Download CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY

CHAPTER 16
ACID-BASE EQUILIBRIA AND
SOLUBILITY EQUILIBRIA
16.3
(a)
This is a weak acid problem. Setting up the standard equilibrium table:
U
CH3COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
0.40
−x
(0.40 − x)
Ka = 1.8 × 10
−5
=
+
+
H (aq)
−
+
CH3COO (aq)
0.00
+x
+x
0.00
+x
+x
[H + ][CH3COO − ]
x2
x2
=
≈
[CH3COOH]
(0.40 − x)
0.40
3
x = [H ] = 2.7 × 10 M
pH = 2.57
(b)
In addition to the acetate ion formed from the ionization of acetic acid, we also have acetate ion formed
from the sodium acetate dissolving.
−
+
CH3COONa(aq) → CH3COO (aq) + Na (aq)
−
+
Dissolving 0.20 M sodium acetate initially produces 0.20 M CH3COO and 0.20 M Na . The sodium
ions are not involved in any further equilibrium (why?), but the acetate ions must be added to the
equilibrium in part (a).
CH3COOH(aq)
Initial (M):
Change (M):
Equilibrium (M):
U
0.40
−x
(0.40 − x)
Ka = 1.8 × 10
−5
+
=
+
H (aq)
0.00
+x
+x
+
−
CH3COO (aq)
0.20
+x
(0.20 + x)
[H + ][CH3COO − ]
x(0.20)
( x)(0.20 + x)
=
≈
[CH3COOH]
(0.40 − x)
0.40
x = [H ] = 3.6 × 10
−5
M
pH = 4.44
Could you have predicted whether the pH should have increased or decreased after the addition of the
sodium acetate to the pure 0.40 M acetic acid in part (a)?
An alternate way to work part (b) of this problem is to use the Henderson-Hasselbalch equation.
pH = pKa + log
[conjugate base]
[acid]
pH = − log(1.8 × 10−5 ) + log
0.20 M
= 4.74 − 0.30 = 4.44
0.40 M
356
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.4
(a)
This is a weak base calculation.
Kb = 1.8 × 10
x = 1.90 × 10
−5
−3
=
[NH +4 ][OH − ]
( x)( x)
=
[NH3 ]
0.20 − x
−
M = [OH ]
pOH = 2.72
pH = 11.28
(b)
Table 15.4 gives the value of Ka for the ammonium ion. Using this and the given concentrations with
the Henderson-Hasselbalch equation gives:
pH = pKa + log
[conjugate base]
acid
(
)
= − log 5.6 × 10−10 + log
(0.20 )
(0.30 )
pH = 9.25 − 0.18 = 9.07
Is there any difference in the Henderson-Hasselbalch equation in the cases of a weak acid and its
conjugate base and a weak base and its conjugate acid?
16.8
16.9
(a)
HCl (hydrochloric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak
base. Therefore, this is not a buffer system.
(b)
NH3 (ammonia) is a weak base, and its conjugate acid, NH4 is a weak acid. Therefore, this is a buffer
system.
(c)
This solution contains both a weak acid, H2PO4 and its conjugate base, HPO4 . Therefore, this is a
buffer system.
(d)
HNO2 (nitrous acid) is a weak acid, and its conjugate base, NO2 (nitrite ion, the anion of the salt
KNO2), is a weak base. Therefore, this is a buffer system.
(e)
H2SO4 (sulfuric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak
base. Therefore, this is not a buffer system.
(f)
HCOOH (formic acid) is a weak acid, and its conjugate base, HCOO (formate ion, the anion of the salt
HCOOK), is a weak base. Therefore, this is a buffer system.
+
−
2−
−
−
+
NH4 (aq)
U NH3(aq) + H+(aq)
Ka = 5.6 × 10
−10
pKa = 9.25
pH =pK a + log
16.10
(a)
[NH3 ]
+
[NH 4 ]
= 9.25 + log
0.15 M
= 8.88
0.35 M
This problem is greatly simplified because the concentration of the weak acid (acetic acid) is equal to
the concentration of its conjugate base (acetate ion). Let’s set up a table of initial concentrations,
change in concentrations, and equilibrium concentrations.
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
CH3COOH (aq)
Initial
Change
Equilibrium
357
U H+ (aq) + CH3COO− (aq)
2.0 M
−x
2.0 M − x
0
+x
x
Ka =
[H + ][CH3COO− ]
[CH3COOH]
Ka =
[H + ](2.0 + x)
[H + ](2.0)
≈
(2.0 − x)
2.0
2.0 M
+x
2.0 M + x
+
Ka = [H ]
Taking the −log of both sides,
pKa = pH
Thus, for a buffer system in which the [weak acid] = [weak base],
pH = pKa
−5
pH = −log(1.8 × 10 ) = 4.74
(b)
Similar to part (a),
pH = pKa = 4.74
Buffer (a) will be a more effective buffer because the concentrations of acid and base components are
ten times higher than those in (b). Thus, buffer (a) can neutralize 10 times more added acid or base
compared to buffer (b).
16.11
H2CO3(aq)
U HCO3−(aq) + H+(aq)
K a1 = 4.2 × 10
−7
pK a1 = 6.38
pH = pKa + log
[HCO3− ]
[H 2 CO3 ]
8.00 = 6.38 + log
log
[HCO 3− ]
[H 2CO 3 ]
[HCO 3− ]
.
= 162
[H 2CO 3 ]
[HCO 3− ]
= 41.7
[H 2CO 3 ]
[H 2CO 3 ]
[HCO 3− ]
= 0.024
358
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.12
Step 1: Write the equilibrium that occurs between H2PO4 and HPO4 . Set up a table relating the initial
concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations.
−
−
U H+ (aq) + HPO42− (aq)
H2PO4 (aq)
Initial (M):
Change (M):
Equilibrium (M):
2−
0.15
−x
0.15 − x
0
+x
x
0.10
+x
0.10 + 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 + ][HPO 24− ]
[H 2 PO −4 ]
You can look up the Ka value for dihydrogen phosphate in Table 15.5 of your text.
6.2 × 10
−8
6.2 × 10
−8
=
( x)(0.10 + x)
(0.15 − x)
≈
( x)(0.10)
(0.15)
+
x = [H ] = 9.3 × 10
−8
M
+
Step 3: Having solved for the [H ], calculate the pH of the solution.
−8
+
pH = −log[H ] = −log(9.3 × 10 ) = 7.03
16.13
Using the Henderson−Hasselbalch equation:
pH = pK a + log
[CH3COO − ]
[CH3COOH]
4.50 = 4.74 + log
[CH3COO − ]
[CH3COOH]
Thus,
[CH3COO − ]
= 0.58
[CH3COOH]
16.14
−
We can use the Henderson-Hasselbalch equation to calculate the ratio [HCO3 ]/[H2CO3]. The HendersonHasselbalch equation is:
pH = pKa + log
[conjugate base]
[acid]
−
For the buffer system of interest, HCO3 is the conjugate base of the acid, H2CO3. We can write:
pH = 7.40 = − log(4.2 × 10−7 ) + log
[HCO3− ]
[H 2 CO3 ]
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
7.40 = 6.38 + log
359
[HCO3− ]
[H 2 CO3 ]
The [conjugate base]/[acid] ratio is:
log
[HCO3− ]
= 7.40 − 6.38 = 1.02
[H 2 CO3 ]
[HCO 3− ]
1
1.02
= 10
= 1.0 × 10
[H 2CO 3 ]
The buffer should be more effective against an added acid because ten times more base is present compared to
acid. Note that a pH of 7.40 is only a two significant figure number (Why?); the final result should only have
two significant figures.
16.15
For the first part we use Ka for ammonium ion. (Why?) The Henderson−Hasselbalch equation is
pH = − log(5.6 × 10−10 ) + log
(0.20 M )
= 9.25
(0.20 M )
For the second part, the acid−base reaction is
+
+
NH3(g) + H (aq) → NH4 (aq)
We find the number of moles of HCl added
 1 L   0.10 mol HCl 
10.0 mL 

 = 0.0010 mol HCl
1L
 1000 mL  

+
The number of moles of NH3 and NH4 originally present are
 1 L   0.20 mol 
65.0 mL 

 = 0.013 mol
 1000 mL   1 L 
+
Using the acid−base reaction, we find the number of moles of NH3 and NH4 after addition of the HCl.
moles NH3 = (0.013 − 0.0010) mol = 0.012 mol NH3
+
moles NH4 = (0.013 + 0.0010) mol = 0.014 mol NH4
+
We find the new pH:
pH = 9.25 + log
16.16
(0.012)
= 9.18
(0.014)
As calculated in Problem 16.10, the pH of this buffer system is equal to pKa.
−5
pH = pKa = −log(1.8 × 10 ) = 4.74
(a)
The added NaOH will react completely with the acid component of the buffer, CH3COOH. NaOH
−
ionizes completely; therefore, 0.080 mol of OH are added to the buffer.
360
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
Step 1: The neutralization reaction is:
−
Initial (mol)
After reaction (mol)
−
→ CH3COO (aq) + H2O (l)
CH3COOH (aq) + OH (aq) 
1.00
0.080
1.00
0.92
0
1.08
Step 2: Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can
convert directly from moles to molar concentration.
CH3COOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
U H+ (aq) + CH3COO− (aq)
0.92
−x
0.92 − x
0
+x
x
1.08
+x
1.08 + x
Write the Ka expression, then solve for x.
Ka =
[H + ][CH3COO− ]
[CH3COOH]
1.8 × 10
−5
=
( x)(1.08 + x)
x(1.08)
≈
(0.92 − x)
0.92
+
x = [H ] = 1.5 × 10
−5
M
+
Step 3: Having solved for the [H ], calculate the pH of the solution.
−5
+
pH = −log[H ] = −log(1.5 × 10 ) = 4.82
The pH of the buffer increased from 4.74 to 4.82 upon addition of 0.080 mol of strong base.
(b)
−
The added acid will react completely with the base component of the buffer, CH3COO . HCl ionizes
+
completely; therefore, 0.12 mol of H ion are added to the buffer
Step 1: The neutralization reaction is:
−
Initial (mol)
After reaction (mol)
+
→ CH3COOH (aq)
CH3COO (aq) + H (aq) 
1.00
0.12
1.00
0.88
0
1.12
Step 2: Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can
convert directly from moles to molar concentration.
CH3COOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
U H+ (aq) + CH3COO− (aq)
1.12
−x
1.12 − x
Write the Ka expression, then solve for x.
Ka =
[H + ][CH3COO− ]
[CH3COOH]
0
+x
x
0.88
+x
0.88 + x
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
1.8 × 10
−5
=
361
x(0.88)
( x)(0.88 + x)
≈
(1.12 − x)
1.12
+
x = [H ] = 2.3 × 10
−5
M
+
Step 3: Having solved for the [H ], calculate the pH of the solution.
−5
+
pH = −log[H ] = −log(2.3 × 10 ) = 4.64
The pH of the buffer decreased from 4.74 to 4.64 upon addition of 0.12 mol of strong acid.
16.17
We write
K a1 = 1.1 × 10
K a2 = 2.5 × 10
−3
−6
pK a1 = 2.96
pK a2 = 5.60
In order for the buffer solution to behave effectively, the pKa of the acid component must be close to the
desired pH. Therefore, the proper buffer system is Na2A/NaHA.
16.18
Recall that to prepare a solution of a desired pH, , we should choose a weak acid with a pKa value close to the
desired pH.
Calculating the pKa for each acid:
−3
For HA,
pKa = −log(2.7 × 10 ) = 2.57
For HB,
pKa = −log(4.4 × 10 ) = 5.36
For HC,
pKa = −log(2.6 × 10 ) = 8.59
−6
−9
The buffer solution with a pKa closest to the desired pH is HC. Thus, HC is the best choice to prepare a
buffer solution with pH = 8.60.
16.21
Since the acid is monoprotic, the number of moles of KOH is equal to the number of moles of acid.
 1 L  0.08133 mol 
Moles acid = 16.4 mL 

