine a weak acid cation with a garbage anion. Some weak acid salts
are NH4Cl, C5H5NHNO3, CH3NH3I, and C2H5NH3ClO4. To determine the pH, we write out the Ka
reaction for the weak acid cation and determine Ka (= Kw/Kb). We set-up the ICE table under the
Ka reaction, and then solve the equilibrium problem to calculate [H+] and, in turn, pH.
For a neutral (pH = 7.0) salt, we combine a garbage cation with a garbage anion. Some examples
are NaCl, KNO3, BaBr2, and Sr(ClO4)2.
CHAPTER 14
ACIDS AND BASES
491
For salts that contain a weak acid cation and a weak base anion, we compare the K a value of the
weak acid cation to the Kb value for the weak base anion. When Ka > Kb, the salt produces an
acidic solution (pH < 7.0). When Kb > Ka, the salt produces a basic solution. And when Ka = Kb,
the salt produces a neutral solution (pH = 7.0).
10.
a. The weaker the XH bond in an oxyacid, the stronger the acid.
b. As the electronegativity of neighboring atoms increases in an oxyacid, the strength of the acid
increases.
c. As the number of oxygen atoms increases in an oxyacid, the strength of the acid increases.
In general, the weaker the acid, the stronger the conjugate base and vice versa.
a. Because acid strength increases as the XH bond strength decreases, conjugate base strength
will increase as the strength of the XH bond increases.
b. Because acid strength increases as the electronegativity of neighboring atoms increases,
conjugate base strength will decrease as the electronegativity of neighboring atoms increases.
c. Because acid strength increases as the number of oxygen atoms increases, conjugate base
strength decreases as the number of oxygen atoms increases.
Nonmetal oxides form acidic solutions when dissolved in water:
SO3(g) + H2O(l) → H2SO4(aq)
Metal oxides form basic solutions when dissolved in water:
CaO(s) + H2O(l) → Ca(OH)2
Questions
16.
When a strong acid (HX) is added to water, the reaction HX + H2O → H3O+ + X basically goes
to completion. All strong acids in water are completely converted into H3O+ and X. Thus, no
acid stronger than H3O+ will remain undissociated in water. Similarly, when a strong base (B) is
added to water, the reaction B + H2O → BH+ + OH basically goes to completion. All bases
stronger than OH- are completely converted into OH- and BH+. Even though there are acids and
bases stronger than H3O+ and OH, in water these acids and bases are completely converted into
H3O+ and OH.
17.
10.78 (4 S.F.); 6.78 (3 S.F.); 0.78 (2 S.F.); A pH value is a logarithm. The numbers to the left of
the decimal point identify the power of ten to which [H+] is expressed in scientific notation, e.g.,
10 11 , 10 7 , 10 1 . The number of decimal places in a pH value identifies the number of
significant figures in [H+]. In all three pH values, the [H+] should be expressed only to two
significant figures because these pH values have only two decimal places.
18.
A Lewis acid must have an empty orbital to accept an electron pair, and a Lewis base must have
an unshared pair of electrons.
492
19.
CHAPTER 14
ACIDS AND BASES
a. These are strong acids like HCl, HBr, HI, HNO3, H2SO4 or HClO4.
b. These are salts of the conjugate acids of the bases in Table 14.3. These conjugate acids are all
weak acids. NH4Cl, CH3NH3NO3, and C2H5NH3Br are three examples. Note that the anions
used to form these salts (Cl, NO3, and Br) are conjugate bases of strong acids; this is
because they have no acidic or basic properties in water (with the exception of HSO4, which
has weak acid properties).
c. These are strong bases like LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and
Ba(OH)2.
d. These are salts of the conjugate bases of the neutrally charged weak acids in Table 14.2. The
conjugate bases of weak acids are weak bases themselves. Three examples are NaClO 2,
KC2H3O2, and CaF2. The cations used to form these salts are Li+, Na+, K+, Rb, Cs+, Ca2+, Sr2+,
and Ba2+ since these cations have no acidic or basic properties in water. Notice that these are
the cations of the strong bases you should memorize.
e. There are two ways to make a neutral salt. The easiest way is to combine a conjugate base of
a strong acid (except for HSO4) with one of the cations from a strong base. These ions have
no acidic/basic properties in water so salts of these ions are neutral. Three examples are NaCl,
KNO3, and SrI2. Another type of strong electrolyte that can produce neutral solutions are salts
that contain an ion with weak acid properties combined with an ion of opposite charge having
weak base properties. If the Ka for the weak acid ion is equal to the Kb for the weak base ion,
then the salt will produce a neutral solution. The most common example of this type of salt is
ammonium acetate, NH4C2H3O2. For this salt, Ka for NH4+ = Kb for C2H3O2 = 5.6 × 10 10 .
This salt at any concentration produces a neutral solution.
20.
Ka × Kb = Kw, log (Ka × Kb) = log Kw
log Ka  log Kb = log Kw, pKa + pKb = pKw = 14.00 (at 25°C)
21.
a. H2O(l) + H2O(l) ⇌ H3O+(aq) + OH(aq) or
H2O(l) ⇌ H+(aq) + OH(aq)
K = Kw = [H+][OH]
b. HF(aq) + H2O(l) ⇌ F(aq) + H3O+(aq) or
[H  ][F ]
HF(aq) ⇌ H+(aq) + F(aq) K = Ka =
[HF]
c. C5H5N(aq) + H2O(l)
22.
⇌ C5H5NH+(aq) + OH(aq)
K = Kb =
[C5 H 5 NH  ][OH  ]
[C 5 H 5 N ]
Only statement a is true (assuming the species is not amphoteric). You cannot add a base to water
and get an acidic pH (pH < 7.0). For statement b, you can have negative pH values. This just
indicates an [H+] > 1.0 M. For statement c, a dilute solution of a strong acid can have a higher pH
than a more concentrated weak acid solution. For statement d, the Ba(OH) 2 solution will have an
CHAPTER 14
23.
ACIDS AND BASES
493
[OH] twice of the same concentration of KOH, but this does not correspond to a pOH value
twice that of the same concentration of KOH (prove it to yourselves).
a. This expression holds true for solutions of strong acids having a concentration greater than
1.0 × 10 6 M. 0.10 M HCl, 7.8 M HNO3, and 3.6 × 10 4 M HClO4 are examples where this
expression holds true.
b. This expression holds true for solutions of weak acids where the two normal assumptions
hold. The two assumptions are that water does not contribute enough H+ to solution to matter
and that the acid is less than 5% dissociated in water (from the assumption that x is small
compared to some number). This expression will generally hold true for solutions of weak
acids having a Ka value less than 1 × 10 4 , as long as there is a significant amount of weak
acid present. Three example solutions are 1.5 M HC2H3O2, 0.10 M HOCl, and 0.72 M HCN.
c. This expression holds true for strong bases that donate 2 OH ions per formula unit. As long
as the concentration of the base is above 5 × 10 7 M, this expression will hold true. Three
examples are 5.0 × 10 3 M Ca(OH)2, 2.1 × 10 4 M Sr(OH)2, and 9.1 × 10 5 M Ba(OH)2.
d. This expression holds true for solutions of weak bases where the two normal assumptions
hold. The assumptions are that the OH contribution from water is negligible and that and that
the base is less than 5% ionized in water (for the 5% rule to hold). For the 5% rule to hold,
you generally need bases with Kb < 1 × 10 4 and concentrations of weak base greater than
0.10 M. Three examples are 0.10 M NH3, 0.54 M C6H5NH2, and 1.1 M C5H5N.
24.
H2CO3 is a weak acid with K a1 = 4.3 × 10 7 and K a 2 = 5.6 × 10 11 . The [H+] concentration in
solution will be determined from the K a1 reaction since K a1 >> K a 2 . Since K a1 << 1, then the
[H+] < 0.10 M; only a small percentage of H2CO3 will dissociate into HCO3 and H+. So statement a best describes the 0.10 M H2CO3 solution. H2SO4 is a strong acid and a very good weak
acid ( K a1 >> 1, K a 2 = 1.2 × 10 2 ). All of the 0.1 M H2SO4 solution will dissociate into 0.10 M
H+ and 0.10 M HSO4. However, since HSO4 is a good weak acid due to the relatively large K a
value, then some of the 0.10 M HSO4 will dissociate into some more H+ and SO42. Therefore,
the [H+] will be greater than 0.10 M, but will not reach 0.20 since only some of 0.10 M HSO4
will dissociate. Statement c is best for a 0.10 M H2SO4 solution.
25.
One reason HF is a weak acid is that the HF bond is unusually strong and is difficult to break.
This contributes significantly to the reluctance of the HF molecules to dissociate in water.
26.
a. Sulfur reacts with oxygen to produce SO2 and SO3. These sulfur oxides both react with water
to produce H2SO3 and H2SO4, respectively. Acid rain can result when sulfur emissions are not
controlled. Note that in general, nonmetal oxides react with water to produce acidic solutions.
b. CaO reacts with water to produce Ca(OH)2, a strong base. A gardener mixes lime (CaO) into
soil in order to raise the pH of the soil. The effect of adding lime is to add Ca(OH)2. Note that
in general, metal oxides react with water to produce basic solutions.
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CHAPTER 14
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Exercises
Nature of Acids and Bases
27.
a. HClO4(aq) + H2O(l) → H3O+(aq) + ClO4(aq). Only the forward reaction is indicated since
HClO4 is a strong acid and is basically 100% dissociated in water. For acids, the dissociation
reaction is commonly written without water as a reactant. The common abbreviation for this
reaction is: HClO4(aq) → H+(aq) + ClO4 (aq). This reaction is also called the Ka reaction as
the equilibrium constant for this reaction is called Ka.
b. Propanoic acid is a weak acid, so it is only partially dissociated in water. The dissociation
reaction is: CH3CH2CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CH2CO2 (aq) or
CH3CH2CO2H(aq) ⇌ H+(aq) + CH3CH2CO(aq).
c. NH4+ is a weak acid. Similar to propanoic acid, the dissociation reaction is:
NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq) or NH4+(aq) ⇌ H+(aq) + NH3(aq)
28.
29.
The dissociation reaction (the Ka reaction) of an acid in water commonly omits water as a
reactant. We will follow this practice. All dissociation reactions produce H+ and the conjugate
base of the acid that is dissociated.
[H  ][CN  ]
a. HCN(aq) ⇌ H+(aq) + CN(aq)
Ka =
[HCN]
b. C6H5OH(aq) ⇌ H+(aq) + C6H5O(aq)
Ka =
[H  ][C 6 H 5 O  ]
[C 6 H 5 OH]
c. C6H5NH3+(aq) ⇌ H+(aq) + C6H5NH2(aq)
Ka =
[H  ][C 6 H 5 NH 2 ]
[C 6 H 5 NH3 ]
An acid is a proton (H+) donor and a base is a proton acceptor. A conjugate acid-base pair differs
by only a proton (H+).
a.
b.
c.
Acid
Base
H2CO3
C5H5NH+
C5H5NH+
H2O
H2O
HCO3
Acid
Base
Al(H2O)63+
HONH3+
HOCl
H2O
H2O
C6H5NH2
30.
a.
b.
c.
Conjugate
Base of Acid
HCO3
C5H5N
C5H5N
Conjugate
Base of Acid
Al(H2O)5(OH)2+
HONH2
OCl-
Conjugate
Acid of Base
H3O+
H3O+
H2CO3
Conjugate
Acid of Base
H3O+
H3O+
C6H5NH3+
CHAPTER 14
31.
ACIDS AND BASES
495
Strong acids have a Ka >> 1 and weak acids have Ka < 1. Table 14.2 in the text lists some Ka
values for weak acids. Ka values for strong acids are hard to determine so they are not listed in the
text. However, there are only a few common strong acids so if you memorize the strong acids,
then all other acids will be weak acids. The strong acids to memorize are HCl, HBr, HI, HNO3,
HClO4 and H2SO4.
a. HClO4 is a strong acid.
b. HOCl is a weak acid (Ka = 3.5 × 10 8 ).
c. H2SO4 is a strong acid.
d. H2SO3 is a weak diprotic acid with Ka1 and Ka2 values less than one.
32.
The beaker on the left represents a strong acid in solution; the acid, HA, is 100% dissociated into
the H+ and Aions. The beaker on the right represents a weak acid in solution; only a little bit of
the acid, HB, dissociates into ions, so the acid exists mostly as undissociated HB molecules in
water.
a.
b.
c.
d.
e.
33.
HNO2: weak acid beaker
HNO3: strong acid beaker
HCl: strong acid beaker
HF: weak acid beaker
HC2H3O2: weak acid beaker
The Ka value is directly related to acid strength. As Ka increases, acid strength increases. For
water, use Kw when comparing the acid strength of water to other species. The Ka values are:
HClO4: strong acid (Ka >> 1); HClO2: Ka = 1.2 × 10 2
NH4+: Ka = 5.6 × 10 10 ; H2O: Ka = Kw = 1.0 × 10 14
From the Ka values, the ordering is: HClO4 > HClO2 > NH4+ > H2O.
34.
Except for water, these are the conjugate bases of the acids in the previous exercise. In general,
the weaker the acid, the stronger the conjugate base. ClO4 is the conjugate base of a strong acid;
it is a terrible base (worse than water). The ordering is: NH3 > ClO2 > H2O > ClO4
35.
a. HCl is a strong acid and water is a very weak acid with Ka = Kw = 1.0 × 10 14 . HCl is a
much stronger acid than H2O.
b. H2O, Ka = Kw = 1.0 × 10 14 ; HNO2, Ka = 4.0 × 10 4 ; HNO2 is a stronger acid than H2O
because Ka for HNO2 > Kw for H2O.
c. HOC6H5 , Ka = 1.6 × 10 10 ; HCN, Ka = 6.2 × 10 10 ; HCN is a stronger acid than HOC6H5
because Ka for HCN > Ka for HOC6H5.
36.
a. H2O; The conjugate bases of strong acids are terrible bases (Kb < 10 14 ).
496
CHAPTER 14
ACIDS AND BASES
b. NO2; The conjugate bases of weak acids are weak bases ( 10 14 < Kb < 1).
c. OC6H5; For a conjugate acid-base pair, Ka × Kb = Kw. From this relationship, the stronger
the acid the weaker the conjugate base (Kb decreases as Ka increases). Because HCN is a
stronger acid than HOC6H5 (Ka for HCN > Ka for HOC6H5), OC6H5will be a stronger base
than CN-.
Autoionization of Water and the pH Scale
37.
38.
39.
At 25°C, the relationship: [H+] [OH] = Kw = 1.0 × 10 14 always holds for aqueous solutions.
When [H+] is greater than 1.0 × 10 7 M, the solution is acidic; when [H+] is less than 1.0 ×
10 7 M, the solution is basic; when [H+] = 1.0 × 10 7 M, the solution is neutral. In terms of
[OH], an acidic solution has [OH] < 1.0 × 10 7 M, a basic solution has [OH] > 1.0 × 10 7 M,
and a neutral solution has [OH] = 1.0 × 10 7 M.
a. [OH] =
K w 1.0  10 14

= 1.0 × 10-7 M; The solution is neutral.
[H  ] 1.0  10 7
b. [OH] =
1.0  10 14
= 12 M; The solution is basic.
8.3  10 16
c. [OH] =
1.0  10 14
= 8.3 × 10 16 M; The solution is acidic.
12
d. [OH] =
1.0  10 14
= 1.9 × 10 10 M; The solution is acidic.
5.4  10 5
a. [H+] =
Kw
1.0  10 14

= 6.7 × 10 15 M; basic
1. 5
[OH  ]
b. [H+] =
1.0  10 14
= 2.8 M; acidic
3.6  10 15
c. [H+] =
1.0  10 14
= 1.0 × 10 7 M; neutral
7
1.0  10
d. [H+] =
1.0  10 14
= 1.4 × 10 11 M; basic
7.3  10  4
a. Because the value of the equilibrium constant increases as the temperature increases, the
reaction is endothermic. In endothermic reactions, heat is a reactant so an increase in
temperature (heat) shifts the reaction to produce more products and increases K in the
process.
b. H2O(l) ⇌ H+(aq) + OH(aq)
Kw = 5.47 × 10 14 = [H+][OH] at 50.°C
In pure water [H+] = [OH], so 5.47 × 10 14 = [H+]2, [H+] = 2.34 × 10 7 M = [OH]
CHAPTER 14
40.
ACIDS AND BASES
a. H2O(l) ⇌ H+(aq) + OH (aq)
497
Kw = 2.92 × 10 14 = [H+] [OH]
In pure water [H+] = [OH], so 2.92 × 10 14 = [H+]2, [H+] = 1.71 × 10 7 M = [OH]
b. pH = log [H+] = log (1.71 × 10 7 ) = 6.767
c. [H+] = Kw/[OH] = 2.92 × 10 14 /0.10 = 2.9 × 10 13 M; pH = log (2.9 × 10 13 ) =12.54
41.
pH = log [H+]; pOH = log [OH-]; At 25°C, pH + pOH = 14.00; For Exercise 13.37:
a. pH = log [H+] = log (1.0 × 10 7 ) = 7.00; pOH = 14.00  pH = 14.00  7.00 = 7.00
b. pH = log (8.3 × 10 16 ) = 15.08; pOH = 14.00  15.08 = 1.08
c. pH = log (12) = 1.08; pOH = 14.00  (1.08) = 15.08
d. pH = log (5.4 × 10 5 ) = 4.27; pOH = 14.00  4.27 = 9.73
Note that pH is less than zero when [H+] is greater than 1.0 M (an extremely acidic solution).
For Exercise 13.38:
a. pOH = log [OH] = log (1.5) = 0.18; pH = 14.00  pOH = 14.00  (0.18) = 14.18
b. pOH = log (3.6 × 10 15 ) = 14.44; pH = 14.00  14.44 = 0.44
c. pOH = log (1.0 × 10 7 ) =7.00; pH = 14.00  7.00 = 7.00
d. pOH = log (7.3 × 10 4 ) = 3.14; pH = 14.00  3.14 = 10.86
Note that pH is greater than 14.00 when [OH] is greater than 1.0 M (an extremely basic
solution).
42.
a. [H+] = 10  pH , [H+] = 10 7.40 = 4.0 × 10 8 M
pOH = 14.00  pH = 14.00  7.40 = 6.60; [OH] = 10  pOH = 10-6.60 = 2.5 × 10 7 M
or [OH] =
K w 1.0  10 14

= 2.5 × 10 7 M; This solution is basic since pH > 7.00.

