Download CHAPTER 15 ACIDS AND BASES

CHAPTER 15
ACIDS AND BASES
15.3
Table 15.2 of the text contains a list of important Brønsted acids and bases. (a) both (why?), (b) base,
(c) acid, (d) base, (e) acid, (f) base, (g) base, (h) base, (i) acid, (j) acid.
15.4
Recall that the conjugate base of a Brønsted acid is the species that remains when one proton has been
removed from the acid.
(a)
(b)
(c)
(d)
(e)
15.5
15.6
−
nitrite ion: NO2
−
hydrogen sulfate ion (also called bisulfate ion): HSO4
−
hydrogen sulfide ion (also called bisulfide ion): HS
−
cyanide ion: CN
−
formate ion: HCOO
In general the components of the conjugate acid−base pair are on opposite sides of the reaction arrow. The
base always has one fewer proton than the acid.
−
The conjugate acid−base pairs are (1) HCN (acid) and CN (base) and (2) CH3COO (base) and
CH3COOH (acid).
(b)
(1) HCO3 (acid) and CO3
(c)
(1) H2PO4 (acid) and HPO4
(d)
(1) HClO (acid) and ClO (base) and (2) CH3NH2 (base) and CH3NH3 (acid).
(e)
(1) H2O (acid) and OH (base) and (2) CO3
−
−
(base) and (2) HCO3 (base) and H2CO3 (acid).
2−
+
(base) and (2) NH3 (base) and NH4 (acid).
+
−
2−
−
(base) and HCO3 (acid).
The conjugate acid of any base is just the base with a proton added.
(a)
(b)
−
(c) HCO3
−
(h) HSO4
(b) H2CO3
(g) H2SO4
(d) H3PO4
−
(i) HSO3
(e) H2PO4
−
The Lewis structures are
-O
15.8
−
2−
−
(a) H2S
2−
(f) HPO4
15.7
−
(a)
O
O
C
C
O
and
H
-O
O
O
C
C
-
O
−
+
H and C2H2O4 can act only as acids, C2HO4 can act as both an acid and a base, and C2O4
only as a base.
2−
The conjugate base of any acid is simply the acid minus one proton.
(a) CH2ClCOO−
−
(f) HSO4
−
(k) HS
−
(b) IO4
2−
(g) SO4
2−
(l) S
(c) H2PO4
−
(h) IO3
−
(m) OCl
−
(d) HPO4
2−
(i) SO3
2−
3−
(e) PO4
(j) NH3
can act
326
CHAPTER 15: ACIDS AND BASES
15.15
[H ] = 1.4 × 10
+
−
[OH ] =
15.16
M
1.0 × 10−14
1.4 × 10
−3
= 7.1 × 10 −12 M
−
[OH ] = 0.62 M
+
[H ] =
15.17
−3
(a)
1.0 × 10 −14
= 1.6 × 10 −14 M
0.62
HCl is a strong acid, so the concentration of hydrogen ion is also 0.0010 M. (What is the concentration
of chloride ion?) We use the definition of pH.
+
pH = −log[H ] = −log(0.0010) = 3.00
(b)
KOH is an ionic compound (why?) and is fully ionized. We first find the concentration of hydrogen ion.
Kw
[H + ] =
[OH − ]
=
1.0 × 10−14
= 1.3 × 10−14 M
0.76
The pH is then found from its defining equation
+
pH = −log[H ] = −log[1.3 × 10
15.18
(a)
] = 13.89
Ba(OH)2 is ionic and fully ionized in water. The concentration of the hydroxide ion is 5.6 × 10
2+
(Why? What is the concentration of Ba ?) We find the hydrogen ion concentration.
Kw
[H + ] =
−
[OH ]
=
1.0 × 10−14
5.6 × 10
−4
−4
M
= 1.8 × 10 −11 M
+
pH = −log[H ] = −log(1.8 × 10
The pH is then:
(b)
−14
−11
) = 10.74
Nitric acid is a strong acid, so the concentration of hydrogen ion is also 5.2 × 10
−4
M. The pH is:
−4
+
pH = −log[H ] = −log(5.2 × 10 ) = 3.28
15.19
15.20
+
+
Since pH = − log[H ], we write [H ] = 10
+
−2.42
+
−11.21
(a)
[H ] = 10
(b)
[H ] = 10
= 3.8 × 10
−3
= 6.2 × 10
M
−12
M
+
−pH
+
−6.96
+
−15.00
(c)
[H ] = 10
(d)
[H ] = 10
= 1.1 × 10
−7
= 1.0 × 10
M
−15
M
For (a) and (b) we can calculate the H concentration using the equation representing the definition of pH.
(a)
+
pH = −log [H ]
+
[H ] = 10
(b)
−5.20
= 6.3 × 10
−6
M
+
pH = −log [H ]
+
[H ] = 10
−16.00
= 1.0 × 10
−16
M
CHAPTER 15: ACIDS AND BASES
(c)
327
+
For part (c), it is probably easiest to calculate the [H ] from the ion product of water.
The ion product of water is applicable to all aqueous solutions. At 25°C,
Kw = 1.0 × 10
−14
−
+
= [H ][OH ]
+
Rearrange the equation to solve for [H ].
+
[H ] =
15.21
pH
<7
>7
=7
+
[H ]
1.0 × 10 −14
−
[OH ]
10
. × 10 −14
=
3.7 × 10
−9
= 2.7 × 10
−6
M
Solution is:
acid
basic
neutral
−7
> 1.0 × 10 M
−7
< 1.0 × 10 M
−7
= 1.0 × 10 M
15.22
(a) acidic
(b) neutral
(c) basic
15.23
The pH can be found using Equation (15.5) of the text.
pH = 14.00 − pOH = 14.00 − 9.40 = 4.60
The hydrogen ion concentration can be found as in Example 15.4 of the text.
+
4.60 = −log[H ]
Taking the antilog of both sides:
+
[H ] = 2.5 × 10
15.24
5.50 mL
FG 1 L IJ FG 0.360 mol IJ
H 1000 mL K H 1 L K
−5
M
= 1.98 × 10 −3 mol KOH
−
KOH is a strong base and therefore ionizes completely. The OH concentration equals the KOH
−
concentration, because there is a 1:1 mole ratio between KOH and OH .
−
[OH ] = 0.360 M
−
pOH = −log[OH ] = 0.444
15.25
−
We can calculate the OH concentration from the pOH.
pOH = 14.00 − pH = 14.00 − 10.00 = 4.00
−
[OH ] = 10
−pOH
= 1.0 × 10
−4
M
−
Since NaOH is a strong base, it ionizes completely. The OH concentration equals the initial concentration of
NaOH.
[NaOH] = 1.0 × 10
−4
mol/L
328
CHAPTER 15: ACIDS AND BASES
So, we need to prepare 546 mL of 1.0 × 10
−4
M NaOH.
This is a factor-label problem. We need to perform the following unit conversions.
mol/L → mol NaOH → grams NaOH
546 mL = 0.546 L
? g NaOH
15.26
 1.0 ×10−4 mol NaOH 
 40.00 g NaOH 
= 
 × (0.546 L soln) × 



1 L soln
 1 mol NaOH 


−3
= 2.2 × 10 g NaOH
Molarity of the HCl solution is:
 1 mol HCl 
 1000 mL 
1
18.4 g HCl 


 = 0.762 M
36.46
g
HCl
662
mL


 1 L 
pH = −log(0.762) = 0.118
15.31
(a)
(d)
(g)
strong acid,
weak acid,
strong acid,
(b)
(e)
(h)
weak acid,
weak acid,
weak acid,
(c)
(f)
(i)
15.32
(a) strong base
15.33
The maximum possible concentration of hydrogen ion in a 0.10 M solution of HA is 0.10 M. This is the case
if HA is a strong acid. If HA is a weak acid, the hydrogen ion concentration is less than 0.10 M. The pH
+
+
corresponding to 0.10 M [H ] is 1.00. (Why three digits?) For a smaller [H ] the pH is larger than 1.00
(why?).