 = 0.00133 mol
1L
 1000 mL 

Molar mass =
16.22
0.2688 g
= 202 g/mol
0.00133 mol
The neutralization reaction is:
→ K2A (aq) + 2 H2O (l)
2 KOH (aq) + H2A (aq) 
The number of moles of H2A reacted is:
 1.00 mol KOH   1 mol H 2 A 
−3
11.1 mL KOH × 

 = 5.55 × 10 mol H 2 A
 1000 mL   2 mol KOH 
362
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
We know that 0.500 g of the diprotic acid were reacted (1/10 of the 250 mL was tested). Divide the number
of grams by the number of moles to calculate the molar mass.
M (H2A) =
16.23
0.500 g H 2 A
= 90.1 g/mol
5.55 × 10 −3 mol H 2 A
The neutralization reaction is:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Since one mole of sulfuric acid combines with two moles of sodium hydroxide, we write:
 0.500 mol H 2SO 4   2 mol NaOH 
mol NaOH = (0.0125 L H2SO4) 
 = 0.0125 mol NaOH

1 L soln

 1 mol H 2SO4 
concentration of NaOH =
16.24
0.0125 mol NaOH
1000 mL
×
= 0.25 M
50.0 mL soln
1L
2HCOOH + Ba(OH)2 → (HCOO)2Ba + 2H2O
Number of moles of HCOOH reacted =
0.883 mol
× (20.4 × 10−3 L) = 0.0180 mol HCOOH
1L
The mole ratio between Ba(OH)2 and HCOOH is 1:2.
Therefore, the molarity of the Ba(OH)2 solution is:
0.0180 mol HCOOH ×
16.25
(a)
1 mol Ba(OH)2
1
= 0.466 M
×
2 mol HCOOH
19.3 × 10−3 L
Since the acid is monoprotic, the moles of acid equals the moles of base added.
→ NaA (aq) + H2O (l)
HA (aq) + NaOH (aq) 
Moles acid = 18.4 mL ×
FG 1 L IJ × FG 0.0633 mol IJ
H 1000 mL K H 1 L K
= 0.00116 mol
We know the mass of the unknown acid in grams and the number of moles of the unknown acid.
Molar mass =
(b)
01276
.
g
2
= 1.10 × 10 g/mol
0.00116 mol
The number of moles of NaOH in 10.0 mL of solution is
10.0 mL ×
FG 1 L IJ × FG 0.0633 mol IJ
H 1000 mL K H 1 L K
= 6.33 × 10
−4
mol
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
363
The neutralization reaction is:
→ NaA (aq) + H2O (l)
HA (aq) + NaOH (aq) 
−4
0.00116
6.33 × 10
0
−4
−4
0
6.33 × 10
5.3 × 10
Initial (mol)
After reaction (mol)
Now, the weak acid equilibrium will be reestablished. The total volume of solution is 35.0 mL.
[HA] =
−
[A ] =
5.3 × 10 −4 mol
= 0.015 M
0.035 L
6.33 × 10 −4 mol
= 0.0181 M
0.035 L
+
We can calculate the [H ] from the pH.
+
[H ] = 10
−pH
= 10
−5.87
HA (aq)
= 1.35 × 10
−6
M
+
U
H (aq)
Initial (M)
Change (M)
0.015
−6
−1.35 × 10
0
−6
+1.35 × 10
Equilibrium (M)
0.015
1.35 × 10
−6
+
−
A (aq)
0.0181
−6
+1.35 × 10
0.0181
Substitute the equilibrium concentrations into the equilibrium constant expression to solve for Ka.
Ka =
16.26
(1.35 × 10 −6 )( 0.0181)
[H + ][A − ]
−6
= 1.6 × 10
=
[HA ]
0.015
The resulting solution is not a buffer system. There is excess NaOH and the neutralization is well past the
equivalence point.
 0.167 mol 
Moles NaOH = 0.500 L 
 = 0.0835 mol
1L


 0.100 mol 
Moles CH3COOH = 0.500 L 
 = 0.0500 mol
1L


After neutralization, the amount of NaOH remaining in 0.0835 − 0.0500 = 0.0335 mol. The volume of the
resulting solution is 1.00 L
−
[OH ] =
+
[Na ] =
0.0335 mol
= 0.0335 M
1.00 L
0.0835 mol
= 0.0835 M
1.00 L
1.0 × 10−14
= 3.0 × 10−13 M
0.0335
0.0500 mol
−
= 0.0500 M
[CH3COO ] =
1.00 L
+
[H ] =
[CH3COOH] =
[H + ][CH3COOH − ]
(3.0 × 10−13 )(0.0500)
=
= 8.3 × 10−10 M
−5
Ka
1.8 × 10
364
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.27
Since the solution volume is doubled at the equivalence point, the concentration of the conjugate acid from
+
the salt, CH3NH3 , is:
0.20 M
= 0.10 M
2
The conjugate acid undergoes hydrolysis.
+
U
CH3NH3 + H2O
CH3NH2 + H3O
0.10 M − x
x
+
x
x2
−11
= 2.3 × 10
0.10 − x
Assuming that, 0.10 − x ≈ 0.10
+
x = [H3O ] = 1.5 × 10
−6
M
pH = 5.82
16.28
Concentration of HCOONa at the equivalence point is 0.050 M, since the solution volume doubles (the
volume of NaOH will equal the volume of HCOOH since the molarities are equal).
−
HCOO + H2O
0.050 − x
U
HCOOH
+
x
−
OH
x
x2
= 5.9 × 10−11
0.050 − x
Assume 0.050 − x ≈ 0.050
x = 1.7 × 10
−6
−
M = [OH ]
pOH = 5.77
pH = 8.23
16.31
16.32
(a)
HCOOH is a weak acid and NaOH is a strong base. Suitable indicators are cresol red and
phenolphthalein.
(b)
HCl is a strong acid and KOH is a strong base. Suitable indicators are all those listed with the
exceptions of thymol blue, bromophenol blue, and methyl orange.
(c)
HNO3 is a strong acid and CH3NH2 is a weak base. Suitable indicators are bromophenol blue, methyl
orange, methyl red, and chlorophenol blue.
CO2 in the air dissolves in the solution:
CO2 + H2O
The carbonic acid neutralizes the NaOH.
U H2CO3
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.33
365
The weak acid equilibrium is
HIn (aq)
U H+ (aq) + In− (aq)
We can write a Ka expression for this equilibrium.
Ka =
[H + ][In − ]
[HIn ]
Rearranging,
[HIn ]
[In − ]
[H + ]
Ka
=
+
From the pH, we can calculate the H concentration.
+
[H ] = 10
[HIn]
−
[In ]
=
−pH
= 10
−4
= 1.0 × 10
−4
M
[H + ]
1.0 × 10 −4
= 100
=
Ka
1.0 × 10 −6
−
Since the concentration of HIn is 100 times greater than the concentration of In , the color of the solution will
be that of HIn, the nonionized formed. The color of the solution will be red.
16.34
−
According to Section 16.5 of the text, when [HIn] ≈ [In ] the indicator color is a mixture of the colors of HIn
−
−
and In . In other words, the indicator color changes at this point. When [HIn] ≈ [In ] we can write:
Ka
[In − ]
=
= 1
[HIn]
[H + ]
+
[H ] = Ka = 2.0 × 10
−6
pH = 5.70
16.41
(a)
The solubility equilibrium is given by the equation
AgI(s)
U Ag+(aq) + I−(aq)
The expression for Ksp is given by
+
−
Ksp = [Ag ][I ]
The value of Ksp can be found in Table 16.2 of the text. If the equilibrium concentration of silver ion is
the value given, the concentration of iodide ion must be
[I − ] =
8.3 × 10 −17
=
= 9.1 × 10 −9 M
[Ag + ] 91
. × 10 −9
Ksp
366
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
(b)
The value of Ksp for aluminum hydroxide can be found in Table 16.2 of the text. The equilibrium
expressions are:
Al(OH)3(s)
3+
U Al3+(aq) + 3OH−(aq)
− 3
Ksp = [Al ][OH ]
Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum
ion is:
Ksp
. × 10 −33
18
[Al 3+ ] =
=
= 7.4 × 10 −8 M
[OH − ]3 ( 2.9 × 10 −9 )3
What is the pH of this solution? Will the aluminum concentration change if the pH is altered?
16.42
In each case, we first calculate the number of moles of compound dissolved in one liter of solution
(the molar solubility).
(a)
 7.3 × 10−2 g SrF2