8
[H ] 4.0  10
b. [H+] = 10 15.3 = 5 × 10 16 M; pOH = 14.00  15.3 = 1.3; [OH] = 10  ( 1.3) = 20 M; basic
c. [H+] = 10  ( 1.0) = 10 M; pOH = 14.0  (1.0) = 15.0; [OH] = 10 15.0 = 1 × 10 15 M; acidic
d. [H+] = 10 3.20 = 6.3 × 10 4 M; pOH = 14.00  3.20 = 10.80; [OH] = 10 10.80 = 1.6 × 10 11 M;
acidic
498
CHAPTER 14
ACIDS AND BASES
e. [OH] = 10 5.0 = 1 × 10 5 M; pH = 14.0  pOH = 14.0  5.0 = 9.0; [H+] = 10 9.0
= 1 × 10 9 M; basic
f.
43.
[OH] = 10 9.60 = 2.5 × 10 5 M; pH = 14.00  9.60 = 4.40; [H+] = 10 4.40 = 4.0 × 10 5 M;
acidic
a. pOH = 14.00  6.88 = 7.12; [H+] = 10 6.88 = 1.3 × 10 7 M
[OH] = 10 7.12 = 7.6 × 10 8 M; acidic
b. [H+] =
1.0  10 14
= 0.12 M; pH = log(0.12) = 0.92
8.4  10 14
pOH = 14.00 – 0.92 = 13.08; acidic
c. pH = 14.00 – 3.11 = 10.89; [H+] = 10 10.89 = 1.3 × 10 11 M
[OH] = 10 3.11 = 7.8 × 10 4 M; basic
d. pH = log (1.0 × 10 7 ) = 7.00; pOH = 14.00 – 7.00 = 7.00
[OH] = 10 7.00 = 1.0 × 10 7 M; neutral
44.
a. pOH = 14.00 – 9.63 = 4.37; [H+] = 10 9.63 = 2.3 × 10 10 M
[OH] = 10 4.37 = 4.3 × 10 5 M; basic
b. [H+] =
1.0  10 14
= 2.6 × 10 9 M; pH = log (2.6 × 10 9 ) = 8.59
6
3.9  10
pOH = 14.00 – 8.59 = 5.41; basic
c. pH = log (0.027) = 1.57; pOH = 14.00 – 1.57 = 12.43
[OH] = 10 12.43 = 3.7 × 10 13 M; acidic
d. pH = 14.0 – 12.2 = 1.8; [H+] = 10 1.8 = 2 × 10 2 M
[OH] = 10 12.2 = 6 × 10 13 M; acidic
45.
pOH = 14.0  pH = 14.0  2.1 = 11.9; [H+] = 10  pH = 10 2.1 = 8 × 10 3 M (1 sig fig)
[OH] =
K w 1.0  10 14

= 1 × 10 12 M or [OH] = 10  pOH = 10 11.9 = 1 × 10 12 M

3
[H ]
8  10
The sample of gastric juice is acidic since the pH is less than 7.00 at 25°C.
CHAPTER 14
46.
ACIDS AND BASES
499
pH = 14.00  pOH = 14.00  5.74 = 8.26; [H+] = 10  pH = 10 8.26 = 5.5 × 10 9 M
[OH] =
K w 1.0  10 14

= 1.8 × 10 6 M or [OH] = 10  pOH = 10 5.74 = 1.8 × 10 6 M

9
[H ] 5.5  10
The solution of baking soda is basic since the pH is greater than 7.00 at 25°C.
Solutions of Acids
47.
All the acids in this problem are strong acids that are always assumed to completely dissociate in
water. The general dissociation reaction for a strong acid is: HA(aq) → H+(aq) + A(aq) where
A is the conjugate base of the strong acid HA. For 0.250 M solutions of these strong acids,
0.250 M H+ and 0.250 M A are present when the acids completely dissociate. The amount of H+
donated from water will be insignificant in this problem since H2O is a very weak acid.
a. Major species present after dissociation = H+, ClO4 and H2O;
pH = log [H+] = -log (0.250) = 0.602
b. Major species = H+, NO3 and H2O; pH = 0.602
48.
Strong acids are assumed to completely dissociate in water: HCl(aq) → H+(aq) + Cl (aq)
a. A 0.10 M HCl solution gives 0.10 M H+ and 0.10 M Cl since HCl completely dissociates.
The amount of H+ from H2O will be insignificant. pH = log [H+] = log (0.10) = 1.00
b. 5.0 M H+ is produced when 5.0 M HCl completely dissociates. The amount of H+ from H2O
will be insignificant. pH = log (5.0) = -0.70 (Negative pH values just indicate very
concentrated acid solutions).
c. 1.0 × 10 11 M H+ is produced when 1.0 × 10 11 M HCl completely dissociates. If you take the
negative log of 1.0 × 10 11 this gives pH = 11.00. This is impossible! We dissolved an acid
in water and got a basic pH. What we must consider in this problem is that water by itself
donates 1.0 × 10 7 M H+. We can normally ignore the small amount of H+ from H2O except
when we have a very dilute solution of an acid (as is the case here). Therefore, the pH is that
of neutral water (pH = 7.00) since the amount of HCl present is insignificant.
49.
Both are strong acids.
0.0500 L × 0.050 mol/L = 2.5 × 10 3 mol HCl = 2.5 × 10 3 mol H+ + 2.5 × 10 3 mol Cl
0.1500 L × 0.10 mol/L = 1.5 × 10 2 mol HNO3 = 1.5 × 10 2 mol H+ + 1.5 × 10 2 mol NO3
[H+] =
(2.5  10 3  1.5  10 2 ) mol
Kw
= 0.088 M; [OH] =
= 1.1 × 10 13 M
0.2000 L
[H  ]
500
CHAPTER 14
[Cl] =
50.
2.5  10 3 mol
1.5  10 2 mol
= 0.013 M; [NO3] =
= 0.075 M
0.2000 L
0.2000 L
90.0 × 10 3 L ×
5.00 mol
= 0.450 mol H+ from HCl
L
30.0 × 10 3 L ×
8.00 mol
= 0.240 mol H+ from HNO3
L
[H+] =
ACIDS AND BASES
0.450 mol  0.240 mol
= 0.690 M; pH = log (0.690) = 0.161
1.00 L
pOH = 14.000  0.161 = 13.839; [OH] = 10 13.839 = 1.45 × 10 14 M
51.
[H+] = 10 1.50 = 3.16 × 10 2 M (carryiung one extra sig fig); M1V1 = M2V2
V1 =
M 2 V2
3.16  10 2 mol / L  1.6 L

= 4.2 × 10 3 L
M1
12 mol / L
To 4.2 mL of 12 M HCl, add enough water to make 1600 mL of solution. The resulting solution
will have [H+] = 3.2 × 10 2 M and pH = 1.50.
52.
[H+] = 10 5.10 = 7.9 × 10 6 M; HNO3(aq) → H+(aq) + NO3(aq)
Because HNO3 is a strong acid, we have a 7.9 × 10 6 M HNO3 solution.
0.2500 L ×
53.
7.9  10 6 mol HNO3
63.02 g HNO3

= 1.2 × 10 4 g HNO3
L
mol HNO3
a. HNO2 (Ka = 4.0 × 10 4 ) and H2O (Ka = Kw = 1.0 × 10 14 ) are the major species. HNO2 is a
much stronger acid than H2O so it is the major source of H+. However, HNO2 is a weak acid
(Ka < 1) so it only partially dissociates in water. We must solve an equilibrium problem to
determine [H+]. In the Solutions Guide, we will summarize the initial, change and equilibrium
concentrations into one table called the ICE table. Solving the weak acid problem:
HNO2
Initial
Change
Equil.
⇌
H+
+
NO2
0.250 M
~0
0
x mol/L HNO2 dissociates to reach equilibrium
x
→
+x
+x
0.250 x
x
x

Ka =
[H  ][ NO 2 ]
x2
= 4.0 × 10 4 =
; If we assume x << 0.250, then:
0.250  x
[HNO2 ]
4.0 × 10 4 ≈
x2
,
0.250
x 
4.0  10  4 (0.250 ) = 0.010 M
CHAPTER 14
ACIDS AND BASES
We must check the assumption:
501
x
0.010
× 100 = 4.0%
 100 
0.250
0.250
All the assumptions are good. The H+ contribution from water ( 10 7 M) is negligible, and x is
small compared to 0.250 (percent error = 4.0%). If the percent error is less than 5% for an
assumption, we will consider it a valid assumption (called the 5% rule). Finishing the
problem: x = 0.010 M = [H+]; pH = log(0.010) = 2.00
b. CH3CO2H (Ka = 1.8 × 10 5 ) and H2O (Ka = Kw = 1.0 × 10 14 ) are the major species.
CH3CO2H is the major source of H+. Solving the weak acid problem:
CH3CO2H
Initial
Change
Equil.
⇌
H+
CH3CO2
+
0.250 M
~0
0
x mol/L CH3CO2H dissociates to reach equilibrium
x
→
+x
+x
0.250  x
x
x

Ka =
[H  ][CH 3CO 2 ]
x2
x2
= 1.8 × 10 5 =

(assuming x << 0.250)
[CH 3CO 2 H]
0.250  x 0.250
x = 2.1 × 10-3 M; Checking assumption:
2.1  10 3
× 100= 0.84%. Assumptions good.
0.250
[H+] = x = 2.1 × 10 3 M; pH = -log (2.1 × 10 3 ) = 2.68
54.
a. HOC6H5 (Ka = 1.6 × 10 10 ) and H2O (Ka = Kw = 1.0 × 10 14 ) are the major species. The major
equilibrium is the dissociation of HOC6H5. Solving the weak acid problem:
HOC6H5
Initial
Change
Equil.
⇌
H+
OC6H5
+
0.250 M
~0
0
x mol/L HOC6H5 dissociates to reach equilibrium
x
→
+x
+x
0.250  x
x
x
Ka = 1.6 × 10 10 =

[ H  ][OC6 H 5 ]
x2
x2
=

(assuming x << 0.250)
[ HOC6 H 5 ]
0.250  x 0.250
x = [H+] = 6.3 × 10 6 M; Checking assumption: x is 2.5 × 10 3 % of 0.250, so assumption is
valid by the 5% rule.
pH = log(6.3 × 10 6 ) = 5.20
b. HCN (Ka = 6.2 × 10 10 ) and H2O are the major species. HCN is the major source of H+.
HCN
Initial
0.250 M
⇌
H+
~0
+
CN
0
502
CHAPTER 14
ACIDS AND BASES
x mol/L HCN dissociates to reach equilibrium
x
→
+x
+x
0.250  x
x
x


2
x2
x
[H ][CN ]
Ka = 6.2 × 10 10 =
=

(assuming x << 0.250)
0.250  x 0.250
[HCN]
Change
Equil.
x = [H+] = 1.2 × 10 5 M; Checking assumption: x is 4.8 × 10-3% of 0.250
Assumptions good. pH = log (1.2 × 10 5 ) = 4.92
1 mol HC2 H 3O 2
60.05 g
0.05000 L
0.0560 g HC2 H 3O 2 
55.
[CH3COOH]0 = [HC2H3O2]0 =
HC2H3O2
Initial
Change
Equil.
⇌
H+
+
C2H3O2
Ka = 1.8 × 10 5
0.0187 M
~0
0
x mol/L HC2H3O2 dissociates to reach equilibrium
x
→
+x
+x
0.0187  x
x
x
Ka = 1.8 × 10 5 =

[H  ][C 2 H 3O 2 ]
x2
x2
=

[HC3 H 3O 2 ]
0.0187  x 0.0187
x = [H+] = 5.8 × 10 4 M; pH = 3.24
Assumptions good (x is 3.1% of 0.0187).
[H+] = [C2H3O2] = [CH3COO] = 5.8 × 10 4 M;
56.
= 1.87 × 10 2 M
[CH3COOH] = 0.0187  5.8 × 10 4
= 0.0181 M
5
14
HC3H5O2 (Ka = 1.3 × 10 ) and H2O (Ka = Kw = 1.0 × 10 ) are the major species present.
HC3H5O2 will be the dominant producer of H+ since HC3H5O2 is a stronger acid than H2O.
Solving the weak acid problem:
HC3H5O2
Initial
Change
Equil.
⇌
H+
+
C3H5O2-
0.100 M
~0
0
x mol/L HC3H5O2 dissociates to reach equilibrium
x
→ +x
+x
0.100  x
x
x
Ka = 1.3 × 10 5 =

[H  ][C3 H 5 O 2 ]
=
[HC3 H 5 O 2 ]
x2
x2

0.100  x 0.100
x = [H+] = 1.1 × 10 3 M; pH = log (1.1 × 10 3 ) = 2.96
Assumption follows the 5% rule (x is 1.1% of 0.100).
CHAPTER 14
ACIDS AND BASES
503
[H+] = [C3H5O2] = 1.1 × 10 3 M; [OH] = Kw/[H+] = 9.1 × 10 12 M
[HC3H5O2] = 0.100  1.1 × 10 3 = 0.099 M
1.1  10 3
[H  ]
Percent dissociation =
× 100 =
= 1.1%
0.100
[HC3 H 5O 2 ] 0
57.
This is a weak acid in water. Solving the weak acid problem:
HF
Initial
Change
Equil.
⇌
H+
F
+
Ka = 7.2 × 10 4
0.020 M
~0
0
x mol/L HF dissociates to reach equilibrium
x
→
+x
+x
0.020  x
x
x
Ka = 7.2 × 10 4 =
x2
x2
[H  ][F ]
=

(assuming x << 0.020)
0.020  x 0.020
[HF]
x = [H+] = 3.8 × 10 3 M; Check assumptions:
x
3.8  10 3
 100 
= 19%
0.020
0.020
The assumption x << 0.020 is not good (x is more than 5% of 0.020). We must solve
x2/(0.020  x) = 7.2 × 10 4 exactly by using either the quadratic formula or by the method of
successive approximations (see Appendix 1.4 of text). Using successive approximations, we
let 0.016 M be a new approximation for [HF]. That is, in the denominator, try x = 0.0038 (the
value of x we calculated making the normal assumption), so 0.020 - 0.0038 = 0.016, then
solve for a new value of x in the numerator.
x2
x2

= 7.2 × 10 4 , x = 3.4 × 10 3
0.020  x 0.016
We use this new value of x to further refine our estimate of [HF], i.e., 0.020  x =
0.020  0.0034 = 0.0166 (carry extra significant figure).
x2
x2

= 7.2 × 10 4 , x = 3.5 × 10 3
0.020  x 0.0166
We repeat until we get an answer that repeats itself. This would be the same answer we
would get solving exactly using the quadratic equation. In this case it is: x = 3.5 × 10 3
So: [H+] = [F] = x = 3.5 × 10 3 M; [OH] = Kw/[H+] = 2.9 × 10 12 M
[HF] = 0.020  x = 0.020 - 0.0035 = 0.017 M; pH = 2.46
Note: When the 5% assumption fails, use whichever method you are most comfortable with
to solve exactly. The method of successive approximations is probably fastest when the
504
CHAPTER 14
ACIDS AND BASES
percent error is less than ~25% (unless you have a calculator that can solve quadratic
equations).
58.
Major species: HIO3, H2O; Major source of H+: HIO3 (a weak acid, Ka = 0.17)
HIO3
Initial
⇌
H+
+
IO3
0.20 M
~0
0
x mol/L HIO3 dissociates to reach equilibrium
x
→
+x
+x
0.20  x
x
x
Change
Equil.
Ka = 0.17 =
x2
x2