(b) weak base
strong acid (first stage of ionization),
weak acid,
weak acid.
(c) weak base
false, they are equal
(e) strong base
(a)
false, the pH is greater than 1.00
15.34
(a)
(c)
false, they are equal
true
15.35
The direction should favor formation of F (aq) and H2O(l). Hydroxide ion is a stronger base than fluoride
ion, and hydrofluoric acid is a stronger acid than water.
15.36
Cl is the conjugate base of the strong acid, HCl. It is a negligibly weak base and has no affinity for protons.
Therefore, the reaction will not proceed from left to right to any measurable extent.
(b)
(d)
(b)
(d) weak base
(c)
true
(d)
false
true, find the value of log(1.00) on your calculator
false, if the acid is strong, [HA] = 0.00 M
−
−
Another way to think about this problem is to consider the possible products of the reaction.
−
−
CH3COOH (aq) + Cl (aq) → HCl (aq) + CH3COO (aq)
The favored reaction is the one that proceeds from right to left. HCl is a strong acid and will ionize
−
completely, donating all its protons to the base, CH3COO .
CHAPTER 15: ACIDS AND BASES
15.41
329
We set up a table for the dissociation
C6H5COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
Ka =
+
U
H (aq)
0.10
−x
(0.10 − x)
−
+
C6H5COO (aq)
0.00
+x
x
0.00
+x
x
[H + ][C6 H5COO − ]
x2
=
= 6.5 × 10−5
[C6 H5COOH]
(0.10 − x)
−5
2
−6
x + (6.5 × 10 )x − (6.5 × 10 ) = 0
Solving the quadratic equation:
x = 2.5 × 10
−3
+
M = [H ]
−3
pH = −log(2.5 × 10 ) = 2.60
This problem could be solved more easily if we could assume that (0.10 − x) ≈ 0.10. If the assumption is
mathematically valid, then it would not be necessary to solve a quadratic equation, as we did above.
Re-solve the problem above, making the assumption. Was the assumption valid? What is our criterion for
deciding?
15.42
Step 1: Calculate the concentration of acetic acid before ionization.
0.0560 g acetic acid ×
1 mol acetic acid
−4
= 9.33 × 10 mol acetic acid
60.05 g acetic acid
9.33 × 10 −4 mol
= 0.0187 M acetic acid
0.0500 L soln
Step 2: Next, recognize that acetic acid is a weak, monoprotic acid. It is not one of the six strong acids, so it
must be a weak acid. Express the equilibrium concentrations of all species in terms of initial
concentrations and a single unknown x, that represents the change in concentration. Let (−x) be the
depletion in concentration (mol/L) of CH3COOH. From the stoichiometry of the reaction, it follows
−
+
that the increase in concentration for both H and CH3COO must be x. Complete a table that lists
the initial concentrations, the change in concentrations, and the equilibrium concentrations.
CH3COOH (aq)
Initial (M):
Change (M):
Equilibrium (M):
0.0187
−x
0.0187 − x
U
+
−
H (aq) + CH3COO (aq)
0
+x
x
0
+x
x
Step 3: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the
value of the equilibrium constant (Ka), solve for x.
Ka =
[ H + ][ CH 3COO − ]
[CH 3COOH ]
330
CHAPTER 15: ACIDS AND BASES
You can look up the Ka value for acetic acid in Table 15.3 of your text.
1.8 × 10
−5
=
( x )( x )
(0.0187 − x )
At this point, we can make an assumption that x is very small compared to 0.0187. Hence,
0.0187 − x ≈ 0.0187
1.8 × 10
−5
=
( x )( x )
0.0187
Solving for x.
x = 5.8 × 10
−4
−
+
M = [H ] = [CH3COO ]
−4
[CH3COOH] = (0.0187 − 5.8 × 10 )M = 0.0181 M
Checking the validity of the assumption,
5.8 × 10 −4
× 100% = 3.1% < 5%
0.0187
The assumption is valid.
15.43
First we find the hydrogen ion concentration.
+
[H ] = 10
−pH
= 10
−6.20
= 6.3 × 10
−7
M
Substituting into the acid ionization constant expression:
[H + ][A − ]
(6.3 × 10−7 )(6.3 × 10−7 )
−11
=
= 4.0 × 10
−
7
[HA]
(0.010 − 6.3 × 10 )
Ka =
+
We have omitted the contribution to [H ] due to water.
15.44
+
A pH of 3.26 corresponds to a [H ] of 5.5 × 10
that:
−4
HCOOH (aq)
Initial (M):
Change (M):
Equilibrium (M):
U
x
−4
−5.5 × 10
x − 5.5 × 10
M. Let the original concentration of formic acid be x so
−4
−
+
H (aq) + HCOO (aq)
0
−4
5.5 × 10
0
−4
5.5 × 10
5.5 × 10
5.5 × 10
−4
−4
Substitute Ka and the equilibrium concentrations into the ionization constant expression to solve for x.
[ H + ][ HCOO − ]
= 17
. × 10 −4
[ HCOOH ]
(5.5 ×10−4 ) 2
( x − 5.5 ×10−4 )
= 1.7 × 10−4
x = [HCOOH] = 2.3 × 10
−3
M
CHAPTER 15: ACIDS AND BASES
15.45
(a)
331
Set up a table showing initial and equilibrium concentrations.
U
C6H5COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
0.20
−x
(0.20 − x)
+
H (aq)
+
0.00
+x
x
−
C6H5COO (aq)
0.00
+x
x
Using the value of Ka from Table 15.3 of the text:
Ka =
[H + ][C6 H5COO − ]
x2
=
= 6.5 × 10−5
[C6 H5COOH]
(0.20 − x)
We assume that x is small so (0.20 − x) ≈ 0.20
x2
= 6.5 × 10 −5
0.20
x = 3.6 × 10
−3
Percent ionization =
(b)
−
+
M = [H ] = [C6H5COO ]
3.6 × 10 −3 M
× 100% = 1.8%
0.20 M
Set up a table as above.
C6H5COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
U
0.00020
−x
(0.00020 − x)
+
H (aq)
0.00000
+x
x
+
−
C6H5COO (aq)
0.00000
+x
x
Using the value of Ka from Table 15.3 of the text:
Ka =
[H + ][C6 H5COO − ]
x2
=
= 6.5 × 10−5
[C6 H5COOH]
(0.00020 − x)
In this case we cannot apply the approximation that (0.00020 − x) ≈ 0.00020 (see the discussion in
Example 15.8 of the text). We obtain the quadratic equation:
2
−5
−8
x + (6.5 × 10 )x − (1.3 × 10 ) = 0
−5
The positive root of the equation is x = 8.6 × 10 M. (Is this less than 5% of the original concentration,
0.00020 M? That is, is the acid more than 5% ionized?) The percent ionization is then:
Percent ionization =
8.6 × 10 −5 M
× 100% = 43%
0.00020 M
332
CHAPTER 15: ACIDS AND BASES
15.46
Percent ionization is defined as:
percent ionization =
ionized acid concentration at equilibrium
× 100%
initial concentration of acid
For a monoprotic acid, HA, the concentration of acid that undergoes ionization is equal to the concentration
−
+
of H ions or the concentration of A ions at equilibrium. Thus, we can write:
percent ionization =
(a)
[H + ]
× 100%
[ HA ]0
First, recognize that hydrofluoric acid is a weak acid. It is not one of the six strong acids, so it must be a
weak acid.
Step 1: Express the equilibrium concentrations of all species in terms of initial concentrations and a single
unknown x, that represents the change in concentration. Let (−x) be the depletion in concentration
(mol/L) of HF. From the stoichiometry of the reaction, it follows that the increase in concentration
−
+
for both H and F must be x. Complete a table that lists the initial concentrations, the change in
concentrations, and the equilibrium concentrations.
