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CHAPTER 16
CHEMICAL EQUILIBRIUM
SOLUTIONS TO REVIEW QUESTIONS
1. At 25 C both tubes would appear the same and contain more molecules in the gaseous state than the
tube at 0 C, and less molecules in the gaseous state than the tube at 80 C.
2. The reaction is endothermic because the increased temperature increases the concentration of product
(NO2) present at equilibrium.
3. In an endothermic process heat is absorbed or used by the system so it should be placed on the reactant
side of a chemical equation. In an exothermic process heat is given off by the system so it belongs on
the product side of a chemical equation.
4. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
5. Free protons (Hþ) do not exist in water because they are hydrated forming H3Oþ.
6. The sum of the pH and the pOH is 14. A solution whose pH is 1 would have a pOH of 15.
7. Acids stronger than acetic acid are: benzoic, cyanic, formic, hydrofluoric, and nitrous acids (all
equilibrium constants are greater than the equilibrium constant for acetic acid). Acids weaker than
acetic acid are: carbolic, hydrocyanic, and hypochlorous acids (all have equilibrium constants smaller
than the equilibrium constant for acetic acid). All have one ionizable hydrogen atom.
8. The order of solubility will correspond to the order of the values of the solubility product constants of
the salts being compared. This occurs because each salt in the comparison produces the same number of
ions (two in this case) for each formula unit of salt that dissolves. This type of comparison would not
necessarily be valid if the salts being compared gave different numbers of ions per formula unit of salt
dissolving. The order is: AgC2H3O2, PbSO4, BaSO4, AgCl, BaCrO4, AgBr, AgI, PbS.
9. (a)
(b)
Ksp MnðOHÞ2 ¼ 2:0 1013 ; Ksp Ag2 CrO4 ¼ 1:9 1012 :
Each salt gives 3 ions per formula unit of salt dissolving. Therefore, the salt with the largest Ksp
(in this case Ag2CrO4) is more soluble.
Ksp BaCrO4 ¼ 8:5 1011 ; Ksp Ag2 CrO4 ¼ 1:9 1012 . Ag2CrO4 has a greater molar solubility
than BaCrO4, even though its Ksp is smaller, because the Ag2CrO4 produces more ions per
formula unit of salt dissolving than BaCrO4.
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- Chapter 16 BaCrO4 ðsÞ fi
Ksp ¼ ½Ba2þ CrO42
Ba2þ þ CrO42
Let y ¼ molar solubility of BaCrO4
Ksp ¼ ½y½y ¼ 8:5 1011
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
y ¼ 8:5 1011 ¼ 9:2 106 mol BaCrO4 =L
Ag2 CrO4 ðsÞ fi
2
Ksp ¼ ½Agþ CrO42
2 Agþ þ CrO42
Let y ¼ molar solubility of Ag2 CrO4
Ksp ¼ ½2y2 ½y ¼ 1:9 1012
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
12
3 1:9 10
¼ 7:8 105 mol Ag2 CrO4 =L
y¼
4
Ag2CrO4 has the greater solubility.
10. HC2 H3 O2 fi
Hþ þ C 2 H3 O
2
Initial Concentrations
HC2 H3 O2
0.10 M
Hþ
1:8 105 M
C2 H3 O2
0.10 M
Added
–––––
0.010 mol
–––––
Concentration After
Equilibrium Shifts
0.11 M
1:9 105 M
0.09 M
The initial concentration of Hþ in the buffer solution is very low 1:8 105 M because of the large
excess of acetate ions. 0.010 mol of HCl is added to one liter of the buffer solution. This will supply
0.010 M Hþ. The added Hþ creates a stress on the right side of the equation. The equilibrium shifts to
the left, using up almost all the added Hþ, reducing the acetate ion by approximately 0.010 M, and
increasing the acetic acid by approximately 0.010 M. The concentration of Hþ will not increase
significantly and the pH is maintained relatively constant.
11. In a saturated sodium chloride solution, the equilibrium is
Naþ ðaqÞ þ Cl ðaqÞ fi
NaClðsÞ
Bubbling in HCl gas increases the concentration of Cl , creating a stress, which will cause the
equilibrium to shift to the right, precipitating solid NaCl.
12. The rate of a reaction increases when the concentration of one of the reactants increases. The increase
in concentration causes the number of collisions between the reactants to increase. The rate of a
reaction, being proportional to the frequency of such collisions, as a result, will increase.
13. If pure HI is placed in a vessel at 700 K, some of it will decompose. Since the reaction is reversible
ðH2 þ I2 fi
2 HIÞ HI molecules will react to produce H2 and I2.
14. An increase in temperature causes the rate of reaction to increase, because it increases the velocity of
the molecules. Faster moving molecules increase the number and effectiveness of the collisions
between molecules resulting in an increase in the rate of the reaction.
15. A catalyst speeds up the rate of a reaction by lowering the activation energy. A catalyst is not used up in
the reaction.
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- Chapter 16 16. A þ B fi
CþD
When A and B are initially mixed, the rate of the forward reaction to produce C and D is at its
maximum. As the reaction proceeds, the rate of production of C and D decreases because the
concentrations of A and B decrease. As C and D are produced, some of the collisions between C and D
will result in the reverse reaction, forming A and B. Finally, an equilibrium is achieved in which the
forward rate exactly equals the reverse rate.
H3 Oþ þ C2 H3 O
17. HC2 H3 O2 þ H2 O fi
2
As water is added (diluting the solution from 1.0 M to 0.10 M), the equilibrium shifts to the right,
yielding a higher percent ionization.