1 L soln


 1 mol SrF2 
−4
 × 
 = 5.8 × 10 mol / L

 125.6 g SrF2 

Step 1: Write the equilibrium reaction. Then, from the equilibrium equation, write the solubility product
expression.
SrF2 (s)
U Sr2+ (aq) + 2 F− (aq)
− 2
2+
Ksp = [Sr ][F ]
Step 2: The molar solubility is the amount of SrF2 that dissolves. From the stoichiometry of the equilibrium
equation, you should find that
2+
and
[Sr ] = [SrF2] = 5.8 × 10
−
−4
[F ] = 2[SrF2] = 1.16 × 10
−3
M
M
2+
Step 3: Substitute the equilibrium concentrations of Sr
calculate Ksp.
2+
− 2
−
and F into the solubility product expression to
−4
−3 2
Ksp = [Sr ][F ] = (5.8 × 10 )(1.16 × 10 ) = 7.8 × 10
(b)
 6.7 × 10−3 g Ag3 PO 4


1 L soln

−10

 1 mol Ag3 PO 4 
−5
 × 
 = 1.6 × 10 mol / L

418.7
g
Ag
PO
3
4


(b) is solved in a similar manner to (a)
The equilibrium equation is:
Ag3PO4 (s)
U 3 Ag+ (aq) + PO43− (aq)
+ 3
3−
−5 3
−5
Ksp = [Ag ] [PO4 ] = [3 × (1.6 × 10 )] (1.6 × 10 ) = 1.8 × 10
−18
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.43
367
For MnCO3 dissolving, we write
U Mn2+(aq) + CO32−(aq)
MnCO3(s)
2−
2+
For every mole of MnCO3 that dissolves, one mole of Mn will be produced and one mole of CO3 will be
2−
2+
produced. If the molar solubility of MnCO3 is s mol/L, then the concentrations of Mn and CO3 are:
2−
2+
[Mn ] = [CO3 ] = s = 4.2 × 10
2+
2−
−6
M
−6 2
2
Ksp = [Mn ][CO3 ] = s = (4.2 × 10 ) = 1.8 × 10
16.44
−11
First, we can convert the solubility of MX in g/L to mol/L.
 4.63 × 10−3 g MX 
 1 mol MX 
−5

 × 
 = 1.34 × 10 mol/L = s (molar solubility)


1
L
soln
346
g
MX




The equilibrium reaction is:
MX (s)
U Mn+ (aq) + Xn− (aq)
Since the mole ratio of MX to each of the ions is 1:1, the equilibrium concentrations of each of the ions can
also be represented by s. Solving for Ksp,
n+
n−
−5 2
2
Ksp = [M ][X ] = s = (1.34 × 10 ) = 1.80 × 10
16.45
−10
The charges of the M and X ions are +3 and −2, respectively (are other values possible?). We first calculate
the number of moles of M2X3 that dissolve in 1.0 L of water
 1 mol 
−19
Moles M 2 X3 = (3.6 × 10−17 g) 
mol
 = 1.3 × 10
288
g


−19
3+
The molar solubility, s, of the compound is therefore 1.3 × 10
M. At equilibrium the concentration of M
2−
must be 2s and that of X must be 3s. (See Table 16.3 of the text.)
3+ 2
2− 3
2
3
5
Ksp = [M ] [X ] = [2s] [3s] = 108s
Since these are equilibrium concentrations, the value of Ksp can be found by simple substitution
5
Ksp = 108s = 108(1.3 × 10
16.46
−19 5
) = 4.0 × 10
−93
Step 1: Write the equilibrium reaction. Then, from the equilibrium equation, write the solubility product
expression.
CaF2 (s)
U Ca2+ (aq) + 2 F− (aq)
2+
− 2
Ksp = [Ca ][F ]
Step 2: A certain amount of calcium fluoride will dissociate in solution. Let’s represent this amount as −s.
−
−
2+
2+
Since one unit of CaF2 yields one Ca ion and two F ions, at equilibrium [Ca ] is s and [F ] is 2s.
We summarize the changes in concentration as follows:
368
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
U Ca2+ (aq) + 2 F− (aq)
CaF2 (s)
Initial (M):
Change (M):
Equilibrium (M):
0
+s
s
−s
0
+2s
2s
Recall, that the concentration of a pure solid does not enter into an equilibrium constant expression.
Therefore, the concentration of CaF2 is not important.
2+
Step 3: Substitute the value of Ksp and the concentrations of Ca
product expression to solve for s, the molar solubility.
2+
−
and F in terms of s into the solubility
− 2
Ksp = [Ca ][F ]
4.0 × 10
−11
4.0 × 10
−11
= (s)(2s)
2
3
= 4s
s = molar solubility = 2.2 × 10
The molar solubility indicates that 2.2 × 10
16.47
−4
−4
mol/L
mol of CaF2 will dissolve in 1 L of an aqueous solution.
Let s be the molar solubility of Zn(OH)2. The equilibrium concentrations of the ions are then
−
2+
[Zn ] = s and [OH ] = 2s
2+
− 2
2
3
Ksp = [Zn ][OH ] = (s)(2s) = 4s = 1.8 × 10
−14
1/ 3
F
18
. × 10 −14 I
= 17
. × 10 −5
s=G
J
4
H
K
−
[OH ] = 2s = 3.4 × 10
−5
M and pOH = 4.47
pH = 14.00 − 4.47 = 9.53
If the Ksp of Zn(OH)2 were smaller by many more powers of ten, would 2s still be the hydroxide ion
concentration in the solution?
16.48
−
First we can calculate the OH concentration from the pH.
pOH = 14.00 − pH
pOH = 14.00 − 9.68 = 4.32
−
[OH ] = 10
−pOH
= 10
−4.32
= 4.8 × 10
−5
M
The equilibrium equation is:
MOH (s)
U M+ (aq) + OH− (aq)
+
−
From the balanced equation we know that [M ] = [OH ]
+
−
−5 2
Ksp = [M ][OH ] = (4.8 × 10 ) = 2.3 × 10
−9
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.49
369
According to the solubility rules, the only precipitate that might form is BaCO3.
2−
2+
Ba (aq) + CO3 (aq) → BaCO3(s)
The number of moles of Ba
2+
present in the original 20.0 mL of Ba(NO3)2 solution is
0.10 mol Ba 2 +
1L
×
= 2.0 × 10−3 mol Ba 2 +
1 L soln
1000 mL
20.0 mL ×
2+
The total volume after combining the two solutions is 70.0 mL. The concentration of Ba
2.0 × 10−3 mol Ba 2 + 1000 mL
×
= 2.9 × 10−2 M
70.0 mL
1 L soln
[Ba 2+ ] =
The number of moles of CO3
2−
2−
present in the original 50.0 mL Na2CO3 solution is
0.10 mol CO32−
1L
×
= 5.0 × 10−3 mol CO32−
1 L soln
1000 mL
50.0 mL ×
The concentration of CO3
in 70 mL is
in the 70.0 mL of combined solution is
[CO32− ] =
5.0 × 10−3 mol CO32− 1000 mL
×
= 7.1 × 10−2 M
70.0 mL
1 L soln
−9
Now we must compare Q and Ksp. From Table 16.2 of the text, the Ksp for BaCO3 is 8.1 × 10 . As for Q,
2−
2+
−2
−2
Q = [Ba ][CO3 ] = (2.9 × 10 )(7.1 × 10 ) = 2.1 × 10
−3
−3
−9
Since (2.1 × 10 ) > (8.1 × 10 ), then Q > Ksp. Therefore, BaCO3 will precipitate.
16.50
The net ionic equation is:
Sr
2+
−
→ SrF2 (s)
(aq) + 2 F (aq) 
Let’s find the limiting reagent in the precipitation reaction.
 1 L  0.060 mol 
−
Moles F = 75 mL × 

 = 0.0045 mol
1L
 1000 mL 

Moles Sr
2+
 1 L   0.15 mol 
= 25 mL × 

 = 0.0038 mol
 1000 mL   1 L 
−
2+
From the stoichiometry of the balanced equation, twice as many moles of F are required to react with Sr .
−
−
This would require 0.0076 mol of F , but we only have 0.0045 mol. Thus, F is the limiting reagent.
370
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
Let’s assume that the above reaction goes to completion. Then, we will consider the equilibrium that is
established when SrF2 partially dissociates into ions.
−
2+
→ SrF2 (s)
Sr (aq) + 2 F (aq) 
0.0038
0.0045
0
−0.00225
−0.0045
+0.00225
0.00155
0
0.00225
Initial (mol)
Change (mol)
After reaction (mol)
Now, let’s establish the equilibrium reaction. The total volume of the solution is 100 mL = 0.100 L. Divide
the above moles by 0.100 L to convert to molar concentration.
SrF2 (s)
Initial (M)
Change (M)
Equilibrium (M)
U Sr2+ (aq) + 2 F− (aq)
0.0225
−x
0.0225 − x
0.0155
+x
0.0155 + x
0
+2x
2x
Write the solubility product expression, then solve for x.
− 2
2+
Ksp = [Sr ][F ]
2.0 × 10
−10
2
= (0.0155 + x)(2x) ≈ (0.0155)(2x)
x = 5.7 × 10
−5
2
M
−
[F ] = 2x = 1.1 × 10
−4
M
2+
[Sr ] = 0.0155 + x = 0.016 M
Both sodium ions and nitrate ions are spectator ions and therefore do not enter into the precipitation reaction.
2(0.0038) mol
= 0.076 M
0.10 L
−
[NO3 ] =
+
[Na ] =
16.51
0.0045 mol
= 0.045 M
0.10 L
(a)
The solubility product expressions for both substances have exactly the same mathematical form and are
therefore directly comparable. The substance having the smaller Ksp (AgI) will precipitate first. (Why?)
(b)
When CuI just begins to precipitate the solubility product expression will just equal Ksp (saturated
+
solution). The concentration of Cu at this point is 0.010 M (given in the problem), so the concentration
of iodide ion must be:
+
−
−
Ksp = [Cu ][I ] = (0.010)[I ] = 5.1 × 10
−
[I ] =
−12
5.1 × 10 −12
−10
M
= 5.1 × 10
0.010
−
Using this value of [I ], we find the silver ion concentration
+
[Ag ] =
K sp
−
[I ]
=
8.3 × 10 −17
5.1 × 10
−10
= 1.6 × 10
−7
M
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
(c)
371
The percent of silver ion remaining in solution is:
1.6 × 10−7 M
−3
× 100% = 0.0016% or 1.6 × 10 %
0.010 M
+
% Ag (aq) =
Is this an effective way to separate silver from copper?
16.52
For Fe(OH)3, Ksp = 1.1 × 10
−36
−
3+
. When [Fe ] = 0.010 M, the [OH ] value is:
3+
− 3
Ksp = [Fe ][OH ]
or
1
 K sp  3
[OH ] = 