HF (aq)
Initial (M):
Change (M):
Equilibrium (M):
0.60
−x
0.60 − x
U
+
−
H (aq) + F (aq)
0
+x
x
0
+x
x
Step 2: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the
value of the equilibrium constant (Ka), solve for x.
Ka =
[ H + ][ F − ]
[ HF]
You can look up the Ka value for hydrofluoric acid in Table 15.3 of your text.
7.1 × 10
−4
=
( x )( x )
(0.60 − x )
At this point, we can make an assumption that x is very small compared to 0.60. Hence,
0.60 − x ≈ 0.60
Oftentimes, assumptions such as these are valid if K is very small. A very small value of K means that a very
small amount of reactants go to products. Hence, x is small. If we did not make this assumption, we would
have to solve a quadratic equation.
7.1 × 10
−4
=
( x )( x )
0.60
Solving for x.
+
x = 0.021 M = [H ]
CHAPTER 15: ACIDS AND BASES
333
+
Step 3: Having solved for the [H ], calculate the percent ionization.
percent ionization =
=
[H + ]
× 100%
[ HF]0
0.021 M
× 100% = 3.5%
0.60 M
(b) − (c) are worked in a similar manner to part (a). However, as the initial concentration of HF becomes
smaller, the assumption that x is very small compared to this concentration will no longer be valid. You must
solve a quadratic equation.
(b)
Ka =
x2
[H + ][F− ]
=
= 7.1× 10−4
[HF]
(0.0046 − x)
−4
2
−6
x + (7.1 × 10 )x − (3.3 × 10 ) = 0
x = 1.5 × 10
−3
M
Percent ionization =
(c)
Ka =
1.5 × 10−3 M
× 100% = 33%
0.0046 M
x2
[H + ][F− ]
=
= 7.1× 10−4
[HF]
(0.00028 − x)
−4
2
−7
x + (7.1 × 10 )x − (2.0 × 10 ) = 0
x = 2.2 × 10
−4
M
Percent ionization =
2.2 × 10−4 M
× 100% = 79%
0.00028 M
As the solution becomes more dilute, the percent ionization increases.
15.47
Given 14% ionization, the concentrations must be:
−
+
[H ] = [A ] = 0.14 × 0.040 M = 0.0056 M
[HA] = (0.040 − 0.0056) M = 0.034 M
The value of Ka can be found by substitution.
Ka =
15.48
C9H8O4 (aq)
(a)
[H + ][A − ] (0.0056)2
=
= 9.2 × 10−4
[HA]
0.034
U H+ (aq) + C9H7O4−
K a = 3.0 × 10−4 =
[H + ][C9 H 7 O 4 − ]
x2
=
[C9 H8O 4 ]
(0.20 − x)
334
CHAPTER 15: ACIDS AND BASES
Assuming (0.20 − x) ≈ 0.20
+
x = [H ] = 7.7 × 10
−3
[C 9 H 7 O 4 − ]
× 100%
[C 9 H 8O 4 ]
Percent ionization =
7.7 × 10 −3 M
× 100% = 3.9%
0.20 M
=
(b)
M
At pH 1.00 the concentration of hydrogen ion is 0.10 M (Why only two significant figures?) This will
tend to suppress the ionization of the weak acid (LeChatelier's principle, Section 14.5). The extra
hydrogen ion shifts the position of equilibrium in the direction of the un-ionized acid, and to twosignificant-figure accuracy, we can safely ignore the contribution of the weak acid to the total hydrogen
ion concentration. The percent ionization of the acid is then
Percent ionization =
[C 9 H 7 O 4 − ]
× 100%
[C 9 H 8O 4 ]
and,
Ka =
[H + ][C9 H 7 O 4 − ]
[C9 H8O 4 ]
Ka
or
[H + ]
=
[C9 H 7 O 4− ]
[C9 H8O 4 ]
Substituting into the equation for percent ionization gives
percent ionization =
=
Ka
[H + ]
× 100%
3.0 × 10 −4
× 100% = 0.30%
0.10
The high acidity of the gastric juices appears to enhance the rate of absorption of unionized aspirin
molecules through the stomach lining. In some cases this can irritate these tissues and cause bleeding.
15.51
(a)
We construct the usual table.
NH3(aq)
Initial (M):
Change (M):
Equilibrium (M):
+
H2O(l)
0.10
−x
(0.10 − x)
Kb =
U
+
NH4 (aq)
0.00
+x
x
[NH 4 + ][OH − ]
x2
=
= 1.8 × 10−5
[NH3 ]
(0.10 − x)
Assuming (0.10 − x) ≈ 0.10, we have:
x2
= 1.8 × 10−5
0.10
x = 1.3 × 10
−3
−
M = [OH ]
+
−
OH (aq)
0.00
+x
x
CHAPTER 15: ACIDS AND BASES
−3
pOH = −log(1.3 × 10 ) = 2.89
pH = 14.00 − 2.89 = 11.11
By following the identical procedure, we can show: (b) pH = 8.96.
15.52
−
Step 1: Calculate the pOH from the pH. Then, calculate the OH concentration from the pOH.
pOH = 14.00 − pH = 14.00 − 10.66 = 3.34
−
or,
pOH = −log[OH ]
−
−pOH = log[OH ]
Taking the antilog of both sides of the equation,
10
−pOH
−
= [OH ]
−
[OH ] = 10
−3.34
= 4.6 × 10
−4
M
Step 2: Complete a table that lists the initial concentrations, the change in concentrations, and the
equilibrium concentrations. Let B represent the weak base.
Initial (M):
Change (M):
Equilibrium (M):
−
+
U
B (aq) + H2O (l)
0.30
−x
0.30 − x
BH (aq) + OH (aq)
0
+x
x
0
+x
x
In this case, x is known.
−
+
x = [BH ] = [OH ] = 4.6 × 10
−4
M
−4
[B] = 0.30 − x = (0.30 − 4.6 × 10 )M = 0.30 M
Step 3: Substitute the equilibrium concentrations into the ionization constant expression to solve for Kb.
15.53
Kb =
[ BH + ][ OH − ]
[ B]
Kb =
(4.6 × 10 −4 ) 2
−7
= 7.1 × 10
(0.30)
+
A pH of 11.22 corresponds to a [H ] of 6.03 × 10
The equilibrium is:
NH3(aq) + H2O(l)
−
−12
−
M and a [OH ] of 1.66 × 10
−3
M.
U NH4+(aq) + OH−(aq)
+
The concentration of [OH ] = [NH4 ] (why?) If we let x equal the original concentration of ammonia
K b = 1.8 × 10−5 =
(1.66 × 10−3 )(1.66 × 10−3 )
x − 1.66 × 10−3
335
336
CHAPTER 15: ACIDS AND BASES
Assuming 1.66 × 10
−3
is small relative to x, then
x = 0.15 M = [NH3]
15.54
The reaction is:
NH3 (aq) + H2O (l)
−
U NH4+ (aq) + OH− (aq)
+
Let [OH ] = [NH4 ] = x
At equilibrium we have:
[NH 4+ ][OH − ]
x2
x2
≈
=
= 1.8 ×10−5
[NH 3 ]
(0.080 − x)
0.080
x = 1.2 × 10
−3
M
+
Percent NH3 present as NH4 =
15.59
1.2 × 10 −3
× 100% = 1.5%
0.080
If Ka1 >> Ka2, we can assume that the equilibrium concentration of hydrogen ion results only from the first
stage of ionization. In the second stage this always leads to an expression of the type:
(c + y )( y )
= K a2
(c − y )
where c represents the equilibrium hydrogen ion concentration found in the first stage. If c >> Ka2, we can
assume (c ± y) ≈ c, and consequently y = Ka2.