18. The statement does not contradict Le Chatelier’s Principle. The previous question deals with the case of
dilution. If pure acetic acid is added to a dilute solution, the reaction will shift to the right, producing
more ions in accordance with Le Chatelier’s Principle. But, the concentration of the un-ionized acetic
acid will increase faster than the concentration of the ions, thus yielding a smaller percent ionization.
19. At different temperatures, the degree of ionization of water varies, being higher at higher temperatures.
Consequently, the pH of water can be different at different temperatures.
20. In pure water, Hþ and OH are produced in equal quantities by the ionization of the water molecules,
Hþ þ OH . Since pH ¼ log½Hþ , and pOH ¼ log½OH , they will always be identical
H2 O fi
for pure water. At 25 C, they each have the value of 7, but at higher temperatures, the degree of
ionization is greater, so the pH and pOH would both be less than 7, but still equal.
21. In water the silver acetate dissociates until the equilibrium concentration of ions is reached. In nitric
acid solution, the acetate ions will react with hydrogen ions to form acetic acid molecules. The HNO3
removes acetate ions from the silver acetate equilibrium allowing more silver acetate to dissolve. If
HC1 is used, a precipitate of silver chloride would be formed, since silver chloride is less soluble than
silver acetate. Thus, more silver acetate would dissolve in HCl than in pure water.
Agþ ðaqÞ þ C2 H3 O2 ðaqÞ
AgC2 H3 O2 ðsÞ fi
22. When the salt, sodium acetate, is dissolved in water, the solution becomes basic. The dissolving
reaction is
H2 O
NaC2 H3 O2 ðsÞ ! Naþ ðaqÞ þ C2 H3 O2 ðaqÞ
The acetate ion reacts with water. The reaction does not go to completion, but some OH ions are
produced and at equilibrium the solution is basic.
C2 H3 O2 ðaqÞ þ H2 Oðl Þ fi
OH ðaqÞ þ HC2 H3 O2 ðaqÞ
23. A buffer solution contains a weak acid or base plus a salt of that weak acid or base, such as dilute acetic
acid and sodium acetate.
HC2 H3 O2 ðaqÞ fi
Hþ ðaqÞ þ C2 H3 O2 ðaqÞ
NaC2 H3 O2 ðaqÞ ! Naþ ðaqÞ þ C2 H3 O2 ðaqÞ
When a small amount of a strong acid (Hþ) is added to this buffer solution, the Hþ reacts with the
acetate ions to form un-ionized acetic acid, thus neutralizing the added acid. When a strong base, OH ,
is added it reacts with un-ionized acetic acid to neutralize the added base. As a result, in both cases, the
approximate pH of the solution is maintained.
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- Chapter 16 24. All four K expressions are equilibrium constants. They describe the ratio between the concentrations of
products and reactants for different types of reactions when at equilibrium. Ka is the equilibrium
constant expression for the ionization of a weak acid, Kb is the equilibrium constant expression for the
ionization of a weak base, Kw is the equilibrium constant expression for the ionization of water and Ksp
is the equilibrium constant expression for a slightly soluble salt.
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- Chapter 16 -
SOLUTIONS TO EXERCISES
1. Reversible systems.
(a)
KMnO4 ðsÞ Ð Kþ ðaqÞ þ MnO
4 ðaqÞ
(b)
CO2 ðsÞ Ð CO2 ðgÞ
2. Reversible systems.
(a)
I2 ðsÞ Ð I2 ðgÞ
(b)
NaNO3 ðsÞ Ð Naþ ðaqÞ þ NO
3 ðaqÞ
3. Equilibrium system.
SiF4 ðgÞ þ 2 H2 OðgÞ þ 103:8 kJ Ð SiO2 ðsÞ þ 4 HFðgÞ
(a)
The reaction is endothermic with heat being absorbed.
(b)
The addition of HF will shift the reaction to the left until equilibrium is reestablished. The
concentrations of SiF4, H2O, and HF will be increased. The concentration of SiO2 will be
decreased.
(c)
The addition of heat will shift the reaction to the right.
4. Equilibrium system.
4 HClðgÞ þ O2 ðgÞ Ð 2 H2 OðgÞ þ 2 Cl2 ðgÞ þ 114:4 kJ
(a)
The reaction is exothermic with heat being released.
(b)
The addition of O2 will shift the reaction to the right until equilibrium is reestablished. The
concentrations of O2, H2O, and O2 will be increased. The concentration of HCl will be decreased.
(c)
The addition of heat will cause the reaction to shift to the left.
5. N2 ðgÞ þ 3 H2 ðgÞ fi
2 NH3 ðgÞ þ 92:5 kJ
Change or stress imposed
on the system at
equilibrium
Changes in number
of moles
Direction of reaction,
left or right, to
reestablish equilibrium
N2
H2
NH3
(a) Add N2
right
I
D
I
(b) Remove H2
left
I
D
D
(c) Decrease volume of reaction vessel
right
D
D
I
(d) Increase temperature
left
I
I
D
I ¼ Increase; D ¼ Decrease; N ¼ No Change;
? ¼ insufficient information to determine
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- Chapter 16 6. N2 ðgÞ þ 3 H2 ðgÞ Ð 2 NH3 ðgÞ þ 92:5 kJ
Change or stress imposed
on the system at
equilibrium
(a) Add NH3
(b) Increase volume of reaction vessel
(c) Add a catalyst
(d) Add H2 and NH3
Direction of reaction,
left or right, to
reestablish equilibrium
Changes in number
of moles
H2
NH3
N2
left
left
no change
?