 [Fe3+ ] 


−
 1.1 × 10−36
[OH ] = 

0.010

−
1
3
−12
M
 = 4.8 × 10


−
This [OH ] corresponds to a pH of 2.68. In other words, Fe(OH)3 will begin to precipitate from this solution
at pH of 2.68.
For Zn(OH)2, Ksp = 1.8 × 10
−14
−
2+
. When [Zn ] = 0.010 M, the [OH ] value is:
1
 Ksp  2
−
[OH ] = 

 [Zn 2+ ] 


 1.8 ×10−14
−
[OH ] = 

 0.010
1
2
−6
 = 1.3 × 10 M


This corresponds to a pH of 8.11. In other words Zn(OH)2 will begin to precipitate from the solution at
pH = 8.11. These results show that Fe(OH)3 will precipitate when the pH just exceeds 2.68 and that Zn(OH)2
will precipitate when the pH just exceeds 8.11. Therefore, to selectively remove iron as Fe(OH)3, the pH
must be greater than 2.68 but less than 8.11.
16.55
First let s be the molar solubility of CaCO3 in this solution.
CaCO3(s)
U Ca2+(aq) + CO32−(aq)
Initial (M):
Change (M):
Equilibrium (M):
2+
0.050
+s
(0.050 + s)
2−
Ksp = [Ca ][CO3 ] = (0.050 + s)s = 8.7 × 10
We can assume 0.050 + s ≈ 0.050, then
s=
8.7 × 10−9
= 1.7 × 10−7 M
0.050
−9
0
+s
s
372
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
The mass of CaCO3 can then be found.
 1 L   1.7 × 10−7 mol   100.1 g CaCO3 
−6
3.0 × 102 mL 

 
 = 5.1 × 10 g CaCO 3

1L
1 mol
 1000 mL  


16.56
(a)
Set up a table to find the equilibrium concentrations in pure water.
PbBr2 (s)
Initial (M)
Change (M)
Equilibrium (M)
U Pb2+ (aq) + 2 Br− (aq)
0
+s
s
−s
2+
0
+2s
2s
− 2
Ksp = [Pb ][Br ]
8.9 × 10
−6
= (s)(2s)
2
s = molar solubility = 0.013 M
(b)
Set up a table to find the equilibrium concentrations in 0.20 M KBr. KBr is a soluble salt that ionizes
−
completely giving a initial concentration of Br = 0.20 M.
PbBr2 (s)
Initial (M)
Change (M)
Equilibrium (M)
U Pb2+ (aq) + 2 Br− (aq)
0
+s
s
−s
2+
0.20
+2s
0.20 + 2s
− 2
Ksp = [Pb ][Br ]
8.9 × 10
−6
8.9 × 10
−6
= (s)(0.20 + 2s)
≈ (s)(0.20)
2
2
s = molar solubility = 2.2 × 10
−4
M
Thus, the molar solubility of PbBr2 is reduced from 0.013 M to 2.2 × 10
−
ion (Br ) effect.
(c)
−4
M as a result of the common
Set up a table to find the equilibrium concentrations in 0.20 M Pb(NO3)2. Pb(NO3)2 is a soluble salt
2+
that dissociates completely giving an initial concentration of [Pb ] = 0.20 M.
PbBr2 (s)
Initial (M):
Change (M):
Equilibrium (M):
U Pb2+ (aq) + 2 Br− (aq)
0.20
+s
0.20 + s
−s
2+
0
+2s
2s
− 2
Ksp = [Pb ][Br ]
8.9 × 10
−6
8.9 × 10
−6
= (0.20 + s)(2s)
≈ (0.20)(2s)
2
2
s = molar solubility = 3.3 × 10
−3
M
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
Thus, the molar solubility of PbBr2 is reduced from 0.013 M to 3.3 × 10
2+
ion (Pb ) effect.
16.57
−3
M as a result of the common
We first calculate the concentration of chloride ion in the solution.
 10.0 g CaCl2   1 mol CaCl2
[Cl− ] = 

 1 L soln   111.0 g CaCl2
U
AgCl(s)
Initial (M):
Change (M):
Equilibrium (M):
  2 mol Cl−
 
  1 mol CaCl2
+
Ag (aq)
0.000
+s
s

 = 0.180 M


−
+
Cl (aq)
0.180
+s
(0.180 + s)
If we assume that (0.180 + s) ≈ 0.180, then
−
+
Ksp = [Ag ][Cl ] = 1.6 × 10
[Ag + ] =
Ksp
[Cl− ]
=
−10
1.6 × 10−10
= 8.9 × 10−10 M = s
0.180
The molar solubility of AgCl is 8.9 × 10
16.58
−10
M.
The equilibrium reaction is:
U Ba2+ (aq) + SO42− (aq)
BaSO4 (s)
For both parts of the problem:
2−
2+
Ksp = [Ba ][SO4 ] = 1.1 × 10
(a)
−10
2−
2+
In pure water, let [Ba ] = [SO4 ] = s
2−
2+
Ksp = [Ba ][SO4 ]
1.1 × 10
−10
2
= s
s = 1.0 × 10
−5
M
The molar solubility of BaSO4 in pure water is 1.0 × 10
(b)
−5
mol/L.
Assuming the molar solubility of BaSO4 to be s, then
2+
[Ba ] = s M
and
2−
[SO4 ] = (1.0 + s) M ≈ 1.0 M
2−
2+
Ksp = [Ba ][SO4 ]
1.1 × 10
−10
= (s)(1.0)
s = 1.1 × 10
−10
373
M
Due to the common ion effect, the molar solubility of BaSO4 decreases to 1.1 × 10
−5
2−
1.0 M SO4 (aq) compared to 1.0 × 10 mol/L in pure water.
−10
mol/L in
374
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.59
When the anion of a salt is a base, the salt will be more soluble in acidic solution because the hydrogen ion
decreases the concentration of the anion (Le Chatelier's principle):
−
+
U HB(aq)
B (aq) + H (aq)
16.60
2−
(a)
BaSO4 will be slightly more soluble because SO4
is a base (although a weak one).
(b)
The solubility of PbCl2 in acid is unchanged over the solubility in pure water because HCl is a strong
−
acid, and therefore Cl is a negligibly weak base.
(c)
Fe(OH)3 will be more soluble in acid because OH is a base.
(d)
CaCO3 will be more soluble in acidic solution because the CO3 ions react with H ions to form CO2
and H2O. The CO2 escapes from the solution, shifting the equilibrium. Although it is not important in
this case, the carbonate ion is also a base.
(b)
(c)
(d)
(e)
SO4 (aq) is a weak base
−
OH (aq) is a strong base
2−
C2O4 (aq) is a weak base
3−
PO4 (aq) is a weak base.
−
2−
+
2−
The solubilities of the above will increase in acidic solution. Only (a), which contains an extremely weak
−
base (I is the conjugate base of the strong acid HI) is unaffected by the acid solution.
16.61
In water:
2+
U
Mg(OH)2
Mg
+ 2OH
s
3
2s
Ksp = 4s = 1.2 × 10
s = 1.4 × 10
In a buffer at pH = 9.0
−4
+
−9
−
[OH ] = 1.0 × 10
−11
−11
M
[H ] = 1.0 × 10
1.2 × 10
−
−5
−5 2
= [s](1.0 × 10 )
s = 0.12 M
16.62
From Table 16.2, the value of Ksp for iron(II) is 1.6 × 10
(a)
−14
.
−
At pH = 8.00, pOH = 14.00 − 8.00 = 6.00, and [OH ] = 1.0 × 10
2+
[Fe ] =
Ksp
[OH − ]2
=
1.6 × 10−14
(1.0 × 10 −6 )2
−6
= 0.016 M
The molar solubility of iron(II) hydroxide at pH = 8.00 is 0.016 M
M
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
(b)
−
At pH = 10.00, pOH = 14.00 − 10.00 = 4.00, and [OH ] = 1.0 × 10
Ksp
2+
[Fe ] =
− 2
[OH ]
=
1.6 × 10−14
−4 2
(1.0 × 10 )
M
= 1.6 × 10−6 M
The molar solubility of iron(II) hydroxide at pH = 10.00 is 1.6 × 10
16.63
−4
375
−6
M.
The solubility product expression for magnesium hydroxide is
2+
− 2
Ksp = [Mg ][OH ] = 1.2 × 10
2+
−11
We find the hydroxide ion concentration when [Mg ] is 1.0 × 10
−10
M
1/ 2
 1.2 × 10−11 
[OH ] = 