Is this conclusion also true for the second stage ionization of a triprotic acid like H3PO4?
15.60
The pH of a 0.040 M HCl solution (strong acid) is 1.40. Follow the procedure for calculating the pH of a
diprotic acid to calculate the pH of the sulfuric acid solution.
+
Step 1: H2SO4 is a strong acid. The first ionization stage goes to completion. Therefore, the H ion
−
concentration and the HSO4 concentration from the first ionization stage equal the initial
concentration of sulfuric acid, 0.040 M.
−
+
H2SO4 (aq) → H (aq) + HSO4 (aq)
−
Step 2: Now, consider the second stage of ionization. HSO4 is a weak acid. Set up a table showing the
concentrations for the second ionization stage. Let x be the change in concentration.
−
HSO4 (aq)
Initial (M):
Change (M):
Equilibrium (M):
0.040
−x
0.040 − x
U
+
H (aq) + SO4
0.040
+x
0.040 + x
2−
0
+x
x
(aq)
CHAPTER 15: ACIDS AND BASES
337
Step 3: Write the ionization constant expression for Ka. Then, solve for x.
Ka =
[H + ][SO 4 2− ]
[HSO 4− ]
1.3 × 10
−2
(0.040 + x )( x )
(0.040 − x)
=
Since Ka is quite large, we cannot make the assumptions that
0.040 − x ≈ 0.040
0.040 + x ≈ 0.040
and
Therefore, we must solve a quadratic equation.
2
x + 0.053x − 5.2 × 10
x =
x =
−0.053 ±
−4
= 0
(0.053) 2 − 4(1)( −5.2 × 10 −4 )
2(1)
−0.053 ± 0.070
2
x = 8.5 × 10
−3
M
x = −0.062 M
or
The second solution is physically impossible because you cannot have a negative concentration. The first
solution is the correct answer.
+
Step 4: Having solved for x, we can calculate the H concentration after equilibrium for the second stage of
+
ionization is reached. Then, we can calculate the pH from the H concentration.
−3
+
[H ] = 0.040 M + x = (0.040 + 8.5 × 10 )M = 0.049 M
pH = −log(0.049) = 1.31
Without doing any calculations, could you have known that the pH of the sulfuric acid would be lower (more
acidic) than that of the hydrochloric acid?
15.61
−
There is no H2SO4 in the solution because HSO4 has no tendency to accept a proton to produce H2SO4.
(Why?) We are only concerned with the ionization
−
HSO4 (aq)
Initial (M):
Change (M):
Equilibrium (M):
+
U
H (aq)
0.20
−x
(0.20 − x)
0.00
+x
+x
K a = 1.3 × 10−2 =
[H + ][SO 4 2− ]
−
[HSO 4 ]
=
( x)( x)
(0.20 − x)
Solving the quadratic equation:
+
2−
−
x = [H ] = [SO4 ] = 0.045 M
−
+
[HSO4 ] = (0.20 − 0.045) M = 0.16 M
2−
SO4 (aq)
0.00
+x
+x
338
CHAPTER 15: ACIDS AND BASES
15.62
For the first stage of ionization:
U
H2CO3 (aq)
Initial (M):
Change (M):
Equilibrium (M):
Ka =
1
0.025
−x
(0.025 − x)
+
H (aq)
−
+
HCO3 (aq)
0.00
+x
x
0.00
+x
x
[H + ][HCO3− ]
x2
=
= 4.2 × 10 −7
[H 2 CO3 ]
(0.025 − x)
Assuming (0.025 − x) ≈ 0.025,
x = 1.0 × 10
−4
M
For the second ionization,
−
HCO3 (aq)
Ka2 =
U H+ (aq) + CO32− (aq)
[H + ][CO32− ]
−
[HCO3 ]
= 4.8 ×10−11
−
Since HCO3 is a very weak acid, there is little ionization at this stage. Therefore we have:
+
−
[H ] = [HCO3 ] = 1.0 × 10
−4
2−
−
M and [CO3 ] = 4.8 × 10
−11
M
15.65
The strength of the H−X bond is the dominant factor in determining the strengths of binary acids. As with the
hydrogen halides (see Section 15.9 of the text), the H−X bond strength decreases going down the column in
Group 6A. The compound with the weakest H−X bond will be the strongest binary acid: H2Se > H2S > H2O.
15.66
All the listed pairs are oxoacids that contain different central atoms whose elements are in the same group of the
periodic table and have the same oxidation number. In this situation the acid with the most electronegative
central atom will be the strongest.
15.67
(a)
H2SO4 > H2SeO4.
(b)
H3PO4 > H3AsO4
The CH2ClCOOH is a stronger acid than CH3COOH. Chlorine is more electronegative than hydrogen and
attracts electrons toward itself, making the O−H bond more polar. The hydrogen atom in CH2ClCOOH is
more easily ionized.
CHAPTER 15: ACIDS AND BASES
15.68
−
−
339
−
The conjugate bases are C6H5O from phenol and CH3O from methanol. The C6H5O is stabilized by
resonance:
O−
O
O
O
−
−
−
−
The CH3O ion has no such resonance stabilization. A more stable conjugate base means an increase in the
strength of the acid.
15.75
There are two possibilities: (i) MX is the salt of a strong acid and a strong base so that neither the cation nor
the anion react with water to alter the pH and (ii) MX is the salt of a weak acid and a weak base with Ka for
the acid equal to Kb for the base. The hydrolysis of one would be exactly offset by the hydrolysis of the other.
15.76
There is an inverse relationship between acid strength and conjugate base strength. As acid strength
decreases, the proton accepting power of the conjugate base increases. In general the weaker the acid, the
stronger the conjugate base. All three of the potassium salts ionize completely to form the conjugate base of
the respective acid. The greater the pH, the stronger the conjugate base, and therefore, the weaker the acid.
The order of increasing acid strength is HZ < HY < HX.
15.77
+
The salt sodium acetate completely dissociates upon dissolution, producing 0.36 M [Na ] and 0.36 M
−
−
[CH3COO ] ions. The [CH3COO ] ions will undergo hydrolysis because they are a weak base.
−
CH3COO (aq)
Initial (M):
Change (M):
Equilibrium (M):
+
H2O(l)
0.36
−x
(0.36 − x)
Kb =
U
CH3COOH(aq)
+
0.00
+x
+x
[CH3COOH][OH − ]
[CH3COO − ]
=
−
OH (aq)
0.00
+x
+x
x2
= 5.6 × 10−10
(0.36 − x)
Assuming (0.36 − x) ≈ 0.36, then
−
x = [OH ] = 1.4 × 10
−5
−5
pOH = −log(1.4 × 10 ) = 4.85
pH = 14.00 − 4.85 = 9.15
15.78
+
−
The salt ammonium chloride completely ionizes upon dissolution, producing 0.42 M [NH4 ] and 0.42 M [Cl ]
+
+
ions. NH4 will undergo hydrolysis because it is a weak acid (NH4 is the conjugate acid of the weak base, NH3).
Step 1: Express the equilibrium concentrations of all species in terms of initial concentrations and a single
unknown x, that represents the change in concentration. Let (−x) be the depletion in concentration
+
(mol/L) of NH4 . From the stoichiometry of the reaction, it follows that the increase in
340
CHAPTER 15: ACIDS AND BASES
+
concentration for both H3O and NH3 must be x. Complete a table that lists the initial
concentrations, the change in concentrations, and the equilibrium concentrations.
+
NH4 (aq) + H2O (l)
Initial (M):
Change (M):
Equilibrium (M):
U
+
NH3 (aq) + H3O (aq)
0.42
−x
(0.42 − x)
0.00
+x
x
0.00
+x
x
+
Step 2: You can calculate the Ka value for NH4 from the Kb value of NH3. The relationship is
Ka × Kb = Kw
or
Ka =
Kw
1.0 × 10−14
−10
= 5.6 × 10
=
5
−
Kb
1.8 × 10
Step 3: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the
value of the equilibrium constant (Ka), solve for x.