I
I
N
?
I ¼ Increase; D ¼ Decrease; N ¼ No Change;
? ¼ insufficient information to determine
7. Direction of shift in equilibrium:
Reaction
Increased
Temperature
(a)
(b)
(c)
right
left
left
Increased Pressure
(Volume Decreases)
right
no change
right
Add
Catalyst
no change
no change
no change
8. Direction of shift in equilibrium:
Reaction
Increased
Temperature
Increased Pressure
(Volume Decreases)
Add
Catalyst
(a)
(b)
(c)
right
left
left
left
left
left
no change
no change
no change
9. Equilibrium shifts
CH4 ðgÞ þ 2 O2 ðgÞ Ð CO2 ðgÞ þ 2 H2 OðgÞ þ 802:3 kJ
(a)
(b)
(c)
(d)
left
none
right
none
10. Equilibrium shifts
2 CO2 ðgÞ þ N2 ðgÞ þ 1095:9 kJ Ð C2 N2 ðgÞ þ 2 O2 ðgÞ
(a)
(b)
(c)
(d)
none
left
right
right
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I
I
N
I
I
D
N
I
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- Chapter 16 -
Keq ¼
½NH3 2 ½H2 O4
½CO2 ½CF4 (c)
Keq ¼
(c)
Keq ¼
Ksp ¼ ½Agþ ½Cl (c)
Ksp ¼ ½Zn2þ ½OH 2
Ksp ¼ Pb2þ CrO42
(d)
2
3
Ksp ¼ ½Ca2þ PO3
4
14. (a)
Ksp ¼ ½Mg2þ CO32
(c)
Ksp ¼ Tl3þ ½OH 3
(b)
Ksp ¼ ½Ca2þ C2 O42
(d)
3 2
Ksp ¼ Pb2þ AsO3
4
11. (a)
2
7
½NO2 ½H2 (b)
½Hþ HCO
3
Keq ¼
½H2 CO3 12. (a)
½Hþ H2 PO
4
Keq ¼
½H3 PO4 (b)
13. (a)
(b)
Keq ¼
½COF2 2
½N2 O5 2
½NO2 4 ½O2 ½CH4 ½H2 S2
½CS2 ½H2 4
15. If the Hþ ion concentration is decreased:
(a) pH is increased
(b) pOH is decreased
(c) ½OH is increased
(d) Kw remains the same. Kw is a constant at a given temperature.
16. If the Hþ ion concentration is increased:
(a) pH is decreased (pH of 1 is more acidic than that of 4)
(b) pOH is increased
(c) ½OH is decreased
(d) Kw remains unchanged. Kw is a constant at a given temperature.
17. The basis for deciding if a salt dissolved in water produces an acidic, a basic, or a neutral solution, is
whether or not the salt reacts with water. Salts that contain an ion derived from a weak acid or base will
produce an acidic or a basic solution.
(c) NaCN basic
(a) CaBr2 neutral
(d) K3PO4 basic
(b) NH4NO3 acidic
18. The basis for deciding if a salt dissolved in water produces an acidic, a basic, or a neutral solution, is
whether or not the salt reacts with water. Salts that contain an ion derived from a weak acid or base will
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- Chapter 16 produce an acidic or a basic solution.
(a) NH4Cl acidic
(b) NaC2H3O2 basic
19. (a)
(b)
(c)
(d)
CuSO4 acidic
KI neutral
NO
2 ðaqÞ þ H2 Oðl Þ Ð OH ðaqÞ þ HNO2 ðaqÞ
þ
NHþ
4 ðaqÞ þ H2 Oðl Þ Ð H3 O ðaqÞ þ NH3 ðaqÞ
20. (a)
2
PO3
4 ðaqÞ þ H2 Oðl Þ Ð HPO4 ðaqÞ þ OH ðaqÞ
(b)
CO2
3 ðaqÞ þ H2 Oðl Þ Ð HCO3 ðaqÞ þ OH ðaqÞ
21. (a)
(b)
22. (a)
(b)
C2 H3 O
2 ðaqÞ þ H2 Oðl Þ Ð HC2 H3 O2 ðaqÞ þ OH ðaqÞ
PO3
3 ðaqÞ þ 3 H2 Oðl Þ Ð H3 PO3 ðaqÞ þ 3 OH ðaqÞ
IO
2 ðaqÞ þ H2 Oðl Þ Ð HIO2 ðaqÞ þ OH ðaqÞ
C2 O2
4 ðaqÞ þ H2 Oðl Þ Ð HC2 O4 ðaqÞ þ OH ðaqÞ
23. When excess acid (Hþ) gets into the blood stream it reacts with HCO
3 to form un-ionized H2CO3, thus
neutralizing the acid and maintaining the approximate pH of the blood.
24. When excess base gets into the blood stream it reacts with Hþ to form water. Then H2CO3 ionizes to
replace Hþ, thus maintaining the approximate pH of the blood.