 1.0 × 10−10 


−
= 0.35 M
−
Therefore the concentration of OH must be slightly greater than 0.35 M.
16.64
We first determine the effect of the added ammonia. Let's calculate the concentration of NH3. This is a
dilution problem.
MIVI = MFVF
(0.60 M)(2.00 mL) = M2(1002 mL)
M2 = 0.0012 M NH3
−5
Ammonia is a weak base (Kb = 1.8 × 10 ).
NH3 + H2O
initial (M):
change (M):
equil. (M):
0.0012
−x
0.0012 − x
Kb =
U NH4+ + OH−
0
+x
x
0
+x
x
[NH 4+ ][OH − ]
x2
=
= 1.8 × 10−5
[NH 3 ]
(0.0012 − x )
−
Solving the resulting quadratic equation gives x = 0.00014, or [OH ] = 0.00014 M
2+
2+
−
This is a solution of iron(II) sulfate, which contains Fe ions. These Fe ions could combine with OH to
precipitate Fe(OH)2. Therefore, we must use Ksp for iron(II) hydroxide. We compute the value of Qc for this
solution.
− 2
−3
−11
2+
2
Q = [Fe ]0[OH ]0 = (1.0 × 10 )(0.00014) = 2.0 × 10
Q is larger than Ksp [Fe(OH)2] = 1.6 × 10
−14
; therefore a precipitate of Fe(OH)2 will form.
376
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.67
First find the molarity of the copper(II) ion
 1 mol 
Moles CuSO 4 = 2.50 g 
 = 0.0157 mol
 159.6 g 
[Cu 2+ ] =
0.0157 mol
= 0.0174 M
0.90 L
As in Example 16.15 of the text, the position of equilibrium will be far to the right. We assume essentially all
the copper ion is complexed with NH3. The NH3 consumed is 4 × 0.0174 M = 0.0696 M. The uncombined
2+
NH3 remaining is (0.30 − 0.0696) M, or 0.23 M. The equilibrium concentrations of Cu(NH3)4 and NH3 are
2+
therefore 0.0174 M and 0.23 M, respectively. We find [Cu ] from the formation constant expression.
Kf =
[Cu ( NH3 )4 2+ ]
[Cu
2+
2+
][ NH3 ]
[Cu ] = 1.2 × 10
16.68
4
−13
= 5.0 × 1013 =
0.0174
[Cu 2+ ][0.23]4
M
−
2+
In solution, Cd ions will complex with CN ions. The concentration of Cd
following equilibrium
Cd
2+
−
(aq) + 4 CN (aq)
U Cd(CN)42−
Kf = 7.1 × 10
2+
will be determined by the
16
2+
Since Kf is so large, this equilibrium lies far to the right. We can safely assume that all the Cd
Step 1: Calculate the initial concentration of Cd
2+
ions.
1 mol Cd(NO3 ) 2
1 mol Cd
×
236.42 g Cd(NO3 ) 2
1 mol Cd(NO3 )2
2+
[Cd ]0 = 0.50 g ×
reacts.
×
1
−3
= 4.2 × 10 M
0.50 L
Step 2: If we assume that the above equilibrium goes to completion, we can write
Initial (M)
After reaction (M)
2+
Step 3: To find the concentration of free Cd
Kf =
[Cd(CN) 24− ]
[Cd 2+ ][CN − ]4
Rearranging,
2+
[Cd ] =
−
2−
→ Cd(CN)4
(aq) + 4 CN (aq) 
−3
4.2 × 10
0.50
0
−3
0
0.48
4.2 × 10
Cd
[Cd(CN)24 − ]
K f [CN − ]4
2+
at equilibrium, use the formation constant expression.
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
377
Substitute the equilibrium concentrations calculated above into the formation constant expression to calculate
2+
the equilibrium concentration of Cd .
[Cd(CN) 2−4 ]
2+
[Cd ] =
4.2 × 10−3
=
− 4
K f [CN ]
16
(7.1 × 10 )(0.48)
−
[CN ] = 0.48 M + (1.1 × 10
2−
−
[Cd(CN)4 ] = (4.2 × 10
16.69
−3
−18
= 1.1 × 10
4
−18
M
M) = 0.48 M
M) − (1.1 × 10
−18
) = 4.2 × 10
−3
M
The reaction
−
Al(OH)3(s) + OH (aq)
U Al(OH)4−(aq)
is the sum of the two known reactions
Al(OH)3(s)
U Al3+(aq) + 3OH−(aq)
U Al(OH)4−(aq)
−
3+
Ksp = 1.8 × 10
Al (aq) + 4OH (aq)
Kf = 2.0 × 10
−33
33
The equilibrium constant is
K = KspKf = (1.8 × 10
−33
33
)(2.0 × 10 ) = 3.6 =
[Al( OH )4 − ]
[OH − ]
−
When pH = 14.00, [OH ] = 1.0 M, therefore
−
−
[Al(OH)4 ] = K[OH ] = 3.6 × 1 = 3.6 M
This represents the maximum possible concentration of the complex ion at pH 14.00. Since this is much
larger than the initial 0.010 M, the complex ion will be the predominant species.
16.70
−
+
+
Silver iodide is only slightly soluble. It dissociates to form a small amount of Ag and I ions. The Ag ions
+
then complex with NH3 in solution to form the complex ion Ag(NH3)2 . The balanced equations are:
AgI (s)
U Ag+ (aq) + I− (aq)
U Ag(NH3)2+ (aq)
+
Ag (aq) + 2 NH3 (aq)
Overall:
AgI (s) + 2 NH3 (aq)
+
Kf =
U Ag(NH3)2+ (aq) + I− (aq)
[Ag(NH3 )2 + ]
initial (M):
change (M):
equilibrium (M)
−s
0.0
+s
s
0.0
+s
s
+
Because Kf is large, we can assume all of the silver ions exist as Ag(NH3)2 . Thus,
+
−
[Ag(NH3)2 ] = [I ] = s
2
−17
= 1.5 × 107
K = Ksp × Kf = 1.2 × 10
U Ag(NH3)2+ (aq) + I− (aq)
1.0
−2s
(1.0 − 2s)
+
[Ag ][NH3 ]
If s is the molar solubility of AgI then,
AgI (s) + 2 NH3 (aq)
−
Ksp = [Ag ][I ] = 8.3 × 10
−9
378
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
We can write the equilibrium constant expression for the above reaction, then solve for s.
K = 1.2 × 10−9 =
s = 3.5 × 10
At equilibrium, 3.5 × 10
16.71
−5
−5
( s )( s )
(1.0 − 2 s )
2
≈
( s )( s )
(1.0)2
M
moles of AgI dissolves in 1 L of 1.0 M NH3 solution.
The balanced equations are:
+
Ag (aq) + 2NH3(aq)
2+
Zn (aq) + 4NH3(aq)
U Ag(NH3)2+(aq)
U Zn(NH3)42+(aq)
−
Zinc hydroxide forms a complex ion with excess OH and silver hydroxide does not; therefore, zinc
hydroxide is soluble in 6 M NaOH.
16.72
(a)
The equations are as follows:
CuI2 (s)
2+
Cu
U Cu2+ (aq) + 2 I− (aq)
(aq) + 4 NH3 (aq)
U [Cu(NH3)4]2+ (aq)
2+
The ammonia combines with the Cu ions formed in the first step to form the complex ion
2+
2+
[Cu(NH3)4] , effectively removing the Cu ions, causing the first equilibrium to shift to the right
(resulting in more CuI2 dissolving).
(b)
Similar to part (a):
AgBr (s)
U Ag+ (aq) + Br− (aq)
−
+
Ag (aq) + 2 CN (aq)
(c)
U [Ag(CN)2]− (aq)
Similar to parts (a) and (b).
HgCl2 (s)
Hg
2+
U Hg2+ (aq) + 2Cl− (aq)
−
(aq) + 4Cl (aq)
U [HgCl4]2−− (aq)
16.75
Silver chloride will dissolve in aqueous ammonia because of the formation of a complex ion. Lead chloride
will not dissolve; it doesn’t form an ammonia complex.
16.76
Since some PbCl2 precipitates, the solution is saturated. From Table 16.2, the value of Ksp for lead(II)
−4
chloride is 2.4 × 10 . The equilibrium is:
PbCl2 (aq)
U Pb2+ (aq) + 2 Cl− (aq)
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
379
We can write the solubility product expression for the equilibrium.
− 2
2+
Ksp = [Pb ][Cl ]
−
Ksp and [Cl ] are known. Solving for the Pb
K sp
[Pb 2+ ] =
16.77
[Cl− ]2
=
2+
concentration,
2.4 ×10−4
(0.15)2
= 0.011 M
Ammonium chloride is the salt of a weak base (ammonia). It will react with strong aqueous hydroxide to
form ammonia (Le Chatelier’s principle).
−
−
NH4Cl(s) + OH (aq) → NH3(g) + H2O(l) + Cl (aq)
The human nose is an excellent ammonia detector. Nothing happens between KCl and strong aqueous NaOH.
+
2+
16.78
Chloride ion will precipitate Ag but not Cu . So, dissolve some solid in H2O and add HCl. If a precipitate
2+
forms, the salt was AgNO3. A flame test will also work. Cu gives a green flame test.
16.79
According to the Henderson−Hasselbalch equation:
pH = pKa + log
If:
[conjugate base]
[acid]
[conjugate base]
= 10 , then:
[acid]
pH = pKa + 1
If:
[conjugate base]
= 01
. , then:
[acid]
pH = pKa − 1
Therefore, the range of the ratio is:
0.1 <
16.80
[conjugate base]
< 10
[acid]