Ka =
[NH 3 ][H 3O + ]
[NH 4 + ]
=
x2
= 5.6 × 10−10
0.42 − x
Assuming (0.42 − x) ≈ 0.42,
+
x = [H ] = 1.5 × 10
−5
M
−5
pH = −log(1.5 × 10 ) = 4.82
Since NH4Cl is the salt of a weak base (aqueous ammonia) and a strong acid (HCl), we expect the solution to
be slightly acidic, which is confirmed by the calculation.
15.79
HCO3
−
U H+ + CO32−
−
HCO3 + H2O
Ka = 4.8 × 10
U H2CO3 + OH−
Kb =
−11
Kw
10
. × 10 −14
=
= 2.4 × 10 −8
−7
Ka
4.2 × 10
−
HCO3 has a greater tendency to hydrolyze than to ionize (Kb > Ka). The solution will be basic (pH > 7).
15.80
The acid and base reactions are:
acid:
HPO4
2−
(aq)
base:
HPO4
2−
(aq) + H2O (l)
2−
U H+ (aq) + PO43− (aq)
−13
U H2PO4− (aq) + OH− (aq)
−
2−
−7
Ka for HPO4 is 4.8 × 10 . Note that HPO4 is the conjugate base of H2PO4 , so Kb is 1.6 × 10 .
Comparing the two K's, we conclude that the monohydrogen phosphate ion is a much stronger proton acceptor
(base) than a proton donor (acid). The solution will be basic.
15.83
Metal ions with high oxidation numbers are unstable. Consequently, these metals tend to form covalent bonds
(rather than ionic bonds) with oxygen. Covalent metal oxides are acidic while ionic metal oxides are basic.
2−
The latter oxides contain the O ion which reacts with water as follows:
O
2−
+ H2O → 2OH
−
CHAPTER 15: ACIDS AND BASES
15.84
15.85
15.86
341
The most basic oxides occur with metal ions having the lowest positive charges (or lowest oxidation
numbers).
(a)
Al2O3 < BaO < K2O
(b)
CrO3 < Cr2O3 < CrO
(a)
2HCl(aq) + Zn(OH)2(s) → ZnCl2(aq) + 2H2O(l)
(b)
2OH (aq) + Zn(OH)2(s) → Zn(OH)4 (aq)
−
2−
Al(OH)3 is an amphoteric hydroxide. The reaction is:
−
−
Al(OH)3(s) + OH (aq) → Al(OH)4 (aq)
This is a Lewis acid-base reaction. Can you identify the acid and base?
15.89
15.90
(a)
Lewis acid; see the reaction with water shown in Section 15.12 of the text.
(b)
Lewis base; water combines with H to form H3O .
(c)
Lewis base.
(d)
Lewis acid; SO2 reacts with water to form H2SO3. Compare to CO2 above. Actually, SO2 can also act
as a Lewis base under some circumstances.
(e)
Lewis base; see the reaction with H to form ammonium ion.
(f)
Lewis base; see the reaction with H to form water.
(g)
Lewis acid; does H have any electron pairs to donate?
(h)
Lewis acid; compare to the example of NH3 reacting with BF3.
+
+
+
+
+
−
AlCl3 is a Lewis acid with an incomplete octet of electrons and Cl is the Lewis base donating a pair of
electrons.
Cl
Al
Cl
15.91
-
Cl
Cl
+
-
Cl
Cl
Al
Cl
Cl
(a)
Both molecules have the same acceptor atom (boron) and both have exactly the same structure (trigonal
planar). Fluorine is more electronegative than chlorine so we would predict based on electronegativity
arguments that boron trifluoride would have a greater affinity for unshared electron pairs than boron
trichloride.
(b)
Since it has the larger positive charge, iron(III) should be a stronger Lewis acid than iron(II).
15.92
By definition Brønsted acids are proton donors, therefore such compounds must contain at least one hydrogen
atom. In Problem 15.89, Lewis acids that do not contain hydrogen, and therefore are not Brønsted acids, are
CO2, SO2, and BCl3. Can you name others?
15.93
The ionization of any acid is an endothermic process. The higher the temperature, the greater the Ka value.
Formic acid will be a stronger acid at 40°C than at 25°C.
342
CHAPTER 15: ACIDS AND BASES
15.94
We first find the number of moles of CO2 produced in the reaction:
0.350 g NaHCO 3
VCO2 =
nCO2 RT
P
FG 1 mol NaHCO 3 IJ FG 1 mol CO2 IJ
H 84.01 g NaHCO 3 K H 1 mol NaHCO3 K
=
= 4.17 × 10 −3 mol CO 2
(4.17 × 10−3 mol)(0.0821 L ⋅ atm / K ⋅ mol)(37.0 + 273) K
= 0.106 L
(1.00 atm)
15.95
Choice (c) because 0.70 M KOH has a higher pH than 0.60 M NaOH. Adding an equal volume of 0.60 M
−
NaOH lowers the [OH ] to 0.65 M, hence lowering the pH.
15.96
If we assume that the unknown monoprotic acid is a strong acid that is 100% ionized, then the [H ]
concentration will be 0.0642 M.
+
pH = −log (0.0642) = 1.19
Since the actual pH of the solution is higher, the acid must be a weak acid.
15.97
+
(a)
For the forward reaction NH4 and NH3 are the conjugate acid and base pair, respectively. For the
−
reverse reaction NH3 and NH2 are the conjugate acid and base pair, respectively.
(b)
H corresponds to NH4 ; OH corresponds to NH2 . For the neutral solution, [NH4 ] = [NH2 ].
+
+
−
−
+
−
15.98
The reaction of a weak acid with a strong base is driven to completion by the formation of water. Irrespective
of whether the strong base is reacting with a strong monoprotic acid or a weak monoprotic acid, the same
number of moles of acid is required to react with a constant number of moles of base. Therefore the volume
of base required to react with the same concentration of acid solutions (either both weak, both strong, or one
strong and one weak) will be the same.
15.99
Ka =
[H + ][A − ]
[HA]
[HA] ≈ 0.1 M
−
[A ] ≈ 0.1 M
Therefore,
+
Ka = [H ] =
Kw
[OH − ]
−
[OH ] = Kw/Ka
15.100 High oxidation state leads to covalent compounds and low oxidation state leads to ionic compounds.
Therefore, CrO is ionic and basic and CrO3 is covalent and acidic.
CHAPTER 15: ACIDS AND BASES
U HCOO− + H+
15.101 HCOOH
+
H + OH
'
=
Kw
−
Ka = 1.7 × 10
U H2 O
343
−4
K’w =
1
1
=
K w 1.0 × 10 −14
−
HCOOH + OH
U HCOO− + H2O
1
−4 
'
= (1.7 × 10 ) 
K = Ka K w
 1.0 × 10−14


10
 = 1.7 × 10


15.102 We can write two equilibria that add up to the equilibrium in the problem.
CH3COOH (aq)
−
+
[H + ][CH 3COO − ]
−5
= 1.8 × 10
[CH 3COOH ]
1
1
3
=
Ka' =
= 2.2 × 10
K a ( HNO 2 ) 4.5 × 10 −4
U H+ (aq) + CH3COO− (aq)
H (aq) + NO2 (aq)
Ka =
U HNO2 (aq)
Ka' =
−
CH3COOH (aq) + NO2 (aq)
U CH3COO− (aq) + HNO2 (aq)
[HNO 2 ]
[H + ][ NO 2− ]
K =
[CH 3COO − ][HNO 2 ]
[CH 3COOH ][ NO 2− ]
= K a × K a'
The equilibrium constant for this sum is the product of the equilibrium constants of the component reactions.