25. (a)
H2 CO3 ðaqÞ Ð Hþ ðaqÞ þ HCO
3 ðaqÞ
Let x ¼ molarity of Hþ
½Hþ ¼ HCO
3 ¼x
½H2 CO3 ¼ 1:25 M x ¼ 1:25 ðsince x is smallÞ
½Hþ HCO
x2
3
Ka ¼
¼
¼ 4:4 107
½H2 CO3 1:25
x2 ¼ ð1:25Þ 4:4 107
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x ¼ ð1:25Þ 4:4 107 ¼ 7:4 104 M ¼ ½Hþ (b)
(c)
pH ¼ log½Hþ ¼ log 7:4 104 M ¼ 3:13
Percent ionization
½Hþ 7:4 104
ð100Þ ¼
ð100Þ ¼ 0:059%
½H2 CO3 1:25
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- Chapter 16 26. (a)
HC3 H5 O2 ðaqÞ Ð Hþ ðaqÞ þ C3 H5 O
2 ðaqÞ
Let x ¼ molarity of Hþ
½Hþ ¼ C3 H5 O
2 ¼ x
½H2 CO3 ¼ 0:025 M x ¼ 0:025 ðsince x is smallÞ
½Hþ C3 H5 O
x2
2
¼
¼ 8:4 104
Ka ¼
½HC3 H5 O2 0:025
x2 ¼ ð0:025Þ 8:4 104
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
x ¼ ð0:025Þ 8:4 104 ¼ 4:6 103 M ¼ ½Hþ (b)
(c)
pH ¼ log½Hþ ¼ log 4:6 103 M ¼ 2:34
Percent ionization
½Hþ 4:6 103
ð100Þ ¼
ð100Þ ¼ 18%
½HC3 H5 O2 0:025
27. HA Ð Hþ þ A
½Hþ ¼ ½A ¼ ð0:025 M Þð0:0045Þ ¼ 1:1 104 M
½HA ¼ 0:025 M 0:00011 M ¼ 0:025 M
2
½Hþ HCO
1:1 104
3
Ka ¼ Keq ¼
¼ 4:8 107
¼
½H2 CO3 0:025
28. HA Ð Hþ þ A
½Hþ ¼ ½A ¼ ð0:500 M Þð0:0068Þ ¼ 3:4 103 M
½HA ¼ 0:500 M 0:0034 M ¼ 0:497 M
2
3:4 103
½Hþ ½A Ka ¼
¼
¼ 2:3 105
½HA
0:497
29. HC6 H5 OðaqÞ Ð Hþ ðaqÞ þ C6 H5 O ðaqÞ
½Hþ ½C6 H5 O ¼ 1:3 1010
½HC6 H5 O
Let x ¼ molarity of Hþ
½HC6 H5 O ¼ initial concentration x ¼ initial concentration
Since Ka is small, the degree of ionization is small. Therefore, the approximation,
initial concentration x ¼ initial concentration is valid.
Ka ¼
(a)
½Hþ ¼ ½C6 H5 O ¼ x ½HC6 H5 O ¼ 1:0 M
ðxÞðxÞ
¼ 1:3 1010
1:0
x2 ¼ 1:3 1010
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
x ¼ ð1:0Þ 1:3 1010 ¼ 1:1 105 M
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- Chapter 16 1:1 105 M
ð100Þ ¼ 0:0011% ionized
1:0 M
pH ¼ log 1:1 105 ¼ 4:96
(b)
½HC6 H5 O ¼ 0:10 M
ðxÞðxÞ
¼ 1:3 1010
0:10
x2 ¼ ð0:10Þ 1:3 1010
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
x ¼ ð0:10Þ 1:3 1010 ¼ 3:6 106 M
3:6 106 M
ð100Þ ¼ 0:0036% ionized
1:0 M
pH ¼ log 3:6 106 ¼ 5:44
(c)
½HC6 H5 O ¼ 0:010 M
ðxÞðxÞ
¼ 1:3 1010
0:010
x2 ¼ ð0:010Þ 1:3 1010
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x ¼ ð0:010Þ 1:3 1010 ¼ 1:1 106 M
1:1 106 M
ð100Þ ¼ 0:011% ionized
0:010 M
pH ¼ log 1:1 106 ¼ 5:96
30. HC7 H5 O2 ðaqÞ Ð Hþ ðaqÞ þ C7 H5 O
2 ðaqÞ
(a) ½Hþ ¼ C7 H5 O
2 ¼ x ½HC7 H5 O2 ¼ 1:0 M
ðxÞðxÞ
¼ 6:3 105
1:0
x2 ¼ ð1:0Þ 6:3 105
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
x ¼ ð1:0Þ 6:3 105 ¼ 7:9 103 M
7:9 103 M
ð100Þ ¼ 0:79% ionized
1:0 M
pH ¼ log 7:9 103 ¼ 2:10
(b)
½HC7 H5 O2 ¼ 0:10 M
ðxÞðxÞ
¼ 6:3 105
0:10
x2 ¼ ð0:10Þ 6:3 105
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x ¼ ð0:10Þ 6:3 105 ¼ 7:9 103 M
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- Chapter 16 7:9 103 M
ð100Þ ¼ 7:9% ionized
0:10 M
pH ¼ log 7:9 103 ¼ 2:10
(c)
½HC7 H5 O2 ¼ 0:010 M
ðxÞðxÞ
¼ 6:3 105
0:010
x2 ¼ ð0:010Þ 6:3 105
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
5
x ¼ ð0:010Þ 6:3 10
¼ 7:9 104 M
7:9 104 M
ð100Þ ¼ 7:9% ionized
0:010 M
pH ¼ log 7:9 104 ¼ 3:10
31.
½Hþ ½A ½HA
þ
First, find the [H ]. This is calculated from the pH expression, pH ¼ log½Hþ ¼ 3:7.