, x = 0.18; Check assumption.
0.20  x 0.20
Assumption is horrible (x is 90% of 0.20). When the assumption is this poor, it is generally
quickest to solve exactly using the quadratic formula (see Appendix 1.4 in text). The method of
successive approximations will require many trials to finally converge on the answer. For this
problem, 5 trials were required. Using the quadratic formula and carrying extra significant
figures:
x2
0.17 =
, x2 = 0.17(0.20  x), x2 + 0.17 x  0.034 = 0
0.020  x
x=
 0.17  [(0.17 ) 2  4(1)(0.034 )]1 / 2  0.17  0.1406

, x = 0.12 or 0.29
2(1)
2
Only x = 0.12 makes sense. x = 0.12 M = [H+]; pH = log (0.12) = 0.92
59.
Major species: HC2H2ClO2 (Ka = 1.35 × 10 3 ) and H2O; Major source of H+: HC2H2ClO2
HC2H2ClO2
Initial
Change
Equil.
⇌
H+
+ C2H2ClO2
0.10 M
~0
0
x mol/L HC2H2ClO2 dissociates to reach equilibrium
x
→ +x
+x
0.10 x
x
x
Ka = 1.35 × 10 3 =
x2
x2

, x = 1.2 × 10 2 M
0.10  x 0.10
Checking the assumptions finds that x is 12% of 0.10 which fails the 5% rule. We must solve
1.35 × 10 3 = x2/(0.10  x) exactly using either the method of successive approximations or
the quadratic equation. Using either method gives x = [H+] = 1.1 × 10 2 M. pH = log [H+] =
log (1.1 × 10 2 ) = 1.96.
CHAPTER 14
ACIDS AND BASES
2 tablets 
60.
[HC9H7O4] =
Initial
Change
Equil.
0.325 g HC9 H 7 O 4 1 mol HC9 H 7 O 4

tablet
180 .15 g
= 0.0152 M
0.237 L
⇌
HC9H7O4
505
+ C9H7O4
H+
0.0152 M
~0
0
x mol/L HC9H7O4 dissociates to reach equilibrium
x
→ +x
+x
0.0152  x
x
x
Ka = 3.3 × 10 4 =

[ H  ] [C 9 H 7 O 4 ]
x2
x2
=
≈
, x = 2.2 × 10 3 M
0.0152
[ HC9 H 7 O 4 ]
0.0152  x
Assumption that 0.0152 – x  0.0152 fails the 5% rule:
2.2  10 3
× 100 = 14%
0.0152
Using successive approximations or the quadratic equation gives an exact answer of
x = 2.1 × 10 3 M.
[H+] = x = 2.1 × 10 3 M; pH = log (2.1 × 10 3 ) = 2.68
61.
a. HCl is a strong acid. It will produce 0.10 M H+. HOCl is a weak acid. Let's consider the
equilibrium:
HOCl
Initial
Change
Equil.
⇌
H+
+
OCl
Ka = 3.5 × 10 8
0.10 M
0.10 M
0
x mol/L HOCl dissociates to reach equilibrium
x
→ +x
+x
0.10  x
0.10 + x
x
Ka = 3.5 × 10 8 =
[H  ][OCl ]
(0.10  x )( x)
=
 x, x = 3.5 × 10 8 M
[HOCl]
0.10  x
Assumptions are great (x is 3.5 × 10-5% of 0.10). We are really assuming that HCl is the only
important source of H+, which it is. The [H+] contribution from HOCl, x, is negligible. Therefore, [H+] = 0.10 M; pH = 1.00
b. HNO3 is a strong acid, giving an initial concentration of H+ equal to 0.050 M. Consider the
equilibrium:
HC2H3O2
Initial
Change
Equil.
⇌
H+
+ C2H3O2
Ka = 1.8 × 10 5
0.50 M
0.050 M
0
x mol/L HC2H3O2 dissociates to reach equilibrium
x
→
+x
+x
0.50  x
0.050 + x
x
506
CHAPTER 14
ACIDS AND BASES

[H  ][C 2 H 3O 2 ]
(0.050  x ) x 0.050 x
=

[HC2 H 3O 2 ]
0.50
0.50  x
Ka = 1.8 × 10 5 =
x = 1.8 × 10 4 ; Assumptions are good (well within the 5% rule).
62.
[H+] = 0.050 + x = 0.050 M and pH = 1.30
HF and HOC6H5 are both weak acids with Ka values of 7.2 × 10 4 and 1.6 × 10 10 , respectively.
Since the Ka value for HF is much greater than the Ka value for HOC6H5, HF will be the dominant
producer of H+ (we can ignore the amount of H+ produced from HOC6H5 since it will be insignificant).
⇌
HF
Initial
Change
Equil.
H+
+
F
1.0 M
~0
0
x mol/L HF dissociates to reach equilibrium
x
→
+x
+x
1.0  x
x
x
Ka = 7.2 × 10 4 =
[H  ][F ]
x2
x2
=
≈
1 .0  x 1 . 0
[HF]
x = [H+] = 2.7 × 10 2 M; pH = log (2.7 × 10 2 ) = 1.57
Assumptions good.
Solving for [OC6H5] using HOC6H5 ⇌ H+ + OC6H5- equilibrium:
Ka = 1.6 × 10
10


[ H  ][OC6 H 5 ] (2.7  10 2 )[OC6 H 5 ]
=
=
, [OC6H5] = 5.9 × 10 9 M
[ HOC6 H 5 ]
1.0
Note that this answer indicates that only 5.9 × 10 9 M HOC6H5 dissociates, which indicates
that HF is truly the only significant producer of H+ in this solution.
63.
In all parts of this problem, acetic acid (HC2H3O2) is the best weak acid present. We must solve a
weak acid problem.
a.
HC2H3O2
Initial
Change
Equil.
⇌
H+
+ C2H3O2
0.50 M
~0
0
x mol/L HC2H3O2 dissociates to reach equilibrium
x
→
+x
+x
0.50  x
x
x
Ka = 1.8 × 10 5 =

[H  ][C 2 H 3O 2 ]
x2
x2
=
≈
[HC2 H 3O 2 ]
0.50  x 0.50
x = [H+] = [C2H3O2-] = 3.0 × 10 3 M
Assumptions good.
CHAPTER 14
ACIDS AND BASES
Percent dissociation =
507
3.0  10 3
[H  ]
× 100 =
× 100 = 0.60%
0.50
[ HC2 H 3O 2 ]0
b. The setups for solutions b and c are similar to solution a except the final equation is slightly
different, reflecting the new concentration of HC2H3O2.
x2
x2
≈
0.050  x 0.050
Ka = 1.8 × 10 5 =
x = [H+] = [C2H3O2] = 9.5 × 10 4 M
% dissociation =
c. Ka = 1.8 × 10 5 =
Assumptions good.
9.5  10 4
× 100 = 1.9%
0.050
x2
x2
≈
0.0050  x 0.0050
x = [H+] = [C2H3O2] = 3.0 × 10 4 M; Check assumptions.
Assumption that x is negligible is borderline (6.0% error). We should solve exactly. Using the
method of successive approximations (see Appendix 1.4 of text):
x2
x2
1.8 × 10 5 =
=
, x = 2.9 × 10 4
0.0047
0.0050  3.0  10  4
Next trial also gives x = 2.9 × 10 4 .
% dissociation =
2.9  10 4
× 100 = 5.8%
5.0  10 3
d. As we dilute a solution, all concentrations decrease. Dilution will shift the equilibrium to the
side with the greater number of particles. For example, suppose we double the volume of an
equilibrium mixture of a weak acid by adding water, then:
 [H  ] eq   [X  ] eq


 2  2

Q= 
 [HX] eq 


 2 





= 1K
a
2
Q < Ka, so the equilibrium shifts to the right or towards a greater percent dissociation.
e. [H+] depends on the initial concentration of weak acid and on how much weak acid dissociates. For solutions a-c the initial concentration of acid decreases more rapidly than the percent
dissociation increases. Thus, [H+] decreases.
64.
a. HNO3 is a strong acid; it is assumed 100% dissociated in solution.
b.
HNO2
⇌
H+
+
NO2
Ka = 4.0 × 10 4
508
CHAPTER 14
Initial
Change
Equil.
ACIDS AND BASES
0.20 M
~0
0
x mol/L HNO2 dissociates to reach equilibrium
x
→
+x
+x
0.20  x
x
x

[ H  ][ NO2 ]
x2
x2
=
≈
[HNO2 ]
0.20  x 0.20
Ka = 4.0 × 10 4 =
x = [H+] = [NO2] = 8.9 × 10 3 M; Assumptions good.
% dissociation =
c.
8.9  10 3
[H  ]
× 100 =
× 100 = 4.5%
0.20
[HNO2 ] 0
HOC6H5
Initial
Change
Equil.
⇌
H+
+
OC6H5
Ka = 1.6 × 10 10
0.20 M
~0
0
x mol/L HOC6H5 dissociates to reach equilibrium
x
→
+x
+x
0.20  x
x
x
Ka = 1.6 × 10 10 =

[ H  ][OC6 H 5 ]
x2
x2
=
≈
[ HOC6 H 5 ]
0.20  x 0.20
x = [H+] = [OC6H5] = 5.7 × 10 6 M; Assumptions good.
% dissociation =
5.7  10 6
× 100 = 2.9 × 10 3 %
0.20
d. For the same initial concentration, the percent dissociation increases as the strength of the
acid increases (as Ka increases).
65.
Let HX symbolize the weak acid. Setup the problem like a typical weak acid equilibrium
problem.
HX
Initial
Change
Equil.
⇌
H+
+
X
0.15 M
~0
0
x mol/L HX dissociates to reach equilibrium
x →
+x
+x
0.15  x
x
x
If the acid is 3.0% dissociated, then x = [H+] is 3.0% of 0.15: x = 0.030 × (0.15 M) =
4.5 × 10 3 M. Now that we know the value of x, we can solve for Ka.
CHAPTER 14
Ka =
ACIDS AND BASES
509
x2
(4.5  10 3 ) 2
[H  ][X  ]
=
=
= 1.4 × 10 4
0.15  x
[HX]
0.15  4.5  10 3
66.
⇌
HX
Initial
Change
Equil.
H+
X
+
I
~0
0
x mol/L HX dissociates to reach equilibrium
x
→
+x
+x
Ix
x
x
where I = [HX]o
From the problem, x = 0.25(I) and I  x = 0.30 M.
I  0.25(I) = 0.30 M, I = 0.40 M and x = 0.25 (0.40 M) = 0.10 M
Ka =
67.
[H  ][X  ]
x2
(0.10 ) 2
=
=
= 0.033
Ix
0.30
[HX]
Setup the problem using the Ka equilibrium reaction for HOCN.
⇌
HOCN
Initial
Change
Equil.
68.
H+
OCN
+
0.0100 M
~0
0
x mol/L HOCN dissociates to reach equilibrium
x
→
+x
+x
0.0100  x
x
x
Ka =
x2
[H  ][OCN ]
=
; pH = 2.77: x = [H+] = 10  pH = 10 2.77 = 1.7 × 10 3 M
0
.
0100

x
[HOCN]
Ka =
(1.7  10 3 ) 2
= 3.5 × 10 4
3
0.0100  1.7  10
HClO4 is a strong acid with [H+] = 0.040 M. This equals the [H+] in the trichloroacetic acid
solution. Now setup the problem using the Ka equilibrium reaction for CCl3CO2H.
CCl3CO2H
Initial
Equil.
0.050 M
0.050  x

Ka =
⇌
H+
~0
x
+
CCl3CO2
0
x
[H  ][CCl3CO 2 ]
x2
=
; x = [H+] = 4.0 × 10 2 M
0.050  x
[CCl3CO 2 H]
510
CHAPTER 14
Ka =
69.
ACIDS AND BASES
(4.0  10 2 ) 2
= 0.16
0.050  (4.0  10  2 )
Major species: HCOOH and H2O; Major source of H+: HCOOH
HCOOH
Initial
⇌
H+
+ HCOO
C
~0
0
where C = [HCOOH]o
x mol/L HCOOH dissociates to reach equilibrium
x
→
+x
+x
Cx
x
x
Change
Equil.
Ka = 1.8 × 10 4 =
1.8 × 10 4 =
1.8 × 10 4 =
x2
[H  ][HCOO ]
=
where x = [H+]
Cx
[HCOOH]
[ H  ]2
; pH = 2.70, so: [H+] = 10 2.70 = 2.0 × 10 3 M
C  [H  ]
(2.0  10 3 ) 2
C  (2.0  10 3 )
, C  (2.0 × 10 3 ) =
4.0  10 6
, C = 2.4 × 10 2 M
1.8  10  4
A 0.024 M formic acid solution will have pH = 2.70.
70.
[HA]o =
1.0 mol
= 0.50 mol/L; Solve using the Ka equilibrium reaction.
2.0 L
HA
Initial
Equil.
Ka =
0.50 M
0.50  x
⇌
H+
A
+
~0
x
0
x
[H  ][A  ]
x2
=
; In this problem, [HA] = 0.45 M so:
0.50  x
[HA]
[HA] = 0.45 M = 0.50 M  x, x = 0.05 M; Ka =
(0.05) 2
= 6 × 10 3
0.45
Solutions of Bases
71.

a. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH(aq)
Kb =
[ NH 4 ][OH  ]
[ NH 3 ]
b. C5H5N(aq) + H2O(l) ⇌C5H5NH+(aq) + OH(aq)
Kb =
[C5 H 5 NH  ][OH  ]
[C 5 H 5 N ]
CHAPTER 14
72.
ACIDS AND BASES
511

a. C6H5NH2(aq) + H2O(l) ⇌C6H5NH3+(aq) + OH(aq)
Kb =

[(CH 3 ) 2 NH 2 ][OH  ]
[(CH 3 ) 2 NH]

NO3 : Kb << Kw since HNO3 is a strong acid. All conjugate bases of strong acids have no base
strength. H2O: Kb = Kw = 1.0 × 10 14 ; NH3: Kb = 1.8 × 10 5 ; C5H5N: Kb = 1.7 × 10 9
b. (CH3)2NH(aq) + H2O(l) ⇌ (CH3)2NH2+(aq) + OH(aq)
73.
[C 6 H 5 NH3 ][OH  ]
[C 6 H 5 NH 2 ]
Kb =
NH3 > C5H5N > H2O > NO3- (As Kb increases, base strength increases.)
74.
Excluding water, these are the conjugate acids of the bases in the previous exercise. In general,
the stronger the base, the weaker the conjugate acid. Note: Even though NH4+ and C5H5NH+ are
conjugate acids of weak bases, they are still weak acids with Ka values between Kw and 1. Prove
this to yourself by calculating the Ka values for NH4+ and C5H5NH+ (Ka = Kw/Kb).
HNO3 > C5H5NH+ > NH4+ > H2O
75.
a. C6H5NH2
b. C6H5NH2
c. OH
d. CH3NH2
The base with the largest Kb value is the strongest base ( K b, C6 H5 NH2 = 3.8 × 10 10 ,
K b, CH 3 NH2 = 4.4 10 4 . OH is the strongest base possible in water.
76.
a. HClO4
b. C6H5NH3+
c. C6H5NH3+
The acid with the largest Ka value is the strongest acid. To calculate Ka values for C6H5NH3+ and
CH3NH3+, use Ka = Kw/Kb where Kb refers to the bases C6H5NH2 or CH3NH2.
77.
NaOH(aq) → Na+(aq) + OH (aq); NaOH is a strong base which completely dissociates into Na+
and OH. The initial concentration of NaOH will equal the concentration of OH donated by
NaOH.
a. [OH] = 0.10 M; pOH = log[OH] = -log(0.10) = 1.00
pH = 14.00  pOH = 14.00  1.00 = 13.00
Note that H2O is also present, but the amount of OH produced by H2O will be insignificant
compared to the 0.10 M OH produced from the NaOH.
b. The [OH] concentration donated by the NaOH is 1.0 × 10 10 M. Water by itself donates
1.0 × 10 7 M. In this problem, water is the major OH-contributor and [OH] = 1.0 × 10 7 M.
pOH = -log (1.0 × 10 7 ) = 7.00; pH = 14.00 - 7.00 = 7.00
c. [OH] = 2.0 M; pOH = -log (2.0) = -0.30; pH = 14.00  (0.30) = 14.30
78.
a. Ca(OH)2 → Ca2+ + 2 OH ; Ca(OH)2 is a strong base and dissociates completely.
512
CHAPTER 14
ACIDS AND BASES
[OH] = 2(0.00040) = 8.0 × 10 4 M; pOH = log [OH] = 3.10; pH = 14.00  pOH = 10.90
b.
25 g KOH 1 mol KOH
= 0.45 mol KOH/L

L
56.11 g KOH
KOH is a strong base, so [OH] = 0.45 M; pOH = -log (0.45) = 0.35; pH = 13.65
c.
150.0 g NaOH
1 mol
= 3.750 M; NaOH is a strong base, so [OH] = 3.750 M.