−5
3
K = Ka × K a' = (1.8 × 10 )(2.2 × 10 ) = 4.0 × 10
15.103 (a)
(b)
−
H
base1
+
H2 O
acid2
→
−
OH
base2
+
−2
H2
acid1
−
H is the reducing agent and H2O is the oxidizing agent.
15.104 In this specific case the Ka of ammonium ion is the same as the Kb of acetate ion. The two are of exactly (to
two significant figures) equal strength. The solution will have pH 7.00.
What would the pH be if the concentration were 0.1 M in ammonium acetate? 0.4 M?
15.105 Kb = 8.91 × 10
Ka =
−6
Kw
= 1.1 × 10 −9
Kb
pH = 7.40
+
[H ] = 10
Ka =
−7.40
= 3.98 × 10
−8
[H + ][conjugate base]
[acid]
344
CHAPTER 15: ACIDS AND BASES
Therefore,
K
[conjugate base]
1.1 × 10−9
= a =
= 0.028
[acid]
[H + ] 3.98 × 10−8
15.106 The fact that fluorine attracts electrons in a molecule more strongly than hydrogen should cause NF3 to be a
poor electron pair donor and a poor base. NH3 is the stronger base.
+
15.107 Because the P−H bond is weaker, there is a greater tendency for PH4 to ionize. Therefore, PH3 is a weaker
base than NH3.
15.108 The autoionization for deuterium-substituted water is: D2O
−
+
[D ][OD ] = 1.35 × 10
U D+ + OD−
−15
(1)
+
pD = −log[D ] = −log 1.35 × 10 −15 = 7.43
(a)
The definition of pD is:
(b)
To be acidic, the pD must be < 7.43.
(c)
Taking −log of both sides of equation (1) above:
−
+
−log[D ] + −log[OD ] = −log(1.35 × 10
−15
)
pD + pOD = 14.87
15.109 (a)
(f)
HNO2
(b) HF
−
2−
HCO3 and CO3
(c)
−
(d)
BF3
NH3
(e)
H2SO3
+
The reactions for (f) are: HCO3 (aq) + H (aq) → CO2(g) + H2O(l)
2−
+
CO3 (aq) + 2H (aq) → CO2(g) + H2O(l)
15.110 First we must calculate the molarity of the trifluoromethane sulfonic acid. (Molar mass = 150.1 g)
Molarity = 0.616 g ×
1 mol
1
×
= 0.0164 M
150.1 g 0.250 L
+
Since trifluoromethane sulfonic acid is a strong acid and is 100% ionized, the [H ] is 0.0165 M.
pH = − log(0.0164) = 1.79
15.111 (a)
+
The Lewis structure of H3O is:
H
O
H
+
H
Note that this structure is very similar to the Lewis structure of NH3. The geometry is pyramidal.
CHAPTER 15: ACIDS AND BASES
(b)
2+
+
345
+
H4O does not exist because the positively charged H3O has no affinity to accept the positive H ion.
2+
If H4O existed, it would have a tetrahedral geometry.
15.112 The reactions are
HF
U H+ + F−
−
F + HF
(1)
U HF2−
(2)
Note that for equation (2), the equilibrium constant is relatively large with a value of 5.2. This means that the
−
equilibrium lies to the right. Applying Le Chatelier’s principle, as HF ionizes in the first step, the F that is
produced is partially removed in the second step. More HF must ionize to compensate for the removal of the
−
+
F , at the same time producing more H .
15.113 The equations are:
Cl2(g) + H2O(l)
U HCl(aq) + HClO(aq)
HCl(aq) + AgNO3(aq)
U AgCl(s) + HNO3(aq)
−
In the presence of OH ions, the first equation is shifted to the right:
+
H (from HCl) + OH
−

→ H2 O
−
Therefore, the concentration of HClO increases. (The ‘bleaching action’ is due to ClO ions.)
15.114 (a)
We must consider both the complete ionization of the strong acid, and the partial ionization of water.
−
+
→ H + A
HA 
H2 O
U H+ + OH−
+
From the above two equations, the [H ] in solution is:
−
+
−
[H ] = [A ] + [OH ]
We can also write:
−
+
[H ][OH ] = Kw
−
[OH ] =
Kw
[H + ]
Substituting into Equation (1):
+
Kw
−
[H ] = [A ] +
[H + ]
+ 2
−
+
+ 2
−
+
[H ] = [A ][H ] + Kw
[H ] − [A ][H ] − Kw = 0
(1)
346
CHAPTER 15: ACIDS AND BASES
Solving a quadratic equation:
+
[H ] =
(b)
[A − ] ± [A − ] 2 + 4 K w
2
For the strong acid, HCl, with a concentration of 1.0 × 10
−7
1 × 10 −7 ±
[Cl− ] ± [Cl− ]2 + 4 K w
+
=
[H ] =
2
+
[H ] = 1.6 × 10
−7
M (or −6.0 × 10
−8
−
M, the [Cl ] will also be 1.0 × 10
(1 × 10 )
−7
2
(
+ 4 1 × 10 −14
−7
M.
)
2
M, which is impossible)
−7
pH = −log[1.6 × 10 ] = 6.80
15.115 Given the equation:
+
U HbO2 + H+
HbH + O2
(a)
From the equilibrium equation, high oxygen concentration puts stress on the left side of the equilibrium
and thus shifts the concentrations to the right to compensate. HbO2 is favored.
(b)
High acid, H concentration, places stress on the right side of the equation forcing concentrations on the
+
left side to increase, thus releasing oxygen and increasing the concentration of HbH .
(c)
Removal of CO2 decreases H (in the form of carbonic acid), thus shifting the reaction to the right.
More HbO2 will form. Breathing into a paper bag increases the concentration of CO2 (re-breathing the
exhaled CO2), thus causing more O2 to be released as explained above.
+
+
15.116 The solution for the first step is standard:
H3PO4(aq)
Initial (M):
Change (M):
Equil. (M):
U
0.100
−x
(0.100 − x)
K a1 =
+
H (aq)
+
0.000
+x
x
−
H2PO4 (aq)
0.000
+x
x
[ H + ][ H 2 PO 4 − ]
x2
=
= 7.5 × 10 −3
0.100 − x
[ H 3PO 4 ]
b
g
In this case we probably cannot say that (0.100 − x) ≈ 0.100 due to the magnitude of Ka. We obtain the
quadratic equation:
−3
2
−4
x + (7.5 × 10 )x − (7.5 × 10 ) = 0
The positive root is x = 0.0239 M. We have:
+
−
[H ] = [H2PO4 ] = 0.0239 M
[H3PO4] = (0.100 − 0.0239) M = 0.076 M
CHAPTER 15: ACIDS AND BASES
For the second ionization:
−
Initial (M):
Change (M):
Equil (M):
+
U
H2PO4 (aq)
H (aq)
0.0239
−y
(0.0239 − y)
2−
+
HPO4 (aq)
0.0239
+y
0.0239 + y
[H + ][HPO 42 − ]
Ka 2 =
=
−
[H 2 PO 4 ]
0.000
+y
y
(0.0239 + y )( y )
(0.0239 − y )
= 6.2 × 10−8
Since Ka 2 is small, we can assume (0.0239 + y) ≈ (0.0239 − y), so that y = 6.2 × 10
−
+
[H ] = [H2PO4 ] = 0.0239 M
2−
[HPO4 ] = Ka 2 = 6.2 × 10
−8
M
We set up the problem for the third ionization in the same manner.