½Hþ ¼ 2 104
HA Ð Hþ þ A
Ka ¼
½Hþ ¼ ½A ¼ 2 104 ½HA ¼ 0:37
2 104 2 104
½Hþ ½A ¼
Ka ¼
¼ 1 107
½HA
0:37
32.
½Hþ ½A ½HA
HA Ð Hþ þ A
Ka ¼
log½Hþ ¼ 2:89
½Hþ ¼ 1:3 103
pH ¼ 2:89
½Hþ ¼ 1:3 103 ¼ ½A ½HA ¼ 0:23
1:3 103 1:3 103
½Hþ ½A ¼ 7:3 106
¼
Ka ¼
0:23
½HA
33.
1:0 M NaOH yields
½OH ¼ 1:0 M ð100% ionizedÞ
pOH ¼ log 1:0 ¼ 0:00
pH ¼ 14 pOH ¼ 14:00
½Hþ ¼
34.
Kw
1:0 1014
¼ 1 1014
¼
½OH 1:0
3:0 M HNO3 yields
½Hþ ¼ 3:0 M ð100% ionizedÞ
pH ¼ log 3:0 ¼ 0:48
pOH ¼ 14 pH ¼ 14 ð0:48Þ ¼ 14:48
½OH ¼
Kw
1 1014
¼
¼ 3:3 1015
½Hþ 3:0
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- Chapter 16 35.
pH þ pOH ¼ 14:0 pOH ¼ 14:0 pH
(a)
0:250 M HBr yields½Hþ ¼ 0:250 Mð100% ionizedÞ
pH ¼ log½Hþ ¼ logð0:250Þ ¼ 0:602 pOH ¼ 14:0 0:602 ¼ 13:4
(b)
0:333 M KOH yields½OH ¼ 0:333 M ð100% ionizedÞ
pOH ¼ log½OHþ ¼ logð0:333Þ ¼ 0:478 pH ¼ 14:0 0:478 ¼ 13:5
(c)
HC2 H3 O2 Ð Hþ þ C2 H3 O
2
0:895 M
x
x
½Hþ C2 H3 C
2
Ka ¼
¼ 1:8 105
½HC2 H3 O2 ðxÞðxÞ
¼ 1:8 105
0:895
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 ¼ ð0:895Þ 1:8 105 x ¼ 1:6 105
x ¼ 4:0 103 ¼ ½Hþ pH ¼ log 4:0 103 ¼ 2:40
pOH ¼ 14:0 2:40 ¼ 11:6
36. pH þ pOH ¼ 14:0 pOH ¼ 14:0 pH
(a)
0:0010 M NaOH yields½OH ¼ 0:0010 M ð100% ionizedÞ
pOH ¼ log½OH ¼ logð0:0010Þ ¼ 3:00 pH ¼ 14:0 3:00 ¼ 11:0
(b)
ðbÞ 0:125 M HCl yields½Hþ ¼ 0:125 M ð100% ionizedÞ
pH ¼ log½Hþ ¼ logð0:125Þ ¼ 0:903 pOH ¼ 14:0 0:903 ¼ 13:1
(c)
HC6 H5 O Ð Hþ þ C6 H5 O
0:0250 M
x
x
½Hþ ½C6 H5 O Ka ¼
¼ 1:3 1010
½HC6 H5 O
ðxÞðxÞ
¼ 1:3 1010
0:0250
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 ¼ ð0:0250Þ 1:3 1010 x ¼ 3:3 1012
x ¼ 1:8 106 ¼ ½Hþ pH ¼ log 1:8 106 ¼ 5:74
pOH ¼ 14:0 5:74 ¼ 8:3
37. Calculate the ½OH ½OH ¼
(a)
(b)
(c)
Kw
½Hþ 1:0 1014
¼ 1:0 1012 M
1:0 102
1:0 1014
½Hþ ¼ 3:2 107
½OH ¼
¼ 3:1 108 M
3:2 107
KOH is a strong base; 1:25 M ¼ ½OH ¼ 1:25M
½Hþ ¼ 1:0 102
½OH ¼
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- Chapter 16 (d)
First find [Hþ] of 0.75 M HC2H3O2 Ka ¼ 1:8 105
½Hþ ¼ C2 H3 O
2 ¼ x ½HC2 H3 O2 ¼ 0:75 M
x2
¼ 1:8 105
0:75
x2 ¼ ð0:75Þ 1:8 105
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x ¼ ð0:75Þ 1:8 105 ¼ 3:7 103 M
½Hþ ¼ 3:7 103
38. Calculate the ½OH .
½OH ¼
Kw
½Hþ 1:0 1014
¼ 2:5 106
4:0 109
1:0 1014
½OH ¼
¼ 8:3 1010
1:2 105
½Hþ ¼ 4:0 109
(b)
½Hþ ¼ 1:2 105
(c)
First find [Hþ] of 1.25 M HCN Ka ¼ 4:0 1010
½Hþ ¼ ½CN ¼ x ½HCN ¼ 1:25 M
x2
¼ 4:8 1010
1:25
x2 ¼ ð1:25Þ 4:0 1010
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
x ¼ ð1:25Þ 4:0 1010 ¼ 2:2 105 M
(d)
40.
1:0 1014
¼ 2:5 106 M
4:0 109
(a)
½Hþ ¼ 2:2 105
39.