L
40.00 g
pOH = log (3.750) = 0.5740 and pH = 14.0000  (0.5740) = 14.5740
Although we are justified in calculating the answer to four decimal places, in reality pH
values are generally measured to two decimal places, sometimes three.
79.
a. Major species: K+, OH, H2O (KOH is a strong base.)
[OH] = 0.015 M, pOH = log (0.015) = 1.82; pH = 14.00  pOH = 12.18
b. Major species: Ba2+, OH, H2O; Ba(OH)2(aq) → Ba2+(aq) + 2 OH (aq); Since each mol of
the strong base Ba(OH)2 dissolves in water to produce two mol OH, then [OH] =
2(0.015 M) = 0.030 M.
pOH = log (0.030) = 1.52; pH =14.00  1.52 = 12.48
80.
a. Major species: Na+, Li+, OH, H2O (NaOH and LiOH are both strong bases.)
[OH] = 0.050 + 0.050 = 0.100 M; pOH = 1.000; pH = 13.000
b. Major species: Ca2+, Rb+, OH, H2O; Both Ca(OH)2 and RbOH are strong bases and
Ca(OH)2 donates 2 mol OH per mol Ca(OH)2.
[OH] = 2(0.0010) + 0.020 = 0.022 M; pOH = log (0.022) = 1.66; pH = 12.34
81.
pOH = 14.00 – 11.56 = 2.44; [OH] = [KOH] = 10 2.44 = 3.6 × 10 3 M
0.8000 L 
82.
3.6  10 3 mol KOH 56.11 g KOH

= 0.16 g KOH
L
mol KOH
pH = 10.50; pOH = 14.00 - 10.50 = 3.50; [OH-] = 10-3.50 = 3.2 × 10-4 M
Sr(OH)2(aq) → Sr2+(aq) + 2 OH(aq); Sr(OH)2 donates two mol OH per mol Sr(OH)2.
[Sr(OH)2] =
3.2  10 4 mol OH 
×
L
 1 mol Sr (OH) 2 
4


 2 mol OH   = 1.6 × 10 M Sr(OH)2


CHAPTER 14
ACIDS AND BASES
513
A 1.6 × 10-4 M Sr(OH)2 solution will produce a pH = 10.50 solution.
83.
NH3 is a weak base with Kb = 1.8 × 10 5 . The major species present will be NH3 and H2O (Kb =
Kw = 1.0 × 10 14 ). Since NH3 has a much larger Kb value compared to H2O, NH3 is the stronger
base present and will be the major producer of OH. To determine the amount of OH produced
from NH3, we must perform an equilibrium calculation.
NH3(aq) + H2O(l)
⇌
NH4+(aq)
OH(aq)
+
Initial
0.150 M
0
~0
x mol/L NH3 reacts with H2O to reach equilibrium
Change
x
→
+x
+x
Equil.
0.150  x
x
x


2
2
[ NH 4 ][OH ]
x
x
Kb = 1.8 × 10 5 =
=
≈
(assuming x << 0.150)
[ NH 3 ]
0.150  x 0.150
x = [OH] = 1.6 × 10 3 M; Check assumptions: x is 1.1% of 0.150 so the assumption 0.150  x ≈
0.150 is valid by the 5% rule. Also, the contribution of OH from water will be insignificant
(which will usually be the case). Finishing the problem, pOH = log [OH] = log (1.6 × 10 3 M)
= 2.80; pH = 14.00  pOH = 14.00  2.80 = 11.20.
84.
Major species: H2NNH2 (Kb = 3.0 × 10 6 ) and H2O (Kb = Kw = 1.0 × 10 14 ); The weak base
H2NNH2 will dominate OH production. We must perform a weak base equilibrium calculation.
H2NNH2 + H2O
Initial
Change
Equil.
⇌
H2NNH3+
+
OH
Kb = 3.0 × 10 6
2.0 M
0
~0
x mol/L H2NNH2 reacts with H2O to reach equilibrium
x
→
+x
+x
2.0  x
x
x
Kb = 3.0 × 10 6 =

[H 2 NNH3 ][OH  ]
x2
x2
=
≈
(assuming x << 2.0)
[H 2 NNH2 ]
2 .0  x 2 . 0
x = [OH] = 2.4 × 10 3 M; pOH = 2.62; pH = 11.38
Assumptions good (x is 0.12% of 2.0).
[H2NNH3+] = 2.4 × 10 3 M; [H2NNH2] = 2.0 M; [H+] = 10 11.38 = 4.2 × 10 12 M
85.
These are solutions of weak bases in water. We must solve the equilibrium weak base problem.
a.
(C2H5)3N + H2O
Initial
Change
Equil.
⇌
(C2H5)3NH+ +
OH
Kb = 4.0 × 10 4
0.20 M
0
~0
x mol/L of (C2H5)3N reacts with H2O to reach equilibrium
x
→
+x
+x
0.20  x
x
x
514
CHAPTER 14
Kb = 4.0 × 10 4 =
ACIDS AND BASES
[(C 2 H 5 ) 3 NH  ][OH  ]
x2
x2
=
≈
, x = [OH] = 8.9 × 10 3 M
[(C 2 H 5 ) 3 N]
0.20
0.20  x
Assumptions good (x is 4.5% of 0.20). [OH] = 8.9 × 10 3 M
[H+] =
1.0  10 14
Kw
=
= 1.1 × 10 12 M; pH = 11.96
8.9  10 3
[OH ]
b.
HONH2 + H2O
Initial
Equil.
⇌
HONH3+
0.20 M
0.20  x
Kb = 1.1 × 10 8 =
+
0
Kb = 1.1 × 10 8
OH
~0
x
x
x2
x2
≈
, x = [OH] = 4.7 × 10 5 M; Assumptions good.
0.20
0.20  x
[H+] = 2.1 × 10 10 M; pH = 9.68
86.
These are solutions of weak bases in water.
a.
C6H5NH2 + H2O
Initial
Change
Equil.
⇌
C6H5NH3+ +
Kb = 3.8 × 10 10
OH
0.20 M
0
~0
x mol/L of C6H5NH2 reacts with H2O to reach equilibrium
x
→
+x
+x
0.20  x
x
x
3.8 × 10 10 =
x2
x2
≈
, x = [OH] = 8.7 × 10 6 M; Assumptions good.
0.20
0.20  x
[H+] = Kw/[OH]= 1.1 × 10 9 M; pH = 8.96
b.
CH3NH2 + H2O
Initial
Equil.
0.20 M
0.20  x
Kb = 4.38 × 10 4 =
⇌
CH3NH3+ +
0
x
Kb = 4.38 × 10 4
OH
~0
x
x2
x2
≈
, x = 9.4 × 10 3 M; Assumptions good.
0.20
0.20  x
[OH] = 9.4 × 10 3 M; [H+] = Kw/[OH] = 1.1 × 10 12 M; pH = 11.96
87.
This is a solution of a weak base in water. We must solve the weak base equilibrium problem.
C2H5NH2
Initial
Change
+ H2O
⇌
C2H5NH3+ +
OH
Kb = 5.6 × 10
0.20 M
0
~0
x mol/L C2H5NH2 reacts with H2O to reach equilibrium
x
→
+x
+x
4
CHAPTER 14
ACIDS AND BASES
Equil.
515
0.20  x
x
x

Kb =
[C 2 H 5 NH3 ][OH  ]
x2
x2
=
≈
(assuming x << 0.20)
0.20
[C 2 H 5 NH 2 ]
0.20  x
x = 1.1 × 10 2 ; Checking assumption:
1.1  10 2
× 100 = 5.5%
0.20
Assumption fails the 5% rule. We must solve exactly using either the quadratic equation or
the method of successive approximations (see Appendix 1.4 of the text). Using successive
approximations and carrying extra significant figures:
x2
x2
4

= 5.6 × 10 , x = 1.0 × 10 2 M (consistent answer)
0.20  0.011
0.189
x = [OH] = 1.0 × 10 2 M; [H+] =
88.
1.0  10 14
Kw
=
= 1.0 × 10 12 M; pH = 12.00
2

1.0  10
[OH ]
⇌
(C2H5)2NH + H2O
Initial
Change
Equil.
(C2H5)2NH2+
+
OH
Kb = 1.3 × 10 3
0.050 M
0
~0
x mol/L (C2H5)2NH reacts with H2O to reach equilibrium
x
→
+x
+x
0.050  x
x
x

[(C 2 H 5 ) 2 NH 2 ][OH  ]
x2
x2
=
≈
0.050
[(C 2 H 5 ) 2 NH]
0.050  x
x = 8.1 × 10 3 ; Assumption is bad (x is 16% of 0.20).
Kb = 1.3 × 10 3 =
Using successive approximations:
x2
, x = 7.4 × 10 3
1.3 × 10 3 =
0.050  0.081
1.3 × 10 3 =
x2
, x = 7.4 × 10 3 (consistent answer)
0.050  0.074
[OH] = x = 7.4 × 10 3 M; [H+] = Kw/[OH] = 1.4 × 10 12 M; pH = 11.85
89.
To solve for percent ionization, just solve the weak base equilibrium problem.
a.
NH3 + H2O
Initial 0.10 M
Equil. 0.10  x
Kb = 1.8 × 10 5 =
⇌
NH4+
+
0
x
OH
Kb = 1.8 × 10 5
~0
x
x2
x2
≈
, x = [OH] = 1.3 × 10 3 M; Assumptions good.
0.10
0.10  x
516
CHAPTER 14
1.3  10 3 M
[OH  ]
× 100 =
× 100 = 1.3%
0.10 M
[ NH3 ]0
Percent ionization =
b.
⇌
NH3 + H2O
Initial
Equil.
NH4+
0.010 M
0.010  x
1.8 × 10 5 =
ACIDS AND BASES
+
0
x
OH
~0
x
x2
x2
, x = [OH] = 4.2 × 10 4 M; Assumptions good.
≈
0.010
0.010  x
4.2  10 4
× 100 = 4.2%
0.010
Percent ionization =
Note: For the same base, the percent ionization increases as the initial concentration of base
decreases.
90.
C5H5N + H2O
⇌
Initial 0.10 M
Equil. 0.10  x
91.
0
~0
x
x
Kb = 1.7 × 10 9 =
x2
x2
≈
, x = [C5H5N] = 1.3 × 10 5 M; Assumptions good.
0.10
0.10  x
%C5H5N reacted =
1.3  10 5 M
× 100 = 1.3 × 10 2 %
0.10 M
Let cod = codeine, C18H21NO3; using the Kb reaction to solve:
cod + H2O
Initial
Change
Equil.
Kb =
⇌
codH+
+
OH
1.7 × 10 3 M
0
~0
x mol/L codeine reacts with H2O to reach equilibrium
x
→
+x
+x
3
1.7 × 10  x
x
x
x2
; pH = 9.59; so: pOH = 14.00  9.59 = 4.41.
1.7  10 3  x
[OH] = x = 10 4.41 = 3.9 × 10 5 M; Kb =
92.
Kb = 1.7 × 10 9
C5H5N+ + OH
HONH2 + H2O
Initial I
Equil. I – x
⇌
(3.9  10 5 ) 2
= 9.2 × 10 7
3
5
1.7  10  3.9  10
HONH3+ + OH
0
x
~0
x
Kb = 1.1 × 10 8
I = [HONH2]0
CHAPTER 14
ACIDS AND BASES
Kb = 1.1 × 10 8 =
517
x2
; From problem: pH = 10.00, so pOH = 4.00, and
Ix
x = [OH] = 1.0 × 10 4 M
1.1 × 10 8 =
(1.0  10 4 ) 2
, I = 0.91 M
I  1.0  10  4
mass HONH2 = 0.2500 L 
0.91 mol HONH2 33.03 g HONH2
= 7.5 g HONH2

L
mol HONH2
Polyprotic Acids
93.
94.
95.
H2SO3(aq) ⇌ HSO3(aq) + H+(aq)
K a1 
HSO3 (aq) ⇌SO32(aq) + H+(aq)
Ka2 
H3C6H5O7(aq) ⇌ H2C6H5O7(aq) + H+(aq)
K a1 
H2C6H5O7(aq) ⇌ HC6H5O72(aq) + H+(aq)
Ka2 
HC6H5O72(aq) ⇌ C6H5O73(aq) + H+(aq)
K a3 

[HSO 3 ][H  ]
[H 2SO 3 ]
[SO32 ][H  ]

[HSO3 ]

[H 2 C 6 H 5 O 7 ][H  ]
[ H 3C 6 H 5 O 7 ]
2
[HC6 H5O7 ][H  ]

[H 2 C6 H 5 O 7 ]
3
[C6 H 5O7 ][H  ]
2
[HC6 H 5O 7 ]
In both these polyprotic acid problems, the dominate equilibrium is the K a1 reaction. The amount
of H+ produced from the subsequent Ka reactions will be minimal since they are all have much
smaller Ka values.
a.
H3PO4
Initial
Change
Equil.
⇌
H+
+
H2PO4
K a1 = 7.5 × 10 3
0.10 M
~0
0
x mol/L H3PO4 dissociates to reach equilibrium
x
→
+x
+x
0.10  x
x
x
K a1 = 7.5 × 10 3 =

[ H  ][ H 2 PO4 ]
x2
x2
, x = 2.7 × 10 2
=
≈
0.10
[ H 3 PO4 ]
0.10  x
Assumption is bad (x is 27% of 0.10). Using successive approximations:
x2
x2
= 7.5 × 10 3 , x = 2.3 × 10-2;
= 7.5 × 10 3 , x = 2.4 × 10 2
0.10
0.10  0.027
(consistent answer)
x = [H+] = 2.4 × 10 2 M; pH = -log (2.4 × 10 2 ) = 1.62
518
CHAPTER 14
b.
Initial
Equil.
H+
0.10 M
0.10  x
K a1 = 4.3 × 10 7 =
96.
⇌
H2CO3
HCO3
+
~0
ACIDS AND BASES
K a1 = 4.3 × 10 7
0
x
x

[H  ][HCO3 ]
x2
x2
=
≈
0.10
[H 2 CO3 ]
0.10  x
x = [H+] = 2.1 × 10 4 M; pH = 3.68; Assumptions good.
The reactions are:
H3AsO4 ⇌ H+ + H2AsO4
K a1 = 5 × 10 3
H2AsO4 ⇌ H+ + HAsO42
K a 2 = 8 × 10 8
HAsO42 ⇌ H+ + AsO43
K a 3 = 6 × 10 10
We will deal with the reactions in order of importance, beginning with the largest K a, K a1 .
H3AsO4
⇌
+
H
+
H2AsO4
Initial
Equil.
0.20 M
0.20  x
~0
5 × 10 3 =
x2
, x = 2.9 × 10 2 = 3 × 10 2 M
0.20  x
x
K a1 = 5 × 10 3 =


[H  ][H 2 AsO4 ]
[H 3 AsO4 ]
0
x
(By using successive approximations
or the quadratic formula.)
[H+] = [H2AsO4] = 3 × 10 2 M; [H3AsO4] = 0.20  0.03 = 0.17 M
2
Because K a 2 =
[H  ][HAsO4 ]
= 8 × 10 8 is much smaller than the K a1 value, very little of

[H 2 AsO4 ]
H2AsO4 (and HAsO42) dissociates compared to H3AsO4. Therefore, [H+] and [H2AsO4 ] will
not change significantly by the K a 2 reaction. Using the previously calculated concentrations of
H+ and H2AsO4 to calculate the concentration of HAsO42:
8 × 10 8 =
2
(3  10 2 ) [HAsO4 ]
, [HAsO42] = 8 × 10 8 M
3  10  2
Assumption that the Ka2 reaction does not change [H+] and [HAsO4] is good. We repeat the
process using K a 3 to get [AsO43].
K a 3 = 6 × 10 10 =
3
[H  ][AsO4 ]
3
(3  10 2 ) [AsO4 ]
=
2
(8  10 8 )
[HAsO4 ]
[AsO43] = 1.6 × 10 15 ≈ 2 × 10 15 M Assumption good.
CHAPTER 14
ACIDS AND BASES
519
So in 0.20 M analytical concentration of H3AsO4:
[H3AsO4] = 0.17 M; [H+] = [H2AsO4] = 3 × 10 2 M
[HAsO42] = 8 × 10 8 M; [AsO43] = 2 × 10 15 M
[OH] = Kw/[H+] = 3 × 10 13 M
97.
The dominant H+ producer is the strong acid H2SO4. A 2.0 M H2SO4 solution produces 2.0 M
HSO4 and 2.0 M H+. However, HSO4- is a weak acid which could also add H+ to the solution.
HSO4
Initial
Change
Equil.
⇌
H+
+
SO42
2.0 M
2.0 M
0
x mol/L HSO4- dissociates to reach equilibrium
x
→
+x
+x
2.0  x
2.0 + x
x
K a 2 = 1.2 × 10 2 =
2
[H  ][SO 4 ]