2−
Initial (M):
Change (M):
Equil. (M):
6.2 × 10
−z
(6.2 × 10
+
U
HPO4 (aq)
H (aq)
−8
−8
− z)
[H + ][PO 43− ]
Ka 3 =
[HPO4
2−
]
3−
+
PO4 (aq)
0.0239
+z
0
+z
0.0239 + z
z
=
(0.0239 + z )( z )
6.2 × 10
−8
−8
−z
= 4.8 × 10−13
We can assume (0.0239 + z) ≈ 0.0239 and (6.2 × 10 ) − z ≈ 6.2 × 10
Therefore,
(0.0239 )( z )
6.2 × 10−8
z = 1.2 × 10
= 4.8 × 10−13
−18
M
The equilibrium concentrations are:
−
+
[H ] = [H2PO4 ] = 0.0239 M
[H3PO4] = 0.076 M
2−
−
[HPO4 ] = 6.2 × 10
3−
−
[PO4 ] = 1.2 × 10
−8
−18
M
M
−8
−8
M. Thus,
347
348
CHAPTER 15: ACIDS AND BASES
15.117 (a)
Number of moles NaOH = M × vol (L) = 0.0568 M × 0.0138 L = 7.84 × 10
−4
mol
If the acid were all dimer, then:
mol of dimer =
mol NaOH
7.84 × 10 −4 mol
−4
=
= 3.92 × 10 mol
2
2
If the acetic acid were all dimer, the pressure that would be exerted would be:
nRT
(392
. × 10 −4 mol )( 0.0821 L ⋅ atm / K ⋅ mol )(324 K )
=
= 0.0290 atm
V
0.360 L
P =
However, the actual pressure is 0.0342 atm. If α mol of dimer dissociates to monomers, then 2α
monomer forms.
(CH3COOH)2
U 2CH3COOH
1−α
2α
The total moles of acetic acid is:
moles dimer + monomer = (1 − α) + 2α = 1 + α
Using partial pressures:
Pobserved = P(1 + α)
0.0342 atm = 0.0290 atm × (1 + α)
α = 0.179
(b)
The equilibrium constant is:
Kp =
2
PCH
3COOH
b
g
P CH COOH
3
2
=
FG 2α IJ 2 b Pobserved g2
H1 + αK
FG 1 − α IJ Pobserved
H1 + αK
+
15.118 0.100 M Na2CO3 → 0.200 M Na + 0.100 M CO3
First stage:
U HCO3−(aq)
0.100 M
xM
K1 =
1−α
2
2−
2−
CO3 (aq) + H2O(l )
4α 2 Pobserved
=
+
−
OH (aq)
x M
Kw
1.0 × 10−14
=
= 2.1 × 10−4
−11
K2
4.8 × 10
K1 = 2.1 × 10
x = 4.6 × 10
−4
−3
=
[ HCO 3 − ][OH − ]
[CO 32 − ]
−
=
−
M = [HCO3 ] = [OH ]
x2
0..100 − x
= 4.53 × 10 −3
CHAPTER 15: ACIDS AND BASES
−
Second stage:
HCO3 (aq)
+
−3
− y) M
(4.6 × 10
K2 =
H2CO3(aq)
+
−
OH (aq)
(4.6 × 10
yM
−3
+ y) M
[ H 2 CO 3 ][OH − ]
[ HCO 3 − ]
2.4 × 10−8 =
y = 2.4 × 10
At equilibrium:
U
H2O(l)
(
y 4.6 × 10−3 + y
4.6 × 10−3 − y
−8
)
M
+
[Na ] = 0.200 M
−
−3
−8
[HCO3 ] = (4.6 × 10 ) M − (2.4 × 10 ) M ≈ 4.6 × 10
[H2CO3] = 2.4 × 10
−
−8
+
−3
15.119 [CO2] = kP = (2.28 × 10
CO2(aq)
(7.30 × 10
+
−3
−3
H2O(l)
1.0 × 10 −14
4.6 × 10
−3
−8
= 2.2 × 10
−12
− x) M
+
H (aq)
+
xM
x2
7.30 × 10−3 − x
x = 5.5 × 10
−5
−3
M
M
mol/L⋅atm)(3.20 atm) = 7.30 × 10
U
M
M
[OH ] = (4.6 × 10 ) M + (2.4 × 10 ] M ≈ 4.6 × 10
[H ] =
−3
−3
M
−
HCO3 (aq)
xM
= 4.2 × 10 −7 ≈
x2
7.30 × 10−3
+
M = [H ]
pH = 4.26
15.120 When NaCN is treated with HCl, the following reaction occurs.
NaCN + HCl → NaCl + HCN
HCN is a very weak acid, and only partially ionizes in solution.
HCN (aq)
U H+(aq) + CN−(aq)
The main species in solution is HCN which has a tendency to escape into the gas phase.
HCN(aq)
U HCN(g)
Since the HCN(g) that is produced is a highly poisonous compound, it would be dangerous to treat NaCN
with acids without proper ventilation.
349
350
CHAPTER 15: ACIDS AND BASES
−4
15.121 When the pH is 10.00, the pOH is 4.00 and the concentration of hydroxide ion is 1.0 × 10 M. The
concentration of HCN must be the same. (Why?) If the concentration of NaCN is x, the table looks like:
−
U
CN (aq) + H2O(l)
x
−4
−1.0 × 10
Initial (M):
Change (M):
HCN(aq)
−4
[HCN][OH − ]
Kb =
−
[CN ]
x = 6.0 × 10
−4
=
OH (aq)
0
−4
+1.0 × 10
0
−4
+1.0 × 10
(1.0 × 10 )
(1.0 × 10 )
−4
(x − 1.0 × 10 )
Equilibrium (M):
−
+
(1.0 × 10−4 )2
−4
( x − 1.0 × 10 )
−4
= 2.0 × 10−5
−
M = [CN ]0
6.0 × 10−4 mol
49.01 g NaCN
× 0.250 L ×
= 7.4 × 10−3 g NaCN
1L
1 mol NaCN
Amount of NaCN =
+
15.122 pH = 2.53 = −log[H ]
+
[H ] = 2.95 × 10
−3
M
The equation representing the ionization of formic acid is:
U
HCOOH(aq)
−3
+
H (aq)
−
+
HCOO (aq)
−3
y − (2.95 × 10 )
−3
(2.95 × 10 )
[H + ][HCOO − ]
−4
=
Ka = 1.7 × 10 =
[HCOOH]
(2.95 × 10 )
(2.95 × 10 )
−3
2
( y − 2.95 × 10 )
−3
y = 0.054 M
There are 0.054 moles of formic acid in 1000 mL of solution. The mass of formic acid in 100 mL is:
0.054 mol formic acid
46.03 g formic acid
×
× 100 mL = 0.25 g formic acid
1000 mL soln
1 mol formic acid
15.123 The equilibrium is established:
CH3COOH(aq)
Initial (M):
Change (M):
Equilibrium (M)
Ka =
U CH3COO−(aq)
0.150
−x
(0.150 − x)
0
+x
x
b
+
+
H (aq)
0.100
+x
(0.100 + x)
g
.
x 0100
+x
[CH 3COO − ][H + ]
.
0100
x
. × 10 −5
=
=
= 18
[CH 3COOH ]
.
.
0150
0150
−x
x = 2.7 × 10
−5
+
M = [H ]
CHAPTER 15: ACIDS AND BASES
−5
351
+
2.7 × 10 M is the [H ] contributed by CH3COOH. HCl is a strong acid that completely ionizes. It
+
contributes a [H ] of 0.100 M to the solution.
−5
+
[H ]total = (0.100 + 2.7 × 10 ) M ≈ 0.100 M
pH = 1.000
The pH is totally determined by the HCl and is independent of the CH3COOH.