We can use the Henderson-Hasselbalch equation to solve for the pH when the indicator is 90% acid /
10% conjugate base and when the indicator is 10% acid / 90% conjugate base.
pH = pK a + log
[conjugate base]
[acid]
Solving for the pH with 90% of the indicator in the HIn form:
pH = 3.46 + log
[10]
= 3.46 − 0.95 = 2.51
[90]
380
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
−
Next, solving for the pH with 90% of the indicator in the In form:
pH = 3.46 + log
[90]
= 3.46 + 0.95 = 4.41
[10]
Thus the pH range varies from 2.51 to 4.41 as the [HIn] varies from 90% to 10%.
16.81
Referring to Figure 16.4, at the half−equivalence point, [weak acid] = [conjugate base]. Using the
Henderson-Hasselbalch equation:
[conjugate base]
pH = pK a + log
[acid]
so,
pH = pKa
16.82
First, calculate the pH of the 2.00 M weak acid (HNO2) solution before any NaOH is added.
Ka =
[H + ][NO 2 − ]
[HNO 2 ]
4.5 × 10 −4 =
x2
2.00 − x
+
x = [H ] = 0.030 M
pH = −log(0.030) = 1.52
Since the pH after the addition is 1.5 pH units greater, the new pH = 1.52 + 1.50 = 3.02.
+
From this new pH, we can calculate the [H ] in solution.
+
[H ] = 10
−pH
= 10
−3.02
= 9.55 × 10
−4
M
When the NaOH is added, we dilute our original 2.00 M HNO2 solution to:
MIVI = MFVF
(2.00 M)(400 mL) = MF(600 mL)
MF = 1.33 M
Since we have not reached the equivalence point, we have a buffer solution. The reaction between HNO2 and
NaOH is:
→ NaNO2 (aq) + H2O (l)
HNO2 (aq) + NaOH (aq) 
Since the mole ratio between HNO2 and NaOH is 1:1, the decrease in [HNO2] is the same as the decrease in
[NaOH].
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
We can calculate the decrease in [HNO2] by setting up the weak acid equilibrium. From the pH of the
−4
+
solution, we know that the [H ] at equilibrium is 9.55 × 10 M.
U H+ (aq) +
HNO2 (aq)
Initial (M)
Change (M)
1.33
−x
0
1.33 − x
Equilibrium (M)
−
NO2 (aq)
9.55 × 10
0
+x
−4
x
We can calculate x from the equilibrium constant expression.
Ka =
[H + ][NO−2 ]
[HNO2 ]
4.5 × 10
−4
=
(9.55 × 10 −4 )( x)
1.33 − x
x = 0.426 M
−
Thus, x is the decrease in [HNO2] which equals the concentration of added OH . However, this is the
concentration of NaOH after it has been diluted to 600 mL. We need to correct for the dilution from
200 mL to 600 mL to calculate the concentration of the original NaOH solution.
MIVI = MFVF
MI(200 mL) = (0.426 M)(600 mL)
[NaOH] = MI = 1.28 M
16.83
−5
The Ka of butyric acid is obtained by taking the antilog of 4.7 which is 2 × 10 . The value of Kb is:
Kb =
16.84
K w 10
. × 10 −14
= 5 × 10 −10
=
Ka
2 × 10 −5
The resulting solution is not a buffer system. There is excess NaOH and the neutralization is well past the
equivalence point. We have a solution of sodium acetate and sodium hydroxide.
FG 0167
.
mol I
H 1 L JK = 0.0835 mol
F 0100
.
mol I
0.500 L G
JK = 0.0500 mol
1
L
H
Moles NaOH = 0.500 L
Moles CH3COOH =
The reaction between sodium hydroxide and acetic acid is:
→ NaCH3COO (aq) + H2O (l)
NaOH (aq) + CH3COOH (aq) 
0.0835
0.0500
0
0.0335
0
0.0500
Initial (mol)
After Reaction (mol)
Since the total volume of the solution is 1.00 L, we can convert the number of moles directly to molarity.
−
[OH ] = 0.0335 M
+
[Na ] = 0.0335 M + 0.0500 M = 0.0835 M
−
[CH3COO ] = 0.0500 M
381
382
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
−
+
We can calculate the [H ] from the [OH ].
Kw
+
[H ] =
[OH − ]
=
1.00 × 10−14
−13
M
= 2.99 × 10
0.0335
Finally, we can calculate the acetic acid concentration from the Ka expression.
Ka =
[H + ][CH3COO− ]
[CH3COOH]
or,
[CH3COOH] =
[H + ][CH3COO− ]
Ka
[CH3COOH] =
16.85
(2.99 × 10 −13 )(0.0500)
1.8 × 10
= 8.3 × 10
−10
M
Most likely the increase in solubility is due to complex ion formation:
Cd(OH)2(s) + 2OH
16.86
−5
The number of moles of Ba
50.0 mL ×
2+
U Cd(OH)42−(aq)
present in the original 50.0 mL of solution is:
1.00 mol Ba 2 +
1L
×
= 0.0500 mol Ba 2 +
1 L soln
1000 mL
The number of moles of SO4
is:
86.4 mL ×
−
2−
present in the original 86.4 mL of solution, assuming complete dissociation,
0.494 mol SO 42 −
1L
= 0.0427 mol SO 42 −
×
1 L soln
1000 mL
The complete ionic equation is:
−
2+
+
2−
Ba (aq) + 2 OH (aq) + 2 H (aq) + SO4 (aq) → BaSO4 (s) + H2O (l)
Initial (mol):
0.0500
0.100
0.0854
0.0427
0
Change (mol) −0.0427
−2(0.0427) −2(0.0427) −0.0427
+0.0427
After rxn (mol) 0.0073
0.015
0
0
0.0427
Thus the mass of BaSO4 formed is:
(0.0427 mol BaSO 4 ) ×
233.4 g BaSO 4
= 9.97 g BaSO4
1 mol BaSO4
−
−
The pH can be calculated from the excess OH in solution. First, calculate the molar concentration of OH .
The total volume of solution is 136.4 mL = 0.1364 L.
[OH − ] =
0.015 mol
= 0.11 M
0.1364 L
pOH = −log(0.11) = 0.96
pH = 14.00 − pOH = 14.00 − 0.96 = 13.04
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.87
383
A solubility equilibrium is an equilibrium between a solid (reactant) and its components (products: ions,
neutral molecules, etc.) in solution. Only (d) represents a solubility equilibrium.
Consider part (b). Can you write the equilibrium constant for this reaction in terms of Ksp for calcium
phosphate?
16.88
First, we calculate the molar solubility of CaCO3.
U Ca2+ (aq) + CO32− (aq)
CaCO3 (s)
initial (M):
change (M):
equil. (M):
?
−s
?−s
0
+s
s
0
+s
s
2−
2+
2
Ksp = [Ca ][CO3 ] = s = 8.7 × 10
s = 9.3 × 10
−5
M = 9.3 × 10
−5
−9
mol/L
The moles of CaCO3 in the kettle is:
116 g
= 1.16 moles CaCO3
100.1 g / mol
The volume of distilled water needed to dissolve 1.16 moles of CaCO3 is:
1.16 mol CaCO3 ×
1L
9.3 × 10
−5
mol CaCO3
= 1.2 × 104 L
The number of times the kettle would have to be filled is:
1.2 × 104 L
= 6.0 × 103 fillings
2.0 L per filling
Note that the very important assumption is made that each time the kettle is filled, the calcium carbonate is
allowed to reach equilibrium before the kettle is emptied.
16.89
Since equal volumes of the two solutions were used, the initial molar concentrations will be halved.
+
[Ag ] =
−
[Cl ] =
012
. M
= 0.060 M
2
2( 014
. M)
= 0.14 M
2
−
+
Let’s assume that the Ag ions and Cl ions react completely to form AgCl (s). Then, we will reestablish the
−
+
equilibrium between AgCl, Ag , and Cl .
+
Initial (M)
Change (M)
After reaction (M)
−
→ AgCl (s)
Ag (aq) + Cl (aq) 
0.060
0.14
0
−0.060
−0.060
+0.060
0
0.080
0.060
384
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
Now, setting up the equilibrium,
Initial (M):
Change (M):
Equilibrium (M):
−
+
U
AgCl(s)
Ag (aq) + Cl (aq)
0.060
−s
0.060 − s
0
+s
s
0.080
+s
0.080 + s
Set up the Ksp expression to solve for s.
−
+
Ksp = [Ag ][Cl ]
1.6 × 10
−10
= (s)(0.080 + s)
s = 2.0 × 10
−9
M
−9
+
[Ag ] = s = 2.0 × 10 M
−
[Cl ] = 0.080 M − s = 0.080 M
014
. M
2+
[Zn ] =
= 0.070 M
2
012
. M
−
= 0.060 M
[NO3 ] =
2
16.90
First we find the molar solubility and then convert moles to grams. The solubility equilibrium for silver
carbonate is:
Ag2CO3(s)
U 2Ag+(aq) + CO32−(aq)
2−
+ 2
2
3
Ksp = [Ag ] [CO3 ] = (2s) (s) = 4s = 8.1 × 10
 8.1 × 10−12
s = 