[HSO4 ]
=
(2.0  x)( x) 2.0 ( x)
≈
, x = 1.2 × 10 2
2.0  x
2.0
Because x is 0.60% of 2.0, the assumption is valid by the 5% rule. The amount of additional H+
from HSO4 is 1.2 × 10 2 . The total amount of H+ present is:
[H+] = 2.0 + 1.2 × 10 2 = 2.0 M; pH = log (2.0) = 0.30
Note: In this problem, H+ from HSO4 could have been ignored. However, this is not always the
case, especially in more dilute solutions of H2SO4.
98.
For H2SO4, the first dissociation occurs to completion. The hydrogen sulfate ion, HSO4, is a
weak acid with K a 2 = 1.2 × 10 2 . We will consider this equilibrium for additional H+ production:
HSO4Initial
Change
Equil.
⇌
H+
+
SO42
0.0050 M
0.0050 M
0
x mol/L HSO4- dissociates to reach equilibrium
x
→
+x
+x
0.0050  x
0.0050 + x
x
(0.0050  x)( x)
≈ x, x = 0.012; Assumption is horrible (240% error).
(0.0050  x)
Using the quadratic formula:
K a 2 = 0.012 =
6.0 × 10 5  0.012 x = x2 + 0.0050 x, x2 + 0.017 x  6.0 × 10 5 = 0
520
CHAPTER 14
x=
ACIDS AND BASES
 0.017  (2.9  10 4  2.4  10 4 )1/ 2  0.017  0.023

, x = 3.0 × 10 3 M
2
2
[H+] = 0.0050 + x = 0.0050 + 0.0030 = 0.0080 M; pH = 2.10
Note: We had to consider both H2SO4 and HSO4 for H+ production in this problem.
Acid-Base Properties of Salts
99.
One difficult aspect of acid-base chemistry is recognizing what types of species are present in
solution, i.e., whether a species is a strong acid, strong base, weak acid, weak base or a neutral
species. Below are some ideas and generalizations to keep in mind that will help in recognizing
types of species present.
a. Memorize the following strong acids: HCl, HBr, HI, HNO3, HClO4 and H2SO4
b. Memorize the following strong bases: LiOH, NaOH, KOH, RbOH, Ca(OH)2, Sr(OH)2 and
Ba(OH)2
c. All weak acids have a Ka value less than 1 but greater than Kw. Some weak acids are in Table
14.2 of the text. All weak bases have a Kb value less than 1 but greater than Kw. Some weak
bases are in Table 14.3 of the text.
d. All conjugate bases of weak acids are weak bases, i.e., all have a Kb value less than 1 but
greater than Kw. Some examples of these are the conjugate bases of the weak acids in Table
14.2 of the text.
e. All conjugate acids of weak bases are weak acids, i.e., all have a Ka value less than 1 but
greater than Kw. Some examples of these are the conjugate acids of the weak bases in Table
14.3 of the text.
f. Alkali metal ions (Li+, Na+, K+, Rb+, Cs+) and heavier alkaline earth metal ions (Ca2+, Sr2+,
Ba2+) have no acidic or basic properties in water.
g. All conjugate bases of strong acids (Cl, Br-, I, NO3, ClO4, HSO4) have no basic properties
in water (Kb << Kw) and only HSO4- has any acidic properties in water.
Let’s apply these ideas to this problem to see what type of species are present. The letters in
parenthesis is/are the generalization(s) above which identifies the species.
KOH: strong base (b)
KCl: neutral; K+ and Cl- have no acidic/basic properties (f and g).
KCN: CN is a weak base, Kb = 1.0 × 10 14 /6.2 × 10 10 = 1.6 × 10 5 (c and d). Ignore K+ (f).
NH4Cl: NH4+ is a weak acid, Ka = 5.6 × 10-10 (c and e). Ignore Cl (g).
HCl: strong acid (a)
The most acidic solution will be the strong acid followed by the weak acid. The most basic solution will be the strong base followed by the weak base. The KCl solution will be between the
acidic and basic solutions at pH = 7.00.
Most acidic → most basic: HCl > NH4Cl > KCl > KCN > KOH
100.
See Exercise 14.99 for some generalizations on acid-base properties of salts. The letters in
parenthesis is/are the generalization(s) listed in Exercise 14.99 which identifies the species.
CaBr2:
neutral; Ca2+ and Br have no acidic/basic properties (f and g).
CHAPTER 14
ACIDS AND BASES
KNO2:
HClO4:
HNO2:
HONH3ClO4:
521
NO2- is a weak base, Kb = 1.0 × 10-14/4.0 × 10-4 = 2.5 × 10-11 (c and d).
Ignore K+ (f).
strong acid (a)
weak acid, Ka = 4.0 × 10 4 (c)
HONH3+ is a weak acid, Ka = 1.0 × 10 14 /1.1 × 10 8 = 9.1 × 10 7 (c and e).
Ignore ClO4 (g). Note that HNO2 has a larger Ka value than HONH3+, so HNO2
is a stronger weak acid than HONH3+.
Using the information above (identity and Ka or Kb values), the ordering is:
101.
most acidic → most basic: HClO4 > HNO2 > HONH3ClO4 > CaBr2 > KNO2
From the Ka values, acetic acid is a stronger acid than hypochlorous acid. Conversely, the
conjugate base of acetic acid, C2H3O2, will be a weaker base than the conjugate base of
hypochlorous acid, OCl. Thus, the hypochlorite ion, OCl, is a stronger base than the acetate ion,
C2H3O2. In general, the stronger the acid, the weaker the conjugate base. This statement comes
from the relationship Kw = Ka × Kb, which holds for all conjugate acid-base pairs.
102.
Since NH3 is a weaker base (smaller Kb value) than CH3NH2, the conjugate acid of NH3 will be a
stronger acid than the conjugate acid of CH3NH2. Thus, NH4+ is a stronger acid than CH3NH3+.
103.
NaN3 → Na+ + N3; Azide, N3, is a weak base since it is the conjugate base of a weak acid. All
conjugate bases of weak acids are weak bases (Kw < Kb < 1). Ignore Na+.
N3 + H2O
Initial
Change
Equil.
⇌
HN3
+
OH
Kb =
Kw
1.0  10 14
=
= 5.3 × 10 10
5
1.9  10
Ka
0.010 M
0
~0
x mol/L of N3 reacts with H2O to reach equilibrium
x
→
+x
+x
0.010  x
x
x
Kb = 5.3 × 10 10 =
[HN3 ][OH  ]

[N3 ]
x = [OH] = 2.3 × 10 6 M; [H+] =
=
x2
x2
≈
(assuming x << 0.010)
0.010
0.010  x
1.0  10 14
= 4.3 × 10 9 M
2.3  10 6
Assumptions good.
[HN3] = [OH] = 2.3 × 10 6 M; [Na+] = 0.010 M; [N3] = 0.010  2.3 × 10 6 = 0.010 M
104.
C2H5NH3Cl → C2H5NH3+ + Cl; C2H5NH3+ is the conjugate acid of the weak base C2H5NH2
(Kb = 5.6 × 10 4 ). As is true for all conjugate acids of weak bases, C2H5NH3+ is a weak acid.
Cl has no basic (or acidic) properties. Ignore Cl. Solving the weak acid problem:
C2H5NH3+
Initial
Change
Equil.
⇌
C2H5NH2
+
H+
Ka = Kw/5.6 × 10 4 = 1.8 × 10 11
0.25 M
0
~0
+
x mol/L C2H5NH3 dissociates to reach equilibrium
x
→
+x
+x
0.25  x
x
x
522
CHAPTER 14
Ka = 1.8 × 10 11 =
[C 2 H 5 NH 2 ][H  ]

=
[C 2 H 5 NH3 ]
ACIDS AND BASES
x2
x2
≈
(assuming x << 0.25)
0.25  x
0.25
x = [H+] = 2.1 × 10 6 M; pH = 5.68; Assumptions good.
[C2H5NH2] = [H+] = 2.1 × 10 6 M; [C2H5NH3+] = 0.25 M; [Cl-] = 0.25 M
[OH] = Kw/[H+] = 4.8 × 10 9 M
105.
a. CH3NH3Cl → CH3NH3+ + Cl: CH3NH3+ is a weak acid. Cl is the conjugate base of a strong
acid. Cl has no basic (or acidic) properties.
CH3NH3+ ⇌ CH3NH2 + H+ Ka =
CH3NH3+
Initial
Change
Equil.
⇌
[CH3 NH2 ][H  ]

[CH3 NH3 ]
CH3NH2
+
=
1.00  10 14
Kw
=
= 2.28 × 10 11
4
Kb
4.38  10
H+
0.10 M
0
~0
+
x mol/L CH3NH3 dissociates to reach equilibrium
x
→
+x
+x
0.10  x
x
x
Ka = 2.28 × 10 11 =
x2
x2
≈
0.10  x
0.10
x = [H+] = 1.5 × 10 6 M; pH = 5.82
(assuming x << 0.10)
Assumptions good.
b. NaCN → Na+ + CN: CN is a weak base. Na+ has no acidic (or basic) properties.
1.0  10 14
K
CN + H2O ⇌ HCN + OH Kb = w =
= 1.6 × 10 5
10
6.2  10
Ka
Initial
Change
Equil.
0.050 M
0
~0
x mol/L CN reacts with H2O to reach equilibrium
x
→ +x
+x
0.050  x
x
x
Kb = 1.6 × 10 5 =
x2
[HCN][OH  ]
x2
=
≈
0.050
0.050  x
[CN  ]
x = [OH] = 8.9 × 10 4 M; pOH = 3.05; pH = 10.95 Assumptions good.
106.
a. KNO2 → K+ + NO2: NO2 is a weak base. Ignore K+.
CHAPTER 14
ACIDS AND BASES
NO2 + H2O
⇌
523
HNO2 + OH
Initial 0.12 M
Equil. 0.12  x
0
~0
x
x
[OH  ][HNO2 ]
Kb = 2.5 × 10 11 =
1.0  10 14
Kw
=
= 2.5 × 10 11
Ka
4.0  10  4
Kb =
=

[ NO2 ]
x2
x2
≈
0.12  x
0.12
x = [OH] = 1.7 × 10 6 M; pOH = 5.77; pH = 8.23
Assumptions good.
b. NaOCl → Na+ + OCl: OCl is a weak base. Ignore Na+.
OCl + H2O
⇌
Initial 0.45 M
Equil. 0.45  x
Kb = 2.9 × 10 7 =
HOCl + OH
0
Kb =
1.0  10 14
Kw
=
= 2.9 × 10 7
8
3.5  10
Ka
~0
x
x
[ HOCl][OH  ]
x2
x2
=
≈
0.45  x
0.45
[OCl ]
x = [OH] = 3.6 × 10 4 M; pOH = 3.44; pH = 10.56 Assumptions good.
c. NH4ClO4 → NH4+ + ClO4: NH4+ is a weak acid. ClO4 is the conjugate base of a strong acid.
ClO4 has no basic (or acidic) properties.
14
NH4+
⇌ NH3 + H+ Ka = K w = 1.0  10 5 = 5.6 × 10 10
1.8  10
Kb
Initial 0.40 M
Equil. 0.40  x
Ka = 5.6 × 10 10 =
0
~0
x
x
[ NH3 ][H  ]