½OH ¼
½OH ¼
½OH ¼
1:0 1014
¼ 4:5 1010 M
2:2 105
NaOH is a strong base; 0:333 M ¼ ½OH ¼ 0:333 M
Calculate the ½Hþ ½Hþ ¼
Kw
½OH (a)
½OH ¼ 1:0 108
½Hþ ¼
1:0 1014
¼ 1:0 106
8
1:0 10
(b)
½OH ¼ 2:0 104
½Hþ ¼
1:0 1014
¼ 5:0 1011
2:0 104
Calculate the ½Hþ (a)
½OH ¼ 4:5 102
(b)
½OH ¼ 5:2 109
Kw
½OH 1:0 1014
½Hþ ¼
¼ 2:2 1013
4:5 102
½Hþ ¼
½Hþ ¼
1:0 1014
¼ 1:9 106
5:2 109
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- Chapter 16 41. The molar solubilities of the salts and their ions are indicated below the formulas in the equilibrium
equations.
(a)
BaSO4 ðsÞ fi
Ba2þ
þ
SO2
4
5
3:9 10
3:9 105
2
Ksp ¼ ½Ba2þ SO42
3:9 105 ¼ 1:5 109
þ
CrO42
2 Ag 5 þ
2 7:8 10
7:8 105
2 Ksp ¼ ½Agþ 2 CrO42
15:6 105 7:8 105 ¼ 1:9 1012
(b) Ag2 CrO4 ðsÞ fi
(c)
First change g=L ! mol=L
0:67 g CaSO4
1 mol
¼ 4:9 103 M CaSO4
136:1 g
L
CaSO4 ðsÞ fi
Ca2þ
þ
SO42
4:9 103
4:9 103
2
2
Ksp ¼ ½Ca2þ SO42 ¼ 4:9 103 ¼ 2:4 105
(d)
First change g=L ! mol=L
0:0019 g AgCl
1 mol
¼ 1:3 105 M AgCl
L
143:4 g
Agþ
þ
Cl
5
1:3 10
1:3 105
2
Ksp ¼ ½Agþ ½Cl ¼ 1:3 105 ¼ 1:7 1010
Ð
AgClðsÞ
42. The molar solubilities of the salts and their ions are indicated below the formulas in the equilibrium
equations.
(a)
Zn2þ
þ
S2
12
3:5 10
3:5 1012
2
Ksp ¼ ½Zn2þ S2 ¼ 3:5 1012 ¼ 1:2 1023
(b)
Pb2þ
þ
2 IO3 5 5
4:0 10
2 4:0 10
2þ 2 2
5
Ksp ¼ Pb
¼ 4:0 10
IO3
8:0 105 ¼ 2:6 1013
(c)
First change g=L ! mol=L
6:73 103 g Ag3 PO4
1 mol
¼ 1:61 105 M Ag3 PO4
418:7 g
L
ZnSðsÞ
Ð
PbðIO3 Þ2 ðsÞ
Ð
þ
PO43
3 Ag 5 þ
3 1:61 10
1:61 105
3 3
Ksp ¼ ½Agþ PO43 ¼ 4:83 105 1:61 105 ¼ 1:81 1018
Ag3 PO4 ðsÞ
Ð
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- Chapter 16 (d)
First change g=L ! mol=L
4
2:33 10 g ZnðOHÞ2
1 mol
¼ 2:34 106 M ZnðOHÞ2
99:41 g
L
Zn2þ
þ
2 OH 6 6
2:34 10
2 2:34 10
2
2
Ksp ¼ ½Zn2þ ½OH ¼ 2:34 106 4:68 106 ¼ 5:13 1017
ZnðOHÞ2 ðsÞ
Ð
43. The molar solubilities of the salts and their ions will be represented in terms of x below their formulas
in the equilibrium equations.
(a)
CaF2
Ð
Ca2þ
x
þ 2 F
2x
Ksp ¼ ½Ca2þ ½F 2 ¼ ðxÞð2xÞ2 ¼ 4x2 ¼ 3:9 1011
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
11
3 3:9 10
x¼
¼ 2:2 104 M
4
(b)
FeðOHÞ3
Ð
Fe3þ
x
þ 3 OH
3x
Ksp ¼ ½Fe3þ ½OH 3 ¼ ðxÞð3xÞ3 ¼ 27x4 ¼ 3:4 1038
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
38
4 3:4 10
x¼
¼ 1:9 1010 M
27
44. The molar solubilities of the salts and their ions will be represented in terms of x below their formulas
in the equilibrium equations.
(a)
Pb2þ þ SO42
x
x
2þ 2
SO4 ¼ ðxÞðxÞ ¼ x2 ¼ 1:3 108
Ksp ¼ Pb
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x ¼ 1:3 108 ¼ 1:1 104 M
(b)
Ba2þ þ CrO42
x
x
2
2þ
Ksp ¼ ½Ba CrO4
¼ ðxÞðxÞ ¼ x2 ¼ 8:5 1011
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x ¼ 8:5 1011 ¼ 9:2 106 M
45. (a)
PbSO4
BaCrO4
Ð
2:2 104 mol CaF2
78:08 g CaF2
ð0:100 LÞ
¼ 1:7 103 g CaF2
L
mol
(b)
Ð
1:9 1010 mol FeðOHÞ3
106:9 FeðOHÞ3
ð0:100 LÞ
¼ 2:0 109 g FeðOHÞ3
L
mol
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- Chapter 16 46. (a)
(b)
1:1 104 PbSO4
303:3 g PbSO4
ð0:100 LÞ
¼ 3:3 103 g PbSO4
L
mol
9:2 106 mol BaCrO4
253:3 g BaCrO4
ð0:100 LÞ
¼ 2:3 104 g BaCrO4
L
mol
47. The molar concentrations of ions, after mixing, are calculated and these concentrations are substituted
into the equilibrium expression. The value obtained is compared to the Ksp of the salt. If the value is
greater than the Ksp, precipitation occurs. If the value is less than the Ksp, no precipitation occurs.