4

−12
1/ 3




= 1.3 × 10−4 M
Converting from mol/L to g/L:
 1.3 × 10 −4 mol   275.8 g 
= 0.036 g/L


 1 L soln   1 mol 


16.91
For Fe(OH)3, Ksp = 1.1 × 10
−36
−
3+
. When [Fe ] = 0.010 M, the [OH ] value is
3+
− 3
Ksp = [Fe ][OH ]
or
1
−
[OH ] =
−
[OH ] =
−
F Ksp I 3
GH [Fe3+ ]JK
1
F 11. × 10−36 I 3
GH 0.010 JK
= 4.8 × 10
−12
M
This [OH ] corresponds to a pH of 2.68. In other words, Fe(OH)3 will begin to precipitate from this solution
at pH of 2.68.
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
For Zn(OH)2, Ksp = 1.8 × 10
−14
385
−
2+
. When [Zn ] = 0.010 M, the [OH ] value is
1
−
F Ksp I 2
GH [Zn 2+ ]JK
−
F 18. × 10−14 I 2
GH 0.010 JK
[OH ] =
1
[OH ] =
= 1.3 × 10
−6
M
This corresponds to a pH of 8.11. In other words Zn(OH)2 will begin to precipitate from the solution at
pH = 8.11. These results show that Fe(OH)3 will precipitate when the pH just exceeds 2.68 and that Zn(OH)2
will precipitate when the pH just exceeds 8.11. Therefore, to selectively remove iron as Fe(OH)3, the pH
must be greater than 2.68 but less than 8.11.
16.92
(a)
−3
−3
To 2.50 × 10 mol HCl (that is, 0.0250 L of 0.100 M solution) is added 1.00 × 10 mol CH3NH2 (that
−3
is, 0.0100 L of 0.100 M solution). After the acid-base reaction, we have 1.50 × 10 mol of HCl
+
remaining. Since HCl is a strong acid, the [H ] will come from the HCl. The total solution volume is
35.0 mL = 0.0350 L.
+
[H ] =
1.50 × 10−3 mol
= 0.0429 M
0.0350 L
pH = 1.37
(b)
−3
When a total of 25.0 mL of CH3NH2 is added, we reach the equivalence point. That is, 2.50 × 10 mol
−3
−3
HCl reacts with 2.50 × 10 mol CH3NH2 to form 2.50 × 10 mol CH3NH3Cl. Since there is a total of
+
50.0 mL of solution, the concentration of CH3NH3 is:
2.50 × 10−3 mol
= 5.00 × 10−2 M
0.0500 L
+
[CH3NH3 ] =
This is a problem involving the hydrolysis of the weak acid CH3NH3
Ka = 2.3 × 10
1.15 × 10
−12
−11
= x
x = 1.07 × 10
−6
=
[CH3 NH 2 ][H + ]
[CH3 NH 3+ ]
2
=
+
x2
(5.00 × 10
−2
−x
)
≈
x2
5.00 × 10−2
+
M = [H ]
pH = 5.97
(c)
−3
When a total of 35.0 mL of 0.100 M CH3NH2 (3.50 × 10 mol) is added to the 25 mL of 0.100 M HCl
−3
−3
−3
(2.50 × 10 mol), the acid-base reaction produces 2.50 × 10 mol CH3NH3Cl with 1.00 × 10 mol of
CH3NH2 in excess. Using the Henderson-Hasselbalch equation:
pH = pKa + log
[conjugate base]
[acid]
pH = −log(2.3 × 10
−11
) + log
(1.00 × 10 −3 )
(2.50 × 10 −3 )
= 10.24
386
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.93
The equilibrium reaction is:
U Pb2+ (aq) + 2 IO3− (aq)
Pb(IO3)2 (aq)
Initial (M)
Change (M)
−2.4 × 10
0
−11
+2.4 × 10
−11
2.4 × 10
Equilibrium (M)
−11
0.10
−11
+2(2.4 × 10 )
≈ 0.10
Substitute the equilibrium concentrations into the solubility product expression to calculate Ksp.
− 2
2+
Ksp = [Pb ][IO3 ]
Ksp = (2.4 × 10
16.94
−11
2
)(0.10) = 2.4 × 10
−13
The precipitate is HgI2.
2+
Hg
−
→ HgI2 (s)
(aq) + 2 I (aq) 
−
With further addition of I , a soluble complex ion is formed and the precipitate redissolves.
−
→ HgI4
HgI2 (s) + 2 I (aq) 
16.95
2+
2−
Ksp = [Ba ][SO4 ] = 1.1 × 10
2−
(aq)
−10
2+
[Ba ] = 1.0 × 10
In 5.0 L, the number of moles of Ba
2+
(5.0 L)(1.0 × 10
−5
M
is
−5
mol/L) = 5.0 × 10
−5
mol Ba
2+
= 5.0 × 10
−5
mol BaSO4
The number of grams of BaSO4 dissolved is
5.0 × 10−5 mol BaSO4 ×
233.4 g BaSO 4
= 0.012 g BaSO4
1 mol BaSO4
In practice, even less BaSO4 will dissolve because the BaSO4 is not in contact with the entire volume of
blood. Ba(NO3)2 is too soluble to be used for this purpose.
16.96
We can use the Henderson-Hasselbalch equation to solve for the pH when the indicator is 95% acid /
5% conjugate base and when the indicator is 5% acid / 95% conjugate base.
pH = pK a + log
[conjugate base]
[acid]
Solving for the pH with 95% of the indicator in the HIn form:
pH = 9.10 + log
[5]
= 9.10 − 1.28 = 7.82
[95]
−
Next, solving for the pH with 95% of the indicator in the In form:
pH = 9.10 + log
[95]
= 9.10 + 1.28 = 10.38
[5]
Thus the pH range varies from 7.82 to 10.38 as the [HIn] varies from 95% to 5%.
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.97
(a)
The solubility product expressions for both substances have exactly the same mathematical form and are
therefore directly comparable. The substance having the smaller Ksp (AgI) will precipitate first. (Why?)
(b)
When CuI just begins to precipitate the solubility product expression will just equal Ksp (saturated
+
solution). The concentration of Cu at this point is 0.010 M (given in the problem), so the concentration
of iodide ion must be:
−
+
−
Ksp = [Cu ][I ] = (0.010)[I ] = 5.1 × 10
−12
51
. × 10 −12
−10
= 5.1 × 10
M
0.010
−
[I ] =
−
Using this value of [I ], we find the silver ion concentration
+
[Ag ] =
(c)
K sp
−
[I ]
=
8.3 × 10 −17
. × 10
51
−10
= 1.6 × 10
−7
M
The percent of silver ion remaining in solution is:
+
% Ag (aq) =
16
. × 10 −7 M
−3
× 100% = 0.0016% or 1.6 × 10 %
0.010 M
Is this an effective way to separate silver from copper?
16.98
387
(a)
We abbreviate the name of cacodylic acid to CacH. We set up the usual table.
U Cac− (aq) + H+ (aq)
CacH (aq)
Initial(M):
Change(M)
Equilibrium(M):
Ka =
0.10
−x
(0.10 − x)
0.00
+x
x
0.00
+x
x
[H + ][Cac− ]
x2
=
= 6.4 × 10−7
0.10 − x
[CacH ]
We assume that (0.10 − x) ≈ 0.10. Then,
x = 2.5 × 10
−4
+
M = [H ]
−4
pH = −log(2.5 × 10 ) = 3.60
(b)
We set up a table for the hydrolysis of the anion:
−
Initial (M):
Change(M):
Equilibrium(M):
Cac (aq) + H2O (l)
0.15
−x
(0.l5 − x)
U CacH (aq) + OH− (aq)
0.00
+x
x
−
The ionization constant, Kb, for Cac is:
Kb =
Kw
1.0 × 10−14
=
= 1.6 × 10−8
−7
Ka
6.4 × 10
0.00
+x
x
388
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
x2
= 1.6 × 10 −8
0.15 − x
x = 4.9 × 10
−5
M
−5
pOH = −log(4.9 × 10 ) = 4.31
pH = 14.00 − 4.31 = 9.69
(c)
Number of moles of CacH from (a) is:
 0.10 mol CacH 
−3
50.0 mL CacH 
 = 5.0 × 10 mol CacH
 1000 mL 
−
Number of moles of Cac from (b) is:
 0.15 mol CacNa 
−3
25.0 mL CacNa 
 = 3.8 × 10 mol CacNa
1000
mL


At this point we have a buffer solution.
pH = pKa + log
16.99
[Cac − ]
3.8 ×10−3
= − log(6.4 × 10−7 ) + log
= 6.07
[CacH]
5.0 × 10−3
+
The initial number of moles of Ag is
+
Ag =
0.010 mol Ag +
1L
−4
+
×
× 50 mL = 5.0 × 10 mol Ag
1L
1000 mL
We can use the counts of radioactivity as being proportional to concentration. Thus, we can use the ratio to
+
determine the quantity of Ag still in solution. However, since our original 50 mL of solution has been
diluted to 500 mL, the counts per mL will be reduced by ten. Our diluted solution would then produce 7402.5
+
counts per minute if no removal of Ag had occurred.
+
The number of moles of Ag that correspond to 44.4 counts are:
44.4 counts ×
5.0 × 10 −4 mol Ag +
−6
+
= 3.0 × 10 mol Ag
7402.5 counts
−
The original moles of IO3 =
0.030 mol IO3−
1L
−3
×
× 100 mL = 3.0 × 10 mol
1L
1000 mL
−
+
The quantity of IO3 remaining after reaction with Ag :
+
(original moles − moles reacted with Ag ) = (3.0 × 10
= 2.5 × 10
−3
mol IO3
−
−3
mol) − (5.0 × 10
−4
mol)
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
389
The total final volume is 500 mL or 0.50 L.
+
[Ag ] =
−
[IO3 ] =
3.0 × 10 −6 mol Ag +
−6
= 6.0 × 10 M
0.50 L
2.5 × 10 −3 mol IO3−
−3
= 5.0 × 10 M
0.50 L
−
+
−6
−3
Ksp = [Ag ][IO3 ] = (6.0 × 10 )(5.0 × 10 ) = 3.0 × 10
16.100 (a)
(b)
−8
MCO3 + 2HCl → MCl2 + H2O + CO2
HCl + NaOH → NaCl + H2O
Moles of HCl reacted with MCO3 = Total moles of HCl − Moles of excess HCl
Total moles of HCl = 20.00 mL ×
1L
0.0800 mol
−3
×
= 1.60 × 10 mol HCl
1000 mL
1L
Moles of excess HCl = 5.64 mL ×
1L
0.1000 mol
×
= 5.64 × 10−4 mol HCl
1000 mL
1L
Moles of HCl reacted with MCO3 = 1.60 × 10
−3
mol − 5.64 × 10
Moles of MCO3 reacted = 1.04 × 10−3 mol HCl ×
Molar mass of MCO3 =
0.1022 g
5.20 × 10−4 mol
−4
= 197 g/mol
Molar mass of M = 197 g/mol − 60.01 g/mol = 137 g/mol
+
−
16.101 (a)
H + OH → H2O
(b)
H + NH3 → NH4
+
K =
(c)
K = 1.0 × 10
14
+
1
1
9
= 1.8 × 10
=
Ka
5.6 × 10 −10
−
−
CH3COOH + OH → CH3COO + H2O
Broken into 2 equations:
−
CH3COOH → CH3COO + H
+
−
H + OH → H2O
K =
+
Ka
1/Kw
Ka
18
. × 10 −5
9
=
= 1.8 × 10
Kw
10
. × 10 −14
−3
mol HCl
1 mol MCO3
−4
= 5.20 × 10 mol MCO3
2 mol HCl
Molar mass of CO3 = 60.01
The metal, M, is Ba!
= 1.04 × 10
390
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
(d)
CH3COOH + NH3 → CH3COONH4
Broken into 2 equations:
−
CH3COOH → CH3COO + H
+
NH3 + H → NH4
K =
Ka
K a'
=
+
Ka
1
+
K a'
18
. × 10 −5
5.6 × 10
−10
= 3.2 × 10
4
16.102 The number of moles of NaOH reacted is:
 0.500 mol NaOH 
−3
15.9 mL NaOH 
 = 7.95 × 10 mol NaOH
1000
mL