[ NH4 ]
=
x2
x2
≈
0.40  x
0.40
x = [H+] = 1.5 × 10 5 M; pH = 4.82;
107.
Assumptions good.
All these salts contain Na+, which has no acidic/basic properties, and a conjugate base of a weak
acid (except for NaCl where Cl is a neutral species.). All conjugate bases of weak acids are
weak bases because the Kb values for these species are between 1 and Kw. To identify the species,
we will use the data given to determine the Kb value for the weak conjugate base. From the Kb
value and data in Table 14.2 of the text, we can identify the conjugate base present by calculating
the Ka value for the weak acid. We will use A as an abbreviation for the weak conjugate base.
A + H2O
⇌
HA
+
OH
524
CHAPTER 14
Initial
0.100 mol/1.00 L
0
~0
x mol/L A reacts with H2O to reach equilibrium
x
→
+x
+x
0.100  x
x
x
Change
Equil.
Kb =
ACIDS AND BASES
[HA][OH  ]
x2
=
; From the problem, pH = 8.07:
0.100  x
[A  ]
pOH = 14.00  8.07 = 5.93; [OH] = x = 10 5.93 = 1.2 × 10 6 M
Kb =
(1.2  10 6 ) 2
= 1.4 × 10 11 = Kb value for the conjugate base of a weak acid.
6
0.100  1.2  10
The Ka value for the weak acid equals Kw/Kb: Ka =
1.0  10 14
= 7.1 × 10 4
1.4  10 11
From Table 14.2 of the text, this Ka value is closest to HF. Therefore, the unknown salt is NaF.
108.
BHCl → BH+ + Cl-; Cl- is the conjugate base of the strong acid HCl, so Cl- has no acidic/basic
properties. BH+ is a weak acid since it is the conjugate acid of a weak base, B. Determining the
Ka value for BH+:
BH+
Initial
Change
Equil.
Ka =
⇌
B
+
H+
0.10 M
0
~0
x mol/L BH+ dissociates to reach equilibrium
x
→
+x
+x
0.10 x
x
x
[ B][H  ]
x2
; From the problem, pH = 5.82:
=
0.10  x
[ BH  ]
[H+] = x = 10 5.82 = 1.5 × 10 6 M; Ka =
(1.5  10 6 ) 2
= 2.3 × 10-11
0.10  1.5  10 6
Kb for the base, B = Kw/Ka = 1.0 × 10 14 /2.3 × 10 11 = 4.3 × 10 4 .
From Table 14.3 of the text, this Kb value is closest to that for CH3NH2, so the unknown salt
is CH3NH3Cl.
109.
Major species present: Al(H2O)63+ (Ka = 1.4 × 10-5), NO3- (neutral) and H2O (Kw = 1.0 × 10 14 );
Al(H2O)63+ is a stronger acid than water so it will be the dominant H+ producer.
Al(H2O)63+
Initial
0.050 M
⇌
Al(H2O)5(OH)2++
0
H+
~0
CHAPTER 14
ACIDS AND BASES
Change
Equil.
525
x mol/L Al(H2O)63+ dissociates to reach equilibrium
x
→
+x
+x
0.050  x
x
x
Ka = 1.4 × 10 5 =
[Al(H 2 O) 5 (OH) 2 ][H  ]
x2
x2
=
≈
0.050
0.050  x
[Al(H 2 O) 36 ]
x = 8.4 × 10 4 M = [H+]; pH = log (8.4 × 10 4 ) = 3.08; Assumptions good.
110.
Major species: Co(H2O)63+ (Ka = 1.0 × 10 5 ), Cl- (neutral) and H2O (Kw = 1.0 × 10 14 );
Co(H2O)63+ will determine the pH since it is a stronger acid than water. Solving the weak acid
problem in the usual manner:
Co(H2O)63+
Initial
Equil.
⇌
Co(H2O)5(OH)2+
0.10 M
0.10  x
Ka = 1.0 × 10 5 =
+
0
H+
Ka = 1.0 × 10 5
~0
x
x
x2
x2
, x = [H+] = 1.0 × 10 3 M
≈
0.10
0.10  x
pH = log (1.0 × 10 3 ) = 3.00; Assumptions good.
111.
Reference Table 14.6 of the text and the solution to Exercise 14.99 for some generalizations on
acid-base properties of salts.
a. NaNO3 → Na+ + NO3 neutral; Neither species has any acidic/basic properties.
b. NaNO2 → Na+ + NO2 basic; NO2 is a weak base and Na+ has no effect on pH.
NO2 + H2O ⇌ HNO2 + OH
Kb =
Kw
K a , HNO2
=
1.0  10 14
= 2.5 × 10 11
4.0  10  4
c. C5H5NHClO4 → C5H5NH+ + ClO4 acidic; C5H5NH+ is a weak acid and ClO4 has no
effect on pH.
C5H5NH+ ⇌ H+ + C5H5N
Ka =
Kw
K b, C5H 5 N
=
1.0  10 14
= 5.9 × 10 6
1.7  10 9
d. NH4NO2 → NH4+ + NO2 acidic; NH4+ is a weak acid (Ka = 5.6 × 10 10 ) and NO2 is a weak
base (Kb = 2.5 × 10 11 ). Because K a , NH  K b, NO , the solution is acidic.
4
2
NH4+ ⇌ H+ + NH3 Ka = 5.6 × 10 10 ; NO2 + H2O ⇌ HNO2 + OH
Kb = 2.5 × 10 11
e. KOCl → K+ + OCl basic; OCl is a weak base and K+ has no effect on pH.
526
CHAPTER 14
OCl + H2O ⇌ HOCl + OH
f.
Kb =
Kw
K a , HOCl
=
ACIDS AND BASES
1.0  10 14
= 2.9 × 10 7
3.5  10 8
NH4OCl → NH4+ + OCl basic; NH4+ is a weak acid and OCl is a weak base. Because
K b, OCl   K a , NH , the solution is basic.
4
NH4+ ⇌ NH3 + H+ Ka = 5.6 × 10 10 ; OCl + H2O ⇌ HOCl + OH
112.
Kb = 2.9 × 10 7
a. KCl → K+ + Cl neutral; K+ and Cl have no effect on pH.
b. NH4C2H3O2 → NH4+ + C2H3O2 neutral; NH4+ is a weak acid and C2H3O2 is a weak base.
Because K a , NH   K b, C2H3O2  , pH = 7.00.
4
NH4+ ⇌ NH3 + H+
Ka =
Kw
1.0  10 14
=
= 5.6 × 10 10
5
K b , NH3
1.8  10
C2H3O2 + H2O ⇌ HC2H3O2 + OH
Kb =
Kw
K b , HC2 H 3O 2
=
1.0  10 14
= 5.6 × 10 10
5
1.8  10
c. CH3NH3Cl → CH3NH3+ + Cl acidic; CH3NH3+ is a weak acid and Cl has no effect
on pH.
CH3NH3+ ⇌ H+ + CH3NH2
Ka =
Kw
K b , CH 3 NH 2
=
1.00  10 14
= 2.28 × 10 11
4
4.38  10
d. KF → K+ + F basic; F- is a weak base and K+ has no effect on pH.
F + H2O ⇌ HF + OH
Kb =
Kw
1.0  10 14
=
= 1.4 × 10 11
K a , HF
7.2  10  4
e. NH4F → NH4+ + F acidic; NH4+ is a weak acid and F is a weak base. Because
K a , NH  K b, F , the solution is acidic.
4
NH4+ ⇌ H+ + NH3 Ka = 5.6 × 10 10 ; F + H2O ⇌ HF + OH Kb = 1.4 × 10 11
f.
CH3NH3CN → CH3NH3+ + CN basic; CH3NH3+ is a weak acid and CN is a weak base.
Because K b, CN   K a , CH NH , the solution is basic.
3
CH3NH3+ ⇌ H+ + CH3NH2
CN + H2O ⇌ HCN + OH
3
Ka = 2.28 × 10-11
Kw
1.0  10 14
Kb =
=
= 1.6 × 10 5
K a , HCN
6.2  10 10
CHAPTER 14
ACIDS AND BASES
527
Relationships Between Structure and Strengths of Acids and Bases
113.
a. HIO3 < HBrO3; As the electronegativity of the central atom increases, acid strength increases.
b. HNO2 < HNO3; As the number of oxygen atoms attached to the central nitrogen atom
increases, acid strength increases.
c. HOI < HOCl; Same reasoning as in a.
d. H3PO3 < H3PO4; Same reasoning as in b.
114.
a. BrO3 < IO3; These are the conjugate bases of the acids in Exercise 14.113a. Since HBrO3 is
the stronger acid, the conjugate base of HBrO3 (BrO3) will be the weaker base. IO3 will be
the stronger base since HIO3 is the weaker acid.
b. NO3 < NO2; These are the conjugate bases of the acids in Exercise 14.113b. Conjugate base
strength is inversely related to acid strength.
c. OCl < OI. These are the conjugate bases of the acids in Exercise 14.113c.
115.
a. H2O < H2S < H2Se; As the strength of the H‒X bond decreases, acid strength increases.
b. CH3CO2H < FCH2CO2H < F2CHCO2H < F3CCO2H; As the electronegativity of neighboring
atoms increases, acid strength increases.
c. NH4+ < HONH3+; Same reason as in b.
d. NH4+ < PH4+; Same reason as in a.
116.
In general, the stronger the acid, the weaker the conjugate base.
a. SeH < SH < OH; These are the conjugate bases of the acids in Exercise 14.115a. The
ordering of the base strength is the opposite of the acids.
b. PH3 < NH3 (See Exercise 14.115d.)
c. HONH2 < NH3 (See Exercise 14.115c.)
117.
In general, metal oxides form basic solutions when dissolved in water and nonmetal oxides form
acidic solutions in water.
a. basic; CaO(s) + H2O(l) → Ca(OH)2(aq); Ca(OH)2 is a strong base.
b. acidic; SO2(g) + H2O(l) → H2SO3(aq); H2SO3 is a weak diprotic acid.
c. acidic; Cl2O(g) + H2O(l) → 2 HOCl(aq); HOCl is a weak acid.
118
a. basic; Li2O(s) + H2O(l) → 2 LiOH(aq); LiOH is a strong base.
528
CHAPTER 14
ACIDS AND BASES
b. acidic; CO2(g) + H2O(l) → H2CO3(aq); H2CO3 is a weak diprotic acid.
c. basic; SrO(s) + H2O(l) → Sr(OH)2(aq); Sr(OH)2 is a strong base.
Lewis Acids and Bases
119.
A Lewis base is an electron pair donor, and a Lewis acid is an electron pair acceptor.
120.
121.
a. Fe3+, acid; H2O, base
b. H2O, acid; CN, base
c. HgI2, acid; I, base
Al(OH)3(s) + 3 H+(aq) → Al3+(aq) + 3 H2O(l) (Brønsted-Lowry base, H+ acceptor)
Al(OH)3(s) + OH (aq) → Al(OH)4(aq)
122.
c. BF3, acid; F, base
b. Ag+, acid; NH3, base
a. B(OH)3, acid; H2O, base
(Lewis acid, electron pair acceptor)
Zn(OH)2(s) + 2 H+(aq) → Zn2+(aq) + 2 H2O(l) (Brønsted-Lowry base)
Zn(OH)2(s) + 2 OH (aq) → Zn(OH)42 (aq)
(Lewis acid)
123.
Fe3+ should be the stronger Lewis acid. Fe3+ is smaller and has a greater positive charge.
Because of this, Fe3+ will be more strongly attracted to lone pairs of electrons as compared to
Fe2+.
124.
The Lewis structures for the reactants and products are:
O
O
C
O
C
O
+
H
O
H
O
H
H
In this reaction, H2O donates a pair of electrons to carbon in CO2, which is followed by a proton
shift to form H2CO3. H2O is the Lewis base, and CO2 is the Lewis acid.
Additional Exercises
125.
At pH = 2.000, [H+] = 10 2.000 = 1.00 × 10 2 M; At pH = 4.000, [H+] = 10 4.000 = 1.00 × 10 4 M
mol H+ present = 0.0100 L ×
0.0100 mol H 
= 1.00 × 10 4 mol H+
L
Let V = total volume of solution at pH = 4.000:
1.00 × 10 4 mol/L =
1.00  10 4 mol H 
, V = 1.00 L
V
Volume of water added = 1.00 L  0.0100 L = 0.99 L = 990 mL
CHAPTER 14
ACIDS AND BASES
529
126.
Conjugate acid-base pairs differ by an H+ in the formula. Pairs in parts a, c, and d are conjugate
acid-base pairs. For part b, HSO4 is the conjugate base of H2SO4. In addition, HSO4 is the
conjugate acid of SO42
127.
a. The initial concentrations are halved since equal volumes of the two solutions are mixed.
⇌
HC2H3O2
Initial
Equil.
+
5.00 × 10 4 M
5.00 × 10 4 + x
0.100 M
0.100  x
Ka = 1.8 × 10 5 =
H+
C2H3O2
0
x
x(5.00  10 4  x)
x(5.00  10 4 )

(0.100  x)
(0.100 )
x = 3.6 × 10 3 ; Assumption is horrible. Using the quadratic formula:
x2 + 5.18 × 10 4 x - 1.8 × 10 6 = 0
x = 1.1 × 10 3 M; [H+] = 5.00 × 10 4 + x = 1.6 × 10 3 M; pH = 2.80
b. x = [C2H3O2] = 1.1 × 10 3 M
128.
Let HSac = saccharin and I = [HSac]o.
⇌
HSac
Initial
Equil.
I
Ix
Ka = 2.0 × 10 12 =
2.0 × 10 12 =
+
~0
x

Ssac
Ka = 1011.70 = 2.0 × 10 12
0
x
x2
; x = [H+] = 10 5.75 = 1.8 × 10 6 M
Ix
(1.8  10 6 ) 2
, I = 1.6 M = [HSac]o.
I  1.8  10 6
100.0 g HC7H4NSO3 
129.
H+
1 mol
1L
1000 mL
= 340 mL


183.19 g 1.6 mol
L
The light bulb is bright because a strong electrolyte is present, i.e., a solute is present that dissolves to produce a lot of ions in solution. The pH meter value of 4.6 indicates that a weak acid is
present. (If a strong acid were present, the pH would be close to zero.) Of the possible substances,
only HCl (strong acid), NaOH (strong base) and NH4Cl are strong electrolytes. Of these three
substances, only NH4Cl contains a weak acid (the HCl solution would have a pH close to zero
and the NaOH solution would have a pH close to 14.0). NH4Cl dissociates into NH4+ and Cl ions
when dissolved in water. Cl is the conjugate base of a strong acid, so it has no basic (or acidic
properties) in water. NH4+, however, is the conjugate acid of the weak base NH3, so NH4+ is a
weak acid and would produce a solution with a pH = 4.6 when the concentration is ~1 M.
530
130.
CHAPTER 14
ACIDS AND BASES
CaO(s) + H2O(l) → Ca(OH)2(aq); Ca(OH)2(aq) → Ca2+(aq) + 2 OH(aq)
0.25 g CaO 
[OH] =
1 mol CaO 1 mol Ca (OH) 2
2 mol OH 


56.08 g
1 mol CaO
mol Ca (OH) 2
= 5.9 × 10 3
1.5 L
[OH] = 5.9 × 10 3 M
pOH = log (5.9 × 10 3 ) = 2.23, pH = 14.00 – 2.23 = 11.77
131.
⇌
HBz
Initial
Change
Equil.
Ka =
H+
Bz-
+
HBz = C6H5CO2H
C
~0
0
x mol/L HBz dissociates to reach equilibrium
x
→
+x
+x
Cx
x
x
C = [HBz]o = concentration of HBz
that dissolves to give saturated
solution.
[H  ][Bz  ]
x2
= 6.4 × 10 5 =
, where x = [H+]
Cx
[HBz]
6.4 × 10 5 =
[ H  ]2
; pH = 2.80; [H+] = 10 2.80 = 1.6 × 10 3 M

C  [H ]
C  1.6 × 10 3 =
(1.6  10 3 ) 2
= 4.0 × 10 2 , C = 4.0 × 10 2 + 1.6 × 10 3 = 4.2 × 10 2 M
6.4  10 5
The molar solubility of C6H5CO2H is 4.2 × 10 2 mol/L.
132.
Because K a 2 for H2S is so small, we can ignore the H+ contribution from the K a 2 reaction.
H2S
Initial
Equil.
0.10 M
0.10  x
K a1 = 1.0 × 10 7 =
⇌
~0
x
K a1 = 1.0 × 10 7
HS
H+
0
x
x2
x2

, x = [H+] = 1.0 × 10 4 ; Assumptions good.
0.10
0.10  x
pH = log (1.0 × 10 4 ) = 4.00
Use the K a 2 reaction to determine [S2].
HS
⇌
H+
+
S2
CHAPTER 14
Initial
Equil.
ACIDS AND BASES
1.0 × 10 4 M
1.0 × 10 4  x
K a 2 = 1.0 × 10 19 =
531
1.0 × 10 4 M
1.0 × 10 4 + x
0
x
(1.0  10 4  x) x 1.0  10 4 x

(1.0  10  4  x)
1.0  10  4
x = [S2] = 1.0 × 10 19 M; Assumptions good.
133.
For H2C6H6O6. K a1 = 7.9 × 10 5 and K a 2 = 1.6 × 10 12 . Because K a1  K a 2 , the amount of H+
produced by the K a 2 reaction will be negligible.
1 mol H 2 C6 H 6 O 6
176 .12 g
= 0.0142 M
0.2000 L
0.500 g 
[H2C6H6O6]o =
H2C6H6O6(aq)
Initial
Equil.
⇌
HC6H6O6(aq) + H+(aq)
0.0142 M
0.0142  x
K a1 = 7.9 × 10 5 =
0
x
K a1 = 7.9 × 10 5
~0
x
x2
x2
, x = 1.1 × 10 3 ; Assumption fails the 5% rule.
≈
0.0142  x
0.0142
Solving by the method of successive approximations:
7.9 × 10 5 =
x2
, x = 1.0 × 10 3 M (consistent answer)
3
0.0142  1.1  10
Since H+ produced by the K a 2 reaction will be negligible, [H+] = 1.0 × 10 3 and pH = 3.00.
134.
[H+]o = 1.0 × 10 2 + 1.0 × 10 2 = 2.0 × 10 2 M from the strong acids HCl and H2SO4.
HSO4 is a good weak acid (Ka = 0.012). However, HCN is a poor weak acid (Ka = 6.2 × 10 10 )
and can be ignored. Calculating the H+ contribution from HSO4:
HSO4
Initial
Equil.
Ka =
0.010 M
0.010  x
⇌
H+
0.020 M
0.020 + x
+
SO42
Ka = 0.012
0
x
x (0.020  x)
x (0.020)
= 0.012 ≈
, x = 0.0060; Assumption poor (60% error).
(0.010  x)
(0.010)
Using the quadratic formula: x2 + 0.032 x  1.2 × 10 4 = 0, x = 3.4 × 10 3 M
[H+] = 0.020 + x = 0.020 + 3.4 × 10 3 = 0.023 M; pH = 1.64
532
135.
CHAPTER 14
ACIDS AND BASES
For this problem we will abbreviate CH2=CHCO2H as Hacr and CH2=CHCO2 as acr.
a. Solving the weak acid problem:
⇌
Hacr
Initial
Equil.
H+
0.10 M
0.10 - x
x2
0.10  x
~0
x
= 5.6 × 10 5 ≈
b. % dissociation =
Ka = 5.6 × 10 5
acr
+
0
x
x2
, x = [H+] = 2.4 × 10 3 M; pH = 2.62; Assumptions good.
0.10
[H  ]
2.4  10 3
× 100 =
× 100 = 2.4%
[ Hacr]0
0.10
c. acr is a weak base and the major source of OH in this solution.
1.0  10 14
K
acr + H2O ⇌ Hacr + OH Kb = w =
5.6  10 5
Ka
Initial 0.050 M
Equil. 0.050  x
Kb =
0
x
~0
x
x2
[ Hacr][OH  ]
x2
10
10
=
1.8
×
=
≈
0.050
0.050  x
[acr ]
x = [OH] = 3.0 × 10 6 M; pOH = 5.52; pH = 8.48
136.
Kb = 1.8 × 10 10
Assumptions good.
From the pH, C7H4ClO2 is a weak base. Use the weak base data to determine Kb for C7H4ClO2
(which we will abbreviate as CB).
CB
Initial
Equil.
+
H2O
⇌
HCB
0.20 M
0.20  x
+
0
x
OH
~0
x
Because pH = 8.65, pOH = 5.35 and [OH] = 10 5.35 = 4.5 × 10 6 M = x.
Kb =
[HCB][OH  ]
x2
(4.5  10 6 ) 2
=
=
= 1.0 × 10 10

6
0.20  x
[CB ]
0.20  4.5  10
Since CB is a weak base, HCB, chlorobenzoic acid, is a weak acid. Solving the weak acid
problem:
HCB
Initial
Equil.
0.20 M
0.20  x
⇌
H+
~0
x
+
CB
0
x
CHAPTER 14
Ka =
ACIDS AND BASES
1.0  10 14
x2
x2
Kw
4
10
=
=
1.0
×
=
≈
0.20
0.20  x
5.6  10 10
Kb
x = [H+] = 4.5 × 10 3 M; pH = 2.35
137.
533
Fe(H2O)63+ + H2O
a.
Initial
Equil.
⇌
0.10 M
0.10  x
Assumptions good.
Fe(H2O)5(OH)2++
H3O+
0
~0
x
x
[Fe(H 2 O) 5 (OH) 2 ][H 3O  ]
x2
x2
3
10
=
6.0
×
=
≈
0.10
0.10  x
[Fe(H 2 O) 36 ]
Ka =
x = 2.4 × 10 2 ; Assumption is poor (x is 24% of 0.10). Using successive approximations:
x2
= 6.0 × 10 3 , x = 0.021
0.10  0.024
x2
x2
= 6.0 × 10 3 , x = 0.022;
= 6.0 × 10 3 , x = 0.022
0.10  0.021
0.10  0.022
x = [H+] = 0.022 M; pH = 1.66
b. Because of the lower charge, Fe2+(aq) will not be as strong an acid as Fe3+(aq). A solution of
iron(II) nitrate will be less acidic (have a higher pH) than a solution with the same concentration of iron(III) nitrate.
138.
See generalizations in Exercise 14.99.
a. HI:
strong acid; HF: weak acid (Ka = 7.2 × 10 4 )
NaF:
F- is the conjugate base of the weak acid HF so F- is a weak base. The Kb value for
F- = Kw/Ka, HF = 1.4 × 10 11 . Na+ has no acidic or basic properties.
NaI:
neutral (pH = 7.0); Na+ and I- have no acidic/basic properties.
To place in order of increasing pH, we place the compounds from most acidic
(lowest pH) to most basic (highest pH). Increasing pH: HI < HF < NaI < NaF.
b. NH4Br:
HBr:
KBr:
NH3:
NH4+ is a weak acid (Ka = 5.6 × 10 10 ) and Br- is a neutral species.
strong acid
neutral; K+ and Br- have no acidic/basic properties
weak base, Kb = 1.8 × 10 5
Increasing pH: HBr < NH4Br < KBr < NH3
most
most
acidic
basic
534
CHAPTER 14
ACIDS AND BASES
c. C6H5NH3NO3: C6H5NH3+ is a weak acid (K w / K b, C6H5 NH2 = 1.0 × 10 14 /3.8 × 10 10
= 2.6 × 10 5 ) and NO3 is a neutral species.
neutral; Na+ and NO3 have no acidic/basic properties.
strong base
weak acid (Ka = 1.6 × 10 10 )
OC6H5 is a weak base (K b  K w / K a , HOC6H5 = 6.3 × 10 5 ) and K+ is a
neutral species.
C6H5NH2:
weak base (Kb = 3.8 × 10 10 )
HNO3:
strong acid
This is a little more difficult than the previous parts of this problem because two weak acids
and two weak bases are present. Between the weak acids, C6H5NH3+ is a stronger weak acid
than HOC6H5 because the Ka value for C6H5NH3+ is larger than the Ka value for HOC6H5.
Between the two weak bases, because the Kb value for OC6H5 is larger than the Kb value for
C6H5NH2, OC6H5 is a stronger weak base than C6H5NH2.
NaNO3:
NaOH:
HOC6H5:
KOC6H5:
Increasing pH: HNO3 < C6H5NH3NO3 < HOC6H5 < NaNO3 < C6H5NH2 < KOC6H5 < NaOH
most
most
acidic
basic
139.
The solution is acidic from HSO4 ⇌ H+ + SO42. Solving the weak acid problem:
HSO4
Initial
Equil.
⇌
0.10 M
0.10  x
1.2 × 10 2 =
H+
SO42
+
~0
x