100: mL 0:010 M Na2 SO4 ! 100: mL 0:010 M SO42
100: mL 0:001 M PbðNO3 Þ2 ! 100: mL 0:001 M Pb2þ
Volume after mixing ¼ 200: mL
Concentrations after mixing : SO42 ¼ 0:0050 M Pb2þ ¼ 0:0005 M
2þ SO42 ¼ 5:0 103 5 104 ¼ 3 106
Pb
Ksp ¼ 1:3 108 which is less than 3 106 , therefore, precipitation occurs.
48. The molar concentrations of ions, after mixing, are calculated and these concentrations are substituted
into the equilibrium expression. The value obtained is compared to the Ksp of the salt. If the value is
greater than the Ksp, precipitation occurs. If the value is less than the Ksp, no precipitation occurs.
50:0 mL 1:0 104 M AgNO3 ! 50:0 mL 1:0 104 M Agþ
100: mL 1:0 104 M NaCl ! 100: mL 1:0 104 M Cl
Volume after mixing ¼ 150: mL
Concentrations after mixing:
50:0 mL
4
þ
1:0 10 M Ag
¼ 3:3 105 M Agþ
150: mL
100: mL
4
1:0 10 M Cl
¼ 6:7 105 M Cl
150: mL
½Agþ ½Cl ¼ 3:3 105 6:7 105 ¼ 2:2 109
Ksp ¼ 1:7 1010 which is less than 2:2 109 , therefore, precipitation occurs.
49. The concentration of Br ¼ 0:10 M in 1.0 L of 0.10 M NaBr. Substitute this Br concentration in the
Ksp expression and solve for the [Agþ] in equilibrium with 0:10 M Br .
Ksp ¼ ½Agþ ½Br ¼ 5:2 1013
½Agþ ¼
5:2 1013 5:2 1013
¼
¼ 5:2 1012 M
½Br 0:10
5:2 1012 mol Agþ
L
1 mol AgBr
ð1:0 LÞ ¼ 5:2 1012 mol AgBr will dissolve
1 mol Agþ
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- Chapter 16 50.
0:10 mol MgBr2
L
2 mol Br
1 mol MgBr2
¼
0:20 mol Br
¼ 0:20 M Br in solution:
L
Substitute the Br concentration in the Ksp expression and solve for [Agþ] in equilibrium with
0.20 M Br .
½Agþ ¼
5:2 1013 5:2 1013
¼
¼ 2:6 1012 M
½Br 0:20
2:6 1012 mol Agþ
L
1 mol AgBr
ð1:0 LÞ ¼ 2:6 1012 mol AgBr will dissolve
1 mol Agþ
51. HC2 H3 O2 fi
Hþ þ C2 H3 O2
½Hþ C2 H3 O2
Ka ¼
¼ 1:8 105
½HC2 H3 O2 !
½
HC
H
O
2
3
2
½Hþ ¼ Ka ½HC2 H3 O2 ¼ 0:20 M
C2 H3 O2
0:20
5
þ
½H ¼ 1:8 10
¼ 3:6 105 M
0:10
pH ¼ log 3:6 105 ¼ 4:44
52. HC2 H3 O2 fi
Hþ þ C2 H3 O2
½Hþ C2 H3 O2
Ka ¼
¼ 1:8 105
½HC2 H3 O2 !
½
HC
H
O
2 3 2
½Hþ ¼ Ka ½HC2 H3 O2 ¼ 0:20 M
C2 H3 O2
0:20
½Hþ ¼ 1:8 105
¼ 1:8 105 M
0:20
pH ¼ log 1:8 105 ¼ 4:74
53.
Initially; the solution of NaCl is neutral:
pH ¼ log 1 107 ¼ 7:0
C2 H3 O2 ¼ 0:10 M
C2 H3 O2 ¼ 0:20 M
½Hþ ¼ 1 107
Final Hþ ¼ 2:0 102 M
pH ¼ log 2:0 102 ¼ 1:70
Change in pH ¼ 7:0 1:70 ¼ 5:3 units in the unbuffered solution
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- Chapter 16 -
54.
Initially, ½Hþ ¼ 1:8 105
pH ¼ log 1:8 105 ¼ 4:74
Ka½HC2 H3 O2 ½Hþ ¼ C2 H3 O
2
Final ½Hþ ¼ 1:9 105
pH ¼ log 1:9 105 ¼ 4:72
Change in pH ¼ 4:74 4:72 ¼ 0:02 units in the buffered solution
55. The concentration of solid salt is not included in the Ksp equilibrium constant because the concentration
of solid does not change. It is constant and part of the Ksp.
56. The energy diagram represents an exothermic reaction because the energy of the products is lower than
the energy of the reactants. This means that energy was given off during the reaction.
57.
Potential energy
E1C16
Activation
energy with
catalyst
Activation
energy
Energy
absorbed
Progress of reaction
58. H2 þ I2 fi
2 HI
The reaction is a 1 to 1 mole ratio of hydrogen to iodine. The data given indicates that hydrogen is the
limiting reactant.