15.124 The balanced equation is:
mol of Mg =
Mg + 2HCl → MgCl2 + H2
g Mg
187
. g Mg
=
= 0.0769 mol
molar mass Mg
24.31 g / mol
From the balanced equation:
mol of HCl required for reaction = 2 × mol Mg = 2 × 0.0769 mol = 0.154 mol HCl
The concentration of HCl:
+
pH = −0.544, thus [H ] = 3.50 M
mol HCl = M × Vol (L) = 3.50 M × 0.0800 L = 0.280 mol HCl
Moles of HCl left after reaction:
total mol HCl − mol HCl reacted = 0.280 mol − 0.154 mol = 0.126 mol HCl
Molarity of HCl left after reaction:
pH = −0.20
M = mol/L = 0.126 mol/0.080 L = 1.58 M
15.125 (a)
The pH of the solution of HA would be lower. (Why?)
(b)
The electrical conductance of the HA solution would be greater. (Why?)
(c)
The rate of hydrogen evolution from the HA solution would be greater. Presumably, the rate of the
reaction between the metal and hydrogen ion would depend on the hydrogen ion concentration
(i.e., this would be part of the rate law). The hydrogen ion concentration will be greater in the HA
solution.
−
15.126 The important equation is the hydrolysis of NO2 :
−
NO2 + H2O
U HNO2 + OH−
−
+
(a)
Addition of HCl will result in the reaction of the H from the HCl with the OH that was present in the
−
solution. The OH will effectively be removed and the equilibrium will shift to the right to compensate
(more hydrolysis).
(b)
Addition of NaOH is effectively addition of more OH which places stress on the right hand side of the
equilibrium. The equilibrium will shift to the left (less hydrolysis) to compensate for the addition of
−
OH .
(c)
Addition of NaCl will have no effect.
(d)
Recall that the percent ionization of a weak acid increases with dilution (see Figure 15.4). The same is
true for weak bases. Thus dilution will cause more hydrolysis, shifting the equilibrium to the right.
−
352
CHAPTER 15: ACIDS AND BASES
15.127 Like carbon dioxide, sulfur dioxide behaves as a Lewis acid by accepting a pair of electrons from the Lewis
base water. The Lewis acid-base adduct rearranges to form sulfurous acid in a manner exactly analogous to
the rearrangement of the carbon dioxide-water adduct to form carbonic acid that is presented on page 637 of
the textbook.
15.128 In Chapter 11, we found that salts with their formal electrostatic intermolecular attractions had low vapor
pressures and thus high boiling points. Ammonia and its derivatives (amines) are molecules with dipoledipole attractions; as long as the nitrogen has one direct N−H bond, the molecule will have hydrogen bonding.
Even so, these molecules will have much higher vapor pressures than ionic species. Thus, if we could convert
the neutral ammonia-type molecules into salts, their vapor pressures, and thus associated odors, would
decrease. Lemon juice contains acids which can react with neutral ammonia-type (amine) molecules to form
ammonium salts.
+
+
NH3 + H → NH4
+
RNH2 + H → RNH3
+
15.129 pH = 10.64
+
[H ] = 2.3 × 10
−11
−
[OH ] = 4.3 × 10
CH3NH2(aq)
+
M
−4
M
U CH3NH3+(aq)
H2O(l)
−4
(x − 4.3 × 10 ) M
4.3 × 10
Kb =
−4
+
−
OH (aq)
4.3 × 10
M
−4
M
[CH3 NH 3+ ][OH − ]
(4.3 × 10−4 )(4.3 × 10−4 )
=
= 4.4 × 10−4
[CH3 NH 2 ]
x − 4.3 × 10−4
−4
4.4 × 10 x − 1.9 × 10
x = 8.4 × 10
−4
−7
= 1.8 × 10
−7
M
The molar mass of CH3NH2 is 31.06 g/mol.
The mass of CH3NH2 in 100.0 mL is:
8.4 × 10
15.130 HCOOH
0.400 − x
U
+
−4
H +
x
31.06 g CH3 NH 2
mol
× 0.100 L ×
= 2.6 × 10−3 g CH 3 NH 2
L
1 mol CH3 NH 2
−
HCOO
x
Total concentration of particles in solution: (0.400 − x) + x + x = 0.400 + x
Assuming the molarity of the solution is equal to the molality, we can write:
∆Tf = K f m
0.758 = (1.86)(0.400 + x)
0.408 = 0.400 + x
+
−
x = 0.00800 = [H ] = [HCOO ]
CHAPTER 15: ACIDS AND BASES
b
(b)
g
0.00800 0.00800
[H + ][HCOO − ]
=
= 1.6 × 10 −4
[HCOOH ]
0.400 − 0.00800
Ka =
15.131 (a)
gb
353
−
NH2 + H2O → NH3 + OH
−
−
N
3−
+ 3H2O → NH3 + 3OH
N
3−
is the stronger base since each ion produces 3 OH ions.
−
15.132 SO2(g) + H2O(l)
U H+(aq) + HSO3−(aq)
6
Recall that 0.12 ppm SO2 would mean 0.12 parts SO2 per 1 million (10 ) parts of air by volume. The number
of particles of SO2 per volume will be directly related to the pressure.
PSO2 =
0.12 parts SO2
atm = 1.2 × 10−7 atm
106 parts air
+
We can now calculate the [H ] from the equilibrium constant expression.
[H + ][HSO3− ]
PSO2
K =
1.3 × 10
−2
x2
=
1.2 × 10 −7
−2
2
−7
x = (1.3 × 10 )(1.2 × 10 )
x = 3.9 × 10
−5
+
M = [H ]
−5
pH = −log(3.9 × 10 ) = 4.40
15.133
[H + ][ClO − ]
= 3.0 × 10−8
[HClO]
+
A pH of 7.8 corresponds to [H ] = 1.6 × 10
−8
M
+
Substitute [H ] into the equation above to solve for the
[ClO − ]
ratio.
[HClO]
[ClO − ]
3.0 × 10−8
=
= 1.9
[HClO]
1.6 × 10−8
−
This indicates that to obtain a pH of 7.8, the [ClO ] must be 1.9 times greater than the [HClO]. We can write:
part ClO−
−
% ClO =
−
part ClO + part HClO
By difference, %HClO = 34%
× 100% =
1.9
× 100% = 66%
1.9 + 1.0
354
CHAPTER 15: ACIDS AND BASES
15.134 In inhaling the smelling salt, some of the powder dissolves in the basic solution. The ammonium ions react
with the base as follows:
−
+
NH4 (aq) + OH (aq) → NH3(aq) + H2O
It is the pungent odor of ammonia that prevents a person from fainting.
15.135 (a)
The overall equation is
→ 2FeCl3(aq) + 3H2O(l)
Fe2O3(s) + 6HCl(aq) 
and the net ionic equation is
+
3+
→ 2Fe (aq) + 3H2O(l)
Fe2O3(s) + 6H (aq) 
+
+
Since HCl donates the H ion, it is the Brønsted acid. Each Fe2O3 unit accepts six H ions; therefore, it
is the Brønsted base.
(b)
The first stage is
−
2+
−
→ Ca (aq) + HCO3 (aq) + Cl (aq)
CaCO3(s) + HCl(aq) 
and the second stage is
−
−
→ CO2(g) + Cl (aq) + H2O(l)
HCl(aq) + HCO3 (aq) 
The overall equation is
→ CaCl2(aq) + H2O(l) + CO2(g)
CaCO3(s) + 2HCl(aq) 
The CaCl2 formed is soluble in water.
(c)
We need to find the concentration of the HCl solution in order to determine its pH. Let's assume a
volume of 1.000 L = 1000 mL. The mass of 1000 mL of solution is:
1000 mL ×
1073
.
g
= 1073 g
1 mL
The number of moles of HCl in a 15 percent solution is:
15% HCl
1 mol HCl
× 1073 g soln = 16
. × 102 g HCl ×
= 4.4 mol HCl
100% soln
36.46 g HCl
Thus, there are 4.4 moles of HCl in one liter of solution, and the concentration is 4.4 M. The pH of the
solution is
pH = −log(4.4) = −0.64
This is a highly acidic solution (note that the pH is negative), which is needed to dissolve large
quantities of rocks in the oil recovery process.