0
x
2
[H  ][SO 4 ]
Ka = 1.2 × 10-2
x2
x2
≈
, x = 0.035
0.10
0.10  x
=
[HSO 4 ]
Assumption is not good (x is 35% of 0.10). Using successive approximations:
x2
x2
=
= 1.2 × 10 2 , x = 0.028
0.10  x
0.10  0.035
x2
x2
= 1.2 × 10 2 , x = 0.029;
= 1.2 × 10 2 , x = 0.029
0.10  0.028
0.10  0.029
x = [H+] = 0.029 M; pH = 1.54
140.
The relevant reactions are:
H2CO3 ⇌ H+ + HCO3
K a1 = 4.3 × 10 7 ; HCO3 ⇌ H+ + CO32 K a 2 = 5.6 × 10 11
CHAPTER 14
ACIDS AND BASES
535
Initially, we deal only with the first reaction (because K a1  K a 2 ) and then let these results
control values of the concentrations in the second reaction.
⇌
H2CO3
Initial
Equil.
H+
0.010 M
0.010  x
K a1 = 4.3 × 10 7 =
~0
x
HCO3
0
x

[H  ][HCO3 ]
x2
x2
=
≈
0.010
[H 2 CO3 ]
0.010  x
x = 6.6 × 10 5 M = [H+] = [HCO3] ;
HCO3Initial
Equil.
+
⇌
6.6 × 10 5 M
6.6 × 10 5  y
Assumptions good.
H+
+
6.6 × 10 5 M
6.6 × 10 5 + y
If y is small, then [H+] = [HCO3] and K a 2 = 5.6 × 10 11 =
y = [CO32] = 5.6 × 10 11 M;
CO32
0
y
2
[H  ][CO3 ]

[HCO3 ]
≈y
Assumptions good.
The amount of H+ from the second dissociation is 5.6 × 10 11 M or:
5.6  10 11
× 100 = 8.5 × 10-5 % H+ from the second dissociation
5
6.6  10
This result justifies our treating the equilibria separately. If the second dissociation contributed a
significant amount of H+, then we would have to treat both equilibria simultaneously. The reaction that occurs when acid is added to a solution of HCO3 is:
HCO3-(aq) + H+(aq) → H2CO3(aq) → H2O(l) + CO2(g)
The bubbles are CO2(g) and are formed by the breakdown of unstable H2CO3 molecules. We
should write H2O(l) + CO2(aq) or CO2(aq) for what we call carbonic acid. It is for convenience,
however, that we write H2CO3(aq).
141.
a. In the lungs, there is a lot of O2 and the equilibrium favors Hb(O2)4. In the cells, there is a
deficiency of O2 , and the equilibrium favors HbH44+.
b. CO2 is a weak acid, CO2 + H2O ⇌ HCO3 + H+. Removing CO2 essentially decreases H+.
Hb(O2)4 is then favored and O2 is not released by hemoglobin in the cells. Breathing into a
paper bag increases CO2 in the blood, thus increasing H+, which shifts the reaction left.
c. CO2 builds up in the blood and it becomes too acidic, driving the equilibrium to the left.
Hemoglobin can't bind O2 as strongly in the lungs. Bicarbonate ion acts as a base in water and
neutralizes the excess acidity.
536
142.
CHAPTER 14
ACIDS AND BASES
a. NH3 + H3O+ ⇌ NH4+ + H2O

K b for NH3
1.8  10 5
[ NH4 ]
1
Keq =
=
= 1.8 × 109


Kw
1.0  10 14
[ NH3 ][H  ] K a for NH4 
b. NO2 + H3O+ ⇌ HNO2 + H2O Keq =
[HNO2 ]


[ NO2 ][H ]

1
1
=
K a for HNO2
4.0  10 4
= 2.5 × 103
c. NH4+ + OH ⇌ NH3 + H2O
Keq =
1
1
=
= 5.6 × 104
K b for NH3
1.8  10 5
d. HNO2 + OH ⇌ H2O + NO2
Keq =
143.

K a for HNO2
[ NO2 ]
[H  ]
4.0  10 4


=
= 4.0 × 1010
Kw
[HNO2 ][OH  ] [H  ]
1.0  10 14
a. H2SO3
b. HClO3
c. H3PO3
NaOH and KOH are soluble ionic compounds composed of Na+ and K+ cations and OH anions.
All soluble ionic compounds dissolve to form the ions from which they are formed. In oxyacids,
the compounds are all covalent compounds in which electrons are shared to form bonds (unlike
ionic compounds). When these compounds are dissolved in water, the covalent bond between
oxygen and hydrogen breaks to form H+ ions.
Challenge Problems
144.
The pH of this solution is not 8.00 because water will donate a significant amount of H+ from the
autoionization of water. You can’t add an acid to water and get a basic pH. The pertinent reactions are:
H2O ⇌ H+ + OH Kw = [H+] [OH] = 1.0 × 10 14
HCl → H+ + Cl
Ka is very large, so we assume that only the forward reaction occurs.
In any solution, the overall net positive charge must equal the overall net negative charge (called
the charge balance). For this problem:
[positive charge] = [negative charge], so [H+] = [OH] + [Cl]
From Kw, [OH] = Kw/[H+], and from 1.0 × 10 8 M HCl, [Cl] = 1.0 × 10 8 M. Substituting into
the charge balance equation:
[H+] =
1.0  10 14
+ 1.0 × 10 8 , [H+]2  1.0 × 10 8 [H+]  1.0 × 10 14 = 0
[H  ]
Using the quadratic formula to solve:
CHAPTER 14
[H+]
ACIDS AND BASES
537
 (1.0  10 8 )  [(1.0  10 8 ) 2  4(1)(1.0  10 14 )]1 / 2
, [H+] = 1.1 × 10 7 M
2(1)
pH = log (1.1 × 10 7 ) = 6.96
145.
Since this is a very dilute solution of NaOH, we must worry about the amount of OH- donated
from the autoionization of water.
NaOH → Na+ + OH
H2O ⇌ H+ + OH Kw = [H+] [OH] = 1.0 × 10 14
This solution, like all solutions, must be charge balanced, that is [positive charge] = [negative
charge]. For this problem, the charge balance equation is:
[Na+] + [H+] = [OH], where [Na+] = 1.0 × 10 7 M and [H+] =
Kw
[OH  ]
Substituting into the charge balance equation:
1.0 × 10 7 +
1.0  10 14
= [OH], [OH]2 - 1.0 × 10 7 [OH]  1.0 × 10 14 = 0
[OH  ]
Using the quadratic formula to solve:
[OH] =
 (1.0  10 7 )  [(1.0  10 7 ) 2  4(1)(1.0  10 14 )]1/ 2
2(1)
[OH] = 1.6 × 10 7 M; pOH = log (1.6 × 10 7 ) = 6.80; pH = 7.20
146.
Ca(OH)2 (s) → Ca2+(aq) + 2 OH(aq)
This is a very dilute solution of Ca(OH)2 so we can't ignore the OH contribution from H2O.
From the dissociation of Ca(OH)2 alone, 2[Ca2+] = [OH]. Including H2O autoionization to H+
and OH, the overall charge balance is:
2[Ca2+] + [H+] = [OH]
2(3.0 × 10 7 M) + Kw/[OH] = [OH], [OH]2 = 6.0 × 10 7 [OH] + Kw
[OH]2  6.0 × 10 7 [OH]  1.0 × 10 14 = 0; Using quadratic formula: [OH] = 6.2 × 10 7 M
147.
HA
Initial
Equil.
⇌
C
C - 1.00 × 10 4
H+
~0
1.00 × 10 4
+
A
0
1.00 × 10 4
Ka = 1.00 × 10 6
C = [HA]o; For pH = 4.000,
x = [H+] = 1.00 × 10 4 M
538
CHAPTER 14
Ka =
ACIDS AND BASES
(1.00  10 4 ) 2
= 1.00 × 10 6 ; Solving: C = 0.0101 M
C  1.00  10  4
The solution initially contains 50.0 × 10 3 L × 0.0101 mol/L = 5.05 × 10 4 mol HA. We then
dilute to a total volume, V, in liters. The resulting pH = 5.000, so [H+] = 1.00 × 10 5 . In the
typical weak acid problem, x = [H+], so:
⇌
HA
Initial
Equil.
Ka =
H+
5.05 × 10 4 mol/V
5.05 × 10 4 /V  1.00 × 10 5
(1.00  10 5 ) 2
5.05  10
4
/ V  1.00  10
5
A
+
~0
1.00 × 10 5
0
1.00 × 10 5
= 1.00 × 10 6 , 1.00 × 10 4 = 5.05 × 10 4 /V - 1.00 × 10 5
V = 4.59 L; 50.0 mL are present initially, so we need to add 4540 mL of water.
148.
Initial
Change
Equil.
HBrO
⇌
H+
+
BrO
1.0 × 10 6 M
~0
0
x mol/L HBrO dissociates to reach equilibrium
x
→
+x
+x
6
1.0 × 10  x
x
x
Ka = 2 × 10 9 =
x2
x2
;
≈
1.0  10 6
1.0  10 6  x
Ka = 2 × 10 9
x = [H+] = 4 × 10 8 M; pH = 7.4
Let’s check the assumptions. This answer is impossible! We can't add a small amount of an acid
to a water and get a basic solution. The highest possible pH for an acid in water is 7.0. In the
correct solution, we would have to take into account the autoionization of water.
149.
Major species present are H2O, C5H5NH+ (Ka = Kw/Kb(C5H5N) = 1.0 × 10 14 /1.7 × 10 9 =
5.9 × 10 6 ) and F (Kb = Kw/Ka(HF) = 1.0 × 10 14 /7.2 × 10 4 = 1.4 × 10 11 ). The reaction to
consider is the best acid present (C5H5NH+) reacting with the best base present (F). Solving for
the equilibrium concentrations:
C5H5NH+(aq)
Initial
Change
Equil.
0.200 M
x
0.200  x
K = K a, C H
5
5 NH

×
+
F(aq)
⇌
0.200 M
x
→
0.200  x
C5H5N(aq) + HF(aq)
0
+x
0
+x
x
x
1
= 5.9 × 10 6 (1/7.2 × 10 4 ) = 8.2 × 10 3
K a , HF
CHAPTER 14
K=
ACIDS AND BASES
539
x2
[C5 H5 N][HF]
3
10
=
8.2
×
=
; Taking the square root of both sides:
(0.200  x) 2
[C5 H 5 NH ][F ]
0.091 =
x
, x = 0.018  0.091 x, x = 0.016 M
0.200  x
From the setup to the problem, x = [C5H5N] = [HF] = 0.016 M and 0.200  x = 0.200 - 0.016 =
0.184 M = [C5H5NH+] = [F]. To solve for the [H+], we can use either the Ka equilibrium for
C5H5NH+ or the Ka equilibrium for HF. Using C5H5NH+ data:
K a, C H
5
5 NH

[C 5 H 5 N][H  ]
(0.016 ) [H  ]

, [H+] = 6.8 × 10 5 M
(0.184 )
[C 5 H 5 NH  ]
= 5.9 × 10 6 =
pH = log(6.8 × 10 5 ) = 4.17
As one would expect, because the Ka for the weak acid is larger than the Kb for the weak base, a
solution of this salt should be acidic.
150.
Major species: NH4+, OCl, and H2O; Ka for NH4+ = 1.0 × 10 14 /1.8 × 10 5 = 5.6 × 10 10 and
Kb for OCl = 1.0 × 10 14 / 3.5 × 10 8 = 2.9 × 10 7 .
Because OCl is a better base than NH4+ is an acid, the solution will be basic. The dominant
equilibrium is the best acid (NH4+) reacting with the best base (OCl) present.
NH4+
Initial
Change
Equil.
OCl
+
0.50 M
–x
0.50 – x
K = K a , NH4 
K = 0.016 =
1
K a , HOCl
NH3
0.50 M
–x
→
0.50 – x
0
+x
x
+ HOCl
0
+x
x
= 5.6 × 10 10 / 3.5 × 10 8 = 0.016
[ NH3 ][HOCl]

⇌

[ NH4 ][OCl ]
=
x(x)
(0.50  x)(0.50  x)
x2
x
 0.016,
= (0.016)1/2 = 0.13, x = 0.058 M
2
0.50  x
(0.50  x)
To solve for the H+, use any pertinent Ka or Kb value. Using Ka for NH4+:
K a , NH  = 5.6 × 10 10 =
4
151.
[ NH3 ][H  ]

[ NH4 ]
=
(0.058 )[H  ]
, [H+] = 4.3 × 10 9 M, pH = 8.37
0.50  0.058
Since NH3 is so concentrated, we need to calculate the OH- contribution from the weak base NH3.
540
CHAPTER 14
NH3 +
⇌
Initial 15.0 M
Equil. 15.0  x
Kb = 1.8 × 10 5 =
NH4+
+
0
Kb = 1.8 × 10 5
OH
0.0100 M
0.0100 + x
x
ACIDS AND BASES
(Assume no volume change.)
x (0.0100 )
x (0.0100  x)

, x = 0.027; Assumption is horrible
15.0  x
15.0
(x is 270% of 0.0100).
Using the quadratic formula:
1.8 × 10 5 (15.0  x) = 0.0100 x + x2, x2 + 0.0100 x  2.7 × 10 4 = 0
x = 1.2 × 10 2 , [OH] = 1.2 × 10 2 + 0.0100 = 0.022 M
152.
For 0.0010% dissociation: [NH4+] = 1.0 × 10 5 (0.050) = 5.0 × 10 7 M
NH3 + H2O ⇌ NH4+ + OH
Kb =
(5.0  10 7 )[OH  ]
= 1.8 × 10 5
7
0.050  5.0  10
Solving: [OH] = 1.8 M; Assuming no volume change:
1.0 L ×
153.
1.8 mol NaOH 40.00 g NaOH
= 72 g of NaOH

L
mol NaOH
5.11 g 0.08206 L atm

 298 K
dRT
L
mol K
Molar mass =
=
= 125 g/mol
P
1.00 atm
1 mol
125 g
= 0.120 M; pH = 1.80, [H+] = 10 1.80 = 1.6 × 10 2 M
0.100 L
1.50 g 
[HA]o =
HA
⇌
H+ + A
Equil. 0.120  x
Ka =
154.
x
x = [H+] = 1.6 × 10 2 M
x
[H  ][A  ]
(1.6  10 2 ) 2
=
= 2.5 × 10 3
0.120  0.016
[HA]
HC2H3O2
Initial 1.00 M
Equil. 1.00  x
1.8 × 10 5 =
⇌
H+ +
0
x
C2H3O2
Ka = 1.8 × 10 5
0
x
x2
x2
≈
, x = [H+] = 4.24 × 10 3 M (using one extra sig fig)
1.00
1.00  x
CHAPTER 14
ACIDS AND BASES
541
pH = log (4.24 × 10 3 ) = 2.37 Assumptions good.
We want to double the pH to 2(2.37) = 4.74 by addition of the strong base NaOH. As is true with
all strong bases, they are great at accepting protons. In fact, they are so good that we can assume
strong bases accept protons 100% of the time. The best acid present will react with the strong
base. This is HC2H3O2. The initial reaction that occurs when strong base is added is:
HC2H3O2 + OH → C2H3O2 + OH
Note that this reaction has the net effect of converting HC2H3O2 into its conjugate base, C2H3O2.
For a pH = 4.74, let’s calculate the ratio of [C2H3O2]/ HC2H3O2] necessary to achieve this pH.
HC2H3O2
⇌ H+