Since two moles of NaOH combine with one mole of oxalic acid, the number of moles of oxalic acid reacted
−3
is 3.98 × 10 mol. This is the number of moles of oxalic acid hydrate in 25.0 mL of solution. In 250 mL, the
−2
number of moles present is 3.98 × 10 mol. Thus the molar mass is:
5.00 g
3.98 × 10−2 mol
= 126 g / mol
From the molecular formula we can write:
2(1.008) g + 2(12.01) g + 4(16.00) g + x(18.02)g = 126 g
Solving for x:
x = 2
16.103 (a)
(b)
Mix 500 mL of 0.40 M CH3COOH with 500 mL of 0.40 M CH3COONa. Since the final volume is
1.00 L, then the concentrations of the two solutions that were mixed must be one-half of their initial
concentrations.
Mix 500 mL of 0.80 M CH3COOH with 500 mL of 0.40 M NaOH. (Note: half of the acid reacts with
all of the base to make a solution identical to that in part (a) above.)
CH3COOH + NaOH → CH3COONa + H2O
(c)
Mix 500 mL of 0.80 M CH3COONa with 500 mL of 0.40 M HCl. (Note: half of the salt reacts with all
of the acid to make a solution identical to that in part (a) above.)
−
+
CH3COO + H → CH3COOH
16.104 (a)
pH = pKa + log
[conjugate base]
[acid]
8.00 = 9.10 + log
[ionized]
[un − ionized]
[un − ionized]
= 12.6
[ionized]
(1)
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
(b)
391
The total concentration of the indicator is:
 0.050 mL phenolphthalein  

1
−4
(2 drops) × 
×
 × (0.060 M ) = 1.2 × 10 M
1 drop

  50 mL soln 
Using equation (1) above and letting y = [ionized]
1.2 × 10−4 − y
= 12.6
y
y = 8.8 × 10
−6
M
16.105 The sulfur-containing air-pollutants (like H2S) reacts with Pb
darkened look.
16.106 (a)
2+
to form PbS, which gives paintings a
Add sulfate. Na2SO4 is soluble, BaSO4 is not.
(b)
Add sulfide. K2S is soluble, PbS is not
(c)
Add iodide. ZnI2 is soluble, HgI2 is not.
16.107 Strontium sulfate is the more soluble of the two compounds. Therefore, we can assume that all of the SO4
ions come from SrSO4.
U Sr2+(aq) + SO42−(aq)
SrSO4(s)
2−
2+
2
Ksp = [Sr ][SO4 ] = s = 3.8 × 10
2−
−
2+
s = [Sr ] = [SO4 ] =
−7
38
. × 10 −7 = 6.2 × 10
−4
M
For BaSO4:
[Ba 2+ ] =
K sp
[SO 4
2−
]
=
1.1 × 10 −10
6.2 × 10
−4
= 1.8 × 10−7 M
16.108 The amphoteric oxides cannot be used to prepare buffer solutions because they are insoluble in water.
16.109 CaSO4
U Ca2+ + SO42−
2
s = 2.4 × 10
s = 4.9 × 10
−5
−3
M
Solubility: 4.9 × 10
−3
mol 136.2 g
×
= 0.67 g/L
L
1 mol
2−
392
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
Ag2SO4
+
U
2Ag + SO4
2s
1.4 × 10
−5
2−
s
3
= 4s
s = 0.015 M
mol 3111
. g
×
= 4.7 g/L
L
1 mol
Solubility: 0.015
Note: Ag2SO4 has a larger solubility.
16.110 The ionized polyphenols have a dark color. In the presence of citric acid from lemon juice, the anions are
converted to the lighter-colored acids.
16.111 H2PO4
−
U H+ + HPO42−
Ka = 6.2 × 10
−8
pKa = 7.20
7.50 = 7.20 + log
[HPO 24− ]
[H 2 PO 4− ]
[HPO 24− ]
[H 2 PO 4− ]
= 2.0
We need to add enough NaOH so that
−
2−
[HPO4 ] = 2[H2PO4 ]
Initially there was
0.200 L × 0.10 mol/L = 0.020 mol NaH2PO4 present.
−
2−
−
For [HPO4 ] = 2[H2PO4 ], we must add enough NaOH to react with 2/3 of the H2PO4 . After reaction
with NaOH, we have
0.020
−
−
mol H2PO4 = 0.0067 mol H2PO4
3
mol HPO4
2−
= 2 × 0.0067 mol = 0.013 mol HPO4
The moles of NaOH reacted is equal to the moles of HPO4
2−
and HPO4 is 1:1.
−
−
OH + H2PO4 → HPO4
VNaOH =
2−
2−
2−
produced because the mole ratio between OH
+ H2 O
0.013 mol
mol NaOH
=
= 0.013 L = 13 mL
M NaOH
10
. mol / L
−
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.112 Assuming the density of water to be 1.00 g/mL, 0.05 g Pb
0.05 g Pb 2+
×
6
1 × 10 g H 2 O
PbSO4
2+
6
per 10 g water is equivalent to 5 × 10
−5
393
2+
g Pb /L
1 g H2O
1000 mL H 2 O
×
= 5 × 10 −5 g Pb 2 + / L
1 mL H 2 O
1 L H2O
U Pb2+ + SO42−
1.6 × 10
−8
2
= s
s = 1.3 × 10
−4
M
The solubility of PbSO4 in g/L is:
1.3 × 10
−4
mol 303.3 g
×
= 4.0 × 10 −2 g / L
L
1 mol
2+
Yes. The [Pb ] exceeds the safety limit of 5 × 10
16.113 (a)
−5
2+
g Pb /L.
The acidic hydrogen is from the carboxyl group.
O
C
(b)
OH
At pH 6.50, Equation (16.4) of the text can be written as:
6.50 = 2.76 + log
[P − ]
[HP]
[P − ]
3
= 5.5 × 10
[HP]
Thus, nearly all of the penicillin G will be in the ionized form. The ionized form is more soluble in
water because it bears a net charge; penicillin G is largely nonpolar and therefore much less soluble in
water. (Both penicillin G and its salt are effective antibiotics.)
(c)
First, the dissolved NaP salt completely dissociates in water as follows:
H O
+
2
→ Na
NaP  
+
0.12 M
−
P
0.12 M
−
We need to concentrate only on the hydrolysis of the P ion.
−
−
Step 1: Let x be the equilibrium concentrations of HP and OH due to the hydrolysis of P ions. We
summarize the changes:
−
P (aq) + H2O(l)
Initial (M):
Change (M):
Equilibrium (M)
0.12
−x
0.12 − x
U HP(aq) + OH−(aq)
0
+x
x
0
+x
x
394
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
Step 2:
K w 1.00 × 10−14
−12
=
= 6.10 × 10
−3
Ka
1.64 × 10
Kb =
[HP][OH − ]
Kb =
[P − ]
6.10 × 10
−12
=
x2
0.12 − x
Assuming that 0.12 − x ≈ 0.12, we write:
6.10 × 10
−12
x = 8.6 × 10
=
−7
x2
0.12
M
Step 3: At equilibrium:
−
[OH ] = 8.6 × 10
−7
M
−7
pOH = −log(8.6 × 10 ) = 6.07
pH = 14.00 − 6.07 = 7.93
−
−
Because HP is a relatively strong acid, P is a weak base. Consequently, only a small fraction of P
undergoes hydrolysis and the solution is slightly basic.
+
16.114 (c) has the highest [H ]
−
F + SbF5 → SbF6
−
−
Removal of F promotes further ionization of HF.
16.115
Fraction of species present
1
CH3COO
CH3COOH
−
0.75
pH = pKa = 4.74
0.5
0.25
0
0
1
2
3
4
5
6
7
pH
8
9
10
11
12
13
14
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.116 (a)
395
2−
This is a common ion (CO3 ) problem.
The dissociation of Na2CO3 is:
2−
+
H O
2
→ 2 Na (aq) + CO3 (aq)
Na2CO3 (s) 
2(0.050 M)
0.050 M
Let s be the molar solubility of CaCO3 in Na2CO3 solution. We summarize the changes as:
CaCO3 (s)
U Ca2+ (aq) + CO32−
Initial (M)
Change (M)
Equil. (M)
0.00
+s
+s
0.050
+s
0.050 + s
2−
2+
Ksp = [Ca ][CO3 ]
8.7 × 10
−9
= s(0.050 + s)
Since s is small, we can assume that 0.050 + s ≈ 0.050
8.7 × 10
−9
= 0.050s
s = 1.7 × 10
−7
M
2+
Thus, the addition of washing soda to permanent hard water removes most of the Ca
the common ion effect.
2+
ions as a result of
−5
is not removed by this procedure, because MgCO3 is fairly soluble (Ksp = 4.0 × 10 ).
(b)
Mg
(c)
The Ksp for Ca(OH)2 is 8.0 × 10 .
−6
Ca(OH)2
U Ca2+ + 2OH−
At equil:
s
Ksp = 8.0 × 10
3
4s = 8.0 × 10
−6
2s
2+
− 2
= [Ca ][OH ]
−6
s = 0.0126 M
−
[OH ] = 2s = 0.0252 M
pOH = −log(0.0252) = 1.60
pH = 12.40
(d)
−
−
2+
The [OH ] calculated above is 0.0252 M. At this rather high concentration of OH , most of the Mg
2+
will be removed as Mg(OH)2. The small amount of Mg remaining in solution is due to the following
equilibrium:
Mg(OH)2 (s)
U Mg2+ (aq) + 2 OH− (aq)
396
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
2+
− 2
Ksp = [Mg ][OH ]
1.2 × 10
2+
−11
2+
= [Mg ](0.0252)
[Mg ] = 1.9 × 10
(e)
Remove Ca
16.117 pH = pKa + log
2+
−8
2
M
first because it is present in larger amounts.
[In − ]
[HIn]
For acid color:
1
10
pH = pKa − log 10
pH = pKa + log
pH = pKa − 1
For base color:
pH = pKa + log
10
1
pH = pKa + 1
Combining these two equations:
pH = pKa ± 1