+ C2H3O2
[H  ][C 2 H 3 O 2 ]
[HC2 H 3 O 2 ]
Ka =
When pH = 4.74, [H+] = 10 4.74 = 1.8 × 10 5 .
Ka = 1.8 × 10
5


1.8  10 5 [C 2 H 3O 2 ] [C 2 H 3O 2 ]
,
=
= 1.0
[HC2 H 3O 2 ]
[HC2 H 3O 2 ]
So for a solution having pH = 4.74, we need to have equal concentrations (equal moles) of
C2H3O2 and HC2H3O2. Therefore, we need to add an amount of NaOH that will convert one-half
of the HC2H3O2 into C2H3O2. This amount is 0.50 M NaOH.
HC2H3O2 + OH → C2H3O2 + H2O
Initial
1.00 M
Change
0.50
After
0.50 M
completion
0.50 M
0.50
0
0
+0.50
0.50 M
From the above stoichiometry problem, adding enough NaOH(s) to produce a 0.50 M OH
solution will convert one-half of the HC2H3O2 into C2H3O2 resulting in a solution with
pH = 4.74.
mass NaOH = 1.0 L 
155.
0.500 mol NaOH 40.00 g NaOH
= 20.0 g NaOH

L
L
PO43 is the conjugate base of HPO42. The Ka value for HPO42 is K a 3 = 4.8 × 10-13.
PO43(aq) + H2O(l) ⇌ HPO42(aq) + OH(aq) Kb =
Kw
1.0  10 14
=
= 0.021
K a3
4.8  10 13
HPO42 is the conjugate base of H2PO4 ( K a 2 = 6.2 × 10 8 ).
HPO42+ H2O
⇌
H2PO4 + OH
Kb =
Kw
1.0  10 14
=
= 1.6 × 10 7
K a1
6.2  10 8
H2PO4 is the conjugate base of H3PO4 ( K a1 = 7.5 × 10 3 ).
542
CHAPTER 14
⇌
H2PO4 + H2O
H3PO4 + OH
Kb =
ACIDS AND BASES
Kw
1.0  10 14
=
= 1.3 × 10 12
K a1
7.5  10 3
From the Kb values, PO43 is the strongest base. This is expected because PO43 is the conjugate
base of the weakest acid (HPO42).
156.
Major species: Na+, PO43 (a weak base), H2O; From the Kb values calculated in Exercise
14.155, the dominant producer of OH is the Kb reaction for PO43. We can ignore the contribution of OH from the Kb reactions for HPO42 and H2PO4 . From Exercise 14.155, Kb for PO43 =
0.021.
PO43 + H2O
Initial
Equil.
⇌
0.10 M
0.10  x
Kb = 0.021 =
HPO42 + OH
0
x
Kb = 0.021
~0
x
x2
; Because Kb is so large, the 5% assumption will not hold. Solving using
0.10  x
the quadratic equation:
x2 + 0.021 x – 0.0021 = 0, x = [OH] = 3.7 × 10 2 M, pOH = 1.43, pH = 12.57
157.
a. NH4(HCO3) → NH4+ + HCO3
Ka for NH4+ =
K w 1.0  10 14
1.0  10 14
10

10
=
5.6
×
;
K
for
HCO
=
=
= 2.3 × 10 8
b
3
K a1
1.8  10 5
4.3  10 7
Solution is basic because HCO3 is a stronger base than NH4+ is as an acid. The acidic
properties of HCO3 were ignored because K a 2 is very small (5.6 × 10 11 ).
b. NaH2PO4 → Na+ + H2PO4; Ignore Na+.

K a 2 for H 2 PO4 = 6.2 × 10 8 ; Kb for H2PO4 =
K w 1.0  10 14
=
= 1.3 × 10 12
K a1
7.5  10 3
Solution is acidic because Ka > Kb.
c. Na2HPO4 → 2 Na+ + HPO42; Ignore Na+.
2
K a 3 for HPO4 = 4.8 × 10 13 ; Kb for HPO42 =
Solution is basic because Kb > Ka.
d. NH4(H2PO4) → NH4+ + H2PO4
K w 1.0  10 14
=
= 1.6 × 10 7
8
K a2
6.2  10
CHAPTER 14
ACIDS AND BASES
543
NH4+ is weak acid and H2PO4 is also acidic (see b). Solution with both ions present will be
acidic.
e. NH4(HCO2) → NH4+ + HCO2; From Appendix 5, Ka for HCO2H = 1.8 × 10 4 .
Ka for NH4+ = 5.6 × 10 10 ; Kb for HCO2 =
K w 1.0  10 14
=
= 5.6 × 10 11
1.8  10  4
Ka
Solution is acidic because NH4+ is a stronger acid than HCO2 is as a base.
158.
a. HCO3 + HCO3
⇌
H2CO3 + CO32
2
Keq =
[H 2 CO3 ][CO3 ]


[HCO3 ][HCO3 ]

Ka
5.6  10 11
[H  ]
2
=
=
= 1.3 × 10 4
7

K
4
.
3

10
[H ]
a1
b. [H2CO3] = [CO32] since the reaction in part a is the principle equilibrium reaction.
c. H2CO3
⇌
2
2 H+ + CO32
Keq =
[H  ]2 [CO3 ]
 K a1  K a 2
[H 2 CO3 ]
Because, [H2CO3] = [CO32] from part b, [H+]2 = K a1  K a 2 .
[H+] = (K a1  K a 2 )1/ 2 or pH =
pK a1  pK a 2
2
d. [H+] = [(4.3 × 10 7 ) × (5.6 × 10 11 )]1/2, [H+] = 4.9 × 10 9 M; pH = 8.31
1 mol
100.0 g
= 2.00 × 10 3 mol/kg ≈ 2.00 × mol/L (dilute solution)
0.5000 kg
0.100 g 
159.
Molality = m =
ΔTf = iKfm, 0.0056°C = i(1.86°C/molal) (2.00 × 10 3 molal), i = 1.5
If i = 1.0, the percent dissociation of the acid = 0% and if i = 2.0, the percent dissociation of the
acid = 100%. Since i = 1.5, the weak acid is 50.% dissociated.
HA
⇌
H+ + A
Ka =
[H  ][A  ]
[HA]
Because the weak acid is 50.% dissociated:
[H+] = [A] = [HA]o × 0.50 = 2.00 × 10 3 M × 0.50 = 1.0 × 10 3 M
[HA] = [HA]o  amount HA reacted = 2.00 × 10 3 M  1.0 × 10 3 M = 1.0 × 10 3 M
544
CHAPTER 14
Ka =
160.
ACIDS AND BASES
[H  ][A  ]
(1.0  10 3 ) (1.0  10 3 )
=
= 1.0 × 10 3
[HA]
1.0  10 3
a. Assuming no ion association between SO42 (aq) and Fe3+(aq), then i = 5 for Fe2(SO4)3.
π = iMRT = 5(0.0500 mol/L)(0.08206 L atm/Kmol)(298 K) = 6.11 atm
b. Fe2(SO4)3(aq) → 2 Fe3+(aq) + 3 SO42(aq)
Under ideal circumstances, 2/5 of π calculated above results from Fe3+ and 3/5 results from
SO42. The contribution to π from SO42 is 3/5 × 6.11 atm = 3.67 atm. Because SO42 is
assumed unchanged in solution, the SO42 contribution in the actual solution will also be 3.67
atm. The contribution to the actual osmotic pressure from the Fe(H2O)63+ dissociation
reaction is 6.73  3.67 = 3.06 atm.
The initial concentration of Fe(H2O)62+ is 2(0.0500) = 0.100 M. The set-up for the weak acid
problem is:
[H  ][Fe(OH)(H 2 O) 52  ]
Fe(H2O)63+ ⇌ H+ + Fe(OH)(H2O)52+
Ka =
[Fe(H 2 O) 36 ]
Initial 0.100 M
~0
0
x mol/L of Fe(H2O)63+ reacts to reach equilibrium
Equil. 0.100  x
x
x
π = iMRT; Total ion concentration = iM =
π
3.06 atm
= 0.125 M

RT 0.8206 L atm / K  mol (298)
0.125 M = 0.100  x + x + x = 0.100 + x, x = 0.025 M
Ka =
[H  ][Fe(OH)(H 2 O) 52  ]
x2
(0.025) 2
(0.025) 2


=
, Ka = 8.3 × 10 3
0.100  x (0.100  0.025)
0.075
[Fe(H 2 O) 36 ]
Integrative Problems
2.14 g NaIO 
161.
[IO] =
IO
+ H2O
Initial 1.03 × 10 2 M
Equil. 1.03 × 10 2  x
Kb =
1 mol NaIO 1 mol IO 

165 .89 g
mol NaIO
= 1.03 × 10 2 M IO
1.25 L
⇌
HIO + OH
0
x
Kb =
[ HIO][OH  ]
[ IO  ]
~0
x
x2
; From the problem, pOH = 14.00 – 11.32 = 2.68
1.03  10 2  x
CHAPTER 14
ACIDS AND BASES
545
[OH] = 10 2.68 = 2.1 × 10 3 M = x; Kb =
162.
(2.1  10 3 ) 2
= 5.4 × 10 4
1.03  10  2  2.1  10 3
10.0 g NaOCN ×
1 mol
= 0.154 mol NaOCN
65.01 g
10.0 g H2C2O4 ×
1 mol
= 0.111 mol H2C2O4
90.04 g
mol NaOCN
0.154 mol
(actual) =
= 1.39
mol H 2SO 4
0.111 mol
The balanced reaction requires a larger 2:1 mole ratio. Therefore, NaOCN in the numerator is
limiting. Since there is a 2:2 mol correspondence between mole NaOCN reacted and mole HNCO
produced, 0.154 mol of HNCO will be produced.
HNCO
Initial
Equil.
⇌
H+
0.154 mol/0.100 L
~0
1.54  x
x
Ka = 1.2 × 10 4 =
NCO
+
Ka = 1.2 × 10 4
0
x
x2
x2

, x = [H+] = 1.4 × 10 2 M
1.54
1.54  x
pH = –log(1.4 × 10 2 ) = 1.85; Assumptions good.
163.
30.0 mg papH  Cl  1000 mL
1g
1 mol papH  Cl 
1 mol papH 




= 0.0792 M
mL so ln
L
1000 mg
378 .85 g
mol papH  Cl 
⇌
papH+
Initial
Equil.
0.0792 M
0.0792  x
Ka = 2.5 × 10 6 =
pap +
0
x
H+
Ka =
~0
x
x2
x2

, x = [H+] = 4.4 × 10 4 M
0.0792
0.0792  x
pH = -log(4.4 × 10 4 ) = 3.36; Assumptions good.
Marathon Problems
Kw
2.1  10 14
=
= 2.5 × 10 6
9
K b , pap
8.33  10
546
164.
CHAPTER 14
ACIDS AND BASES
To determine the pH of solution A, the Ka value for HX must be determined. Use solution B to
determine Kb for X, which can then be used to calculate Ka for HX (Ka = Kw/Kb).
Solution B:
X

+
⇌
H2O
HX
+

OH
[HX][OH  ]
Kb =
[X  ]
Initial
Change
Equil.
0.0500 M
0
~0
x
→
+x
+x
0.0500  x
x
x
2
x
Kb =
; From the problem, pH = 10.02, so pOH = 3.98 and [OH] = x = 10 3.98
0.0500  x
Kb =
(10 3.98 ) 2
= 2.2 × 10 7
0.0500  10 3.98
Solution A:
H a , HX  K w / K
b, X 
⇌
HX
Initial
Change
Equil.
0.100 M
x
0.100  x
Ka = 4.5 × 10 8 =
= 1.0 × 10 14 /2.2 × 10 7 = 4.5 × 10 8

→
H+
~0
+x
x
+
X
Ka = 4.5 × 10 8 =
[H  ][X  ]
[HX]
0
+x
x
x2
x2

, x = [H+] = 6.7 × 10 5 M
0.100  x
0.100
Assumptions good (x is 6.7 × 10 2 % of 0.100); pH = 4.17
Solution C:
Major species: HX (Ka = 4.5 × 10 8 ), Na+, OH; The OH from the strong base is exceptional at
accepting protons. OH- will react with the best acid present (HX) and we can assume that OH
will react to completion with HX, i.e., until one (or both) of the reactants runs out. Since we have
added one volume of substance to another, we have diluted both solutions from their initial
concentrations. What hasn’t changed is the moles of each reactant. So let’s work with moles of
each reactant initially.
mol HX = 0.0500 L ×
mol OH = 0.0150 L ×
0.100 mol HX
= 5.00 × 10 3 mol HX
L
0.250 mol NaOH 1 mol OH 

= 3.75 × 10 3 mol OH
L
mol NaOH
CHAPTER 14
ACIDS AND BASES
547
Now let’s determine what is remaining in solution after OH reacts completely with HX. Note
that OH is the limiting reagent.
HX
Initial
Change
After completion
+
5.00 × 10 3 mol
3.75 × 10 3
1.25 × 10 3 mol
OH
→
X
+
H2O
3.75 × 10 3 mol
0

3
3
3.75 × 10
→ +3.75 × 10
+3.75 × 10 3
0
3.75 × 10 3 mol

After reaction, the solution contains HX, X, Na+ and H2O. The Na+ (like most +1 metal ions)
has no effect on the pH of water. However, HX is a weak acid and its conjugate base, X , is a
weak base. Since both Ka and Kb reactions refer to these species, we could use either reaction to
solve for the pH; we will use the Kb reaction. To solve the equilibrium problem using the Kb
reaction, we need to convert to concentration units since Kb is in concentration units of mol/L.
[HX] =
1.25  10 3 mol
3.75  10 3 mol
= 0.0192 M; [X] =
= 0.0577 M
(0.0500  0.0150 ) L
0.0650 L
[OH] = 0 (We reacted all of it to completion.)
X
Initial
Change
Equil.
+
H2O
⇌
HX
+
OH
Kb = 2.2 × 10 7
0.0577 M
0.0192 M
0

x mol/L of X reacts to reach equilibrium
x
→
+x
+x
0.0577  x
0.0192 + x
x
Kb = 2.2 × 10 7 =
x = [OH] =
(0.0192  x)( x)
(0.0192 ) x
(assuming x is << 0.0192)

(0.0577  x)
(0.0577 )
2.2  10 7 (0.0577 )
= 6.6 × 10 7 M
0.0192
Assumptions great (x is 3.4 × 10 3 % of
0.0192).
[OH] = 6.6 × 10 7 M, pOH = 6.18, pH = 14.00  6.18 = 7.82 = pH of solution C
The combination is 417782.
165.
a. Strongest acid from group I = HCl; Weakest base from group II = NaNO2
0.20 M HCl + 0.20 M NaNO2; Major species = H+, Cl, Na+, NO2, H2O; Let the added
protons from H+ react completely with the best base present, NO2. Since strong acids are
great at what they do (donate protons), assume the H+ reacts to completion.
548
CHAPTER 14
H+
+
Initial
0.10 M
After
0
→ HNO2
0.10 M
0
0
HNO2
Initial
Change
Equil.
NO2
0.10 M
x
0.10  x
x2
= 4.0 × 10-4;
0.10  x
⇌
ACIDS AND BASES
(Molarities are halved due to dilution.)
0.10 M
H+ + NO2
0
+x
x
0
+x
x
Solving: x = [H+] = 6.3 × 10 3 M , pH = 2.20
b. Weakest acid from group I = (C2H5)3NHCl; Best base from group II = KOI
OI
Initial
Equil.
0.10 M
0.10  x
+ (C2H5)3NH+
0.10 M
0.10  x
⇌
HOI + (C2H5)3N
0
x
0
x
K a for (C 2 H 5 ) 3 NH 
1.0  10 14
1

=
= 1.25 (carrying extra sig fig)
4
K a for HOI
4.0  10
2  10 11
K=
x2
x
= 1.25,
= 1.12, x = 0.053; [HOI] = 0.053 M and [OI] = 0.047 M
2
(0.10  x)
0.10  x
(extra sig fig)
HOI
⇌ H+
+ OI Ka = 2 × 10 11 ; Solve for H+ and pH:
2 × 10 11 =
[H  ](0.047 M )
, [H+] = 2.3 × 10 11 M, pH = 10.64 = 10.6
(0.053 M )
c. Ka for (C2H5)3NH+ =
1.0  10 14
1.0  10 14
11

10
=
2.5
×
;
K
for
NO
=
= 2.5 × 10 11
b
2
4
4
4.0  10
4.0  10
Because Ka = Kb, mixing (C2H5)3NHCl with NaNO2 will result in a solution with pH = 7.00.
We suggest you prove it by doing a calculation similar to that done in part b.