2 mol HI
ð2:10 mol H2 Þ
¼ 4:20 mol HI
1 mol H2
59. H2 þ I2 fi
2 HI
(a) 2.00 mol H2 and 2.00 mol I2 will produce 4.00 mol HI assuming 100% yield. However, at 79%
yield you get
4:00 mol HI 0:79 ¼ 3:16 mol HI
(b) The addition of 0.27 mol I2 makes the iodine present in excess and the 2.00 mol H2 the limiting
reactant. The yield increases to 85%.
2 mol HI
ð2:00 mol H2 Þ
ð0:85Þ ¼ 3:4 mol HI
1 mol H2
There will be 15% unreacted H2 and I2 plus the extra I2 added.
ð0:15Þð2:0 mol H2 Þ ¼ 0:30 mol H2 present; also 0:30 mol I2 :
In addition to the 0.30 mol of unreacted I2, will be the 0.27 mol I2 added.
0:27 mol þ 0:30 mol ¼ 0:57 mol I2 present:
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- Chapter 16 -
(c)
K¼
½HI2
½H2 ½I2 The formation of 3.16 mol HI required the reaction of 1.58 mol I2 and 1.58 mol H2. At
equilibrium, the concentrations are:
2:00 1:58 ¼ 0:42 mol H2 ¼ 0:42 mol I2
3:16 mol HI;
2
Keq ¼
ð3:16Þ
¼ 57
ð0:42Þð0:42Þ
In the calculation of the equilibrium constant, the actual number of moles of reactants and
products present at equilibrium can be used in the calculation in place of molar concentrations.
This occurs because the reaction is gaseous and the liters of HI produced equals the sum of the
liters of H2 and I2 reacting. In the equilibrium expression, the volumes will cancel.
60. H2 þ I2 fi
2 HI
1 mol
ð64:0 g HIÞ
¼ 0:500 mol HI present
127:9 g
1 mol I2
ð0:500 mol HIÞ
¼ 0:250 mol I2 reacted
2 mol HI
1 mol H2
ð0:500 mol HIÞ
¼ 0:250 mol H2 reacted
2 mol HI
1 mol
ð6:00 g H2 Þ
¼ 2:98 mol H2 initially present
2:016 g
1 mol
ð2:00 g I2 Þ
¼ 0:788 mol I2 initially present
253:8 g
At equilibrium, moles present are:
0:500 mol HI;
2:98 0:250 ¼ 2:73 mol H2
0:788 0:250 ¼ 0:538 mol I2
61. PCl3 ðgÞ þ Cl2 ðgÞ Ð PCl5 ðgÞ
Keq ¼
½PCl5 ½PCl3 ½Cl2 The concentrations are:
0:22 mol
¼ 0:011 M
20: L
0:10 mol
PCl3 ¼
¼ 0:0050 M
20: L
1:50 mol
¼ 0:075 M
Cl2 ¼
20: L
0:011
Keq ¼
¼ 29
ð0:0050Þð0:075Þ
PCl5 ¼
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- Chapter 16 62. 100 C 30 C ¼ 70 C temperature increase. This increase is equal to seven 10 C increments.
The reaction rate will be increased by 27 ¼ 128 times.
63.
NHþ
4 fi
0:30 M
NH3 þ Hþ
y
Keq ¼ 5:6 1010
y
þ
y ¼ ½H ¼ ½NH3 ½NH3 ½Hþ ½y½y
¼
¼ 5:6 1010
½0:30
NH4þ
y2 ¼ 5:6 1010 ð0:30Þ ¼ 1:7 1010
y ¼ ½Hþ ¼ 1:3 105
pH ¼ log 1:3 105 ¼ 4:89
ðon acidic solutionÞ
64. pH of an acetic acid-acetate buffer
HC2 H3 O2 fi
Hþ þ C2 H3 O2
½Hþ C2 H3 O2
½Hþ ½0:20
¼ 1:8 105
¼
Ka ¼
0:30
½HC2 H3 O2 1:8 105 ð0:30Þ
þ
¼ 2:7 105
H ¼
0:20
pH ¼ log 2:7 105 ¼ 4:57
65. Concentration of Ba2+ in solution
BaCl2 ðaqÞ þ Na2 CrO4 ðaqÞ Ð BaCrO4 ðsÞ þ NaClðaqÞ
BaCrO4 Ð Ba2þ þ CrO42
Ksp ¼ 8:5 1011
Determine the moles of Ba2+ and CrO42 in solution
0:10 mol
ð0:050 LÞ ¼ 0:0050 mol Ba2þ
L
0:15 mol
ð0:050 LÞ ¼ 0:0075 mol CrO42
L
Excess CrO42 in solution ¼ 0:0025 mol 2:5 103 after 0.005 mol BaCrO4 precipitate.
Concentration of CrO42 in solution (total volume ¼ 100 mL)
2:5 103 mol CrO42
¼ 2:5 102 M CrO42
0:100 L
Now using the Ksp, calculate the Ba2+ remaining in solution.
½Ba2þ CrO42 ¼ ½Ba2þ 2:5 102 ¼ 8:5 1011
½Ba2þ ¼
8:5 1011
¼ 3:4 109 mol=L
2:5 102
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- Chapter 16 66.
Hypochlorous acid
HOCl fi
Equilibrium concentrations:
Hþ þ OCl
½Hþ ¼ ½OCl ¼ 5:9 105 M
½HOCl ¼ 0:1 5:9 105 ¼ 0:10 M neglecting 5:9 105
5:9 105 5:9 105
½Hþ ½OCl ¼ 3:5 108
¼
Ka ¼
0:10
½HOCl
Propanoic acid
HC3 H5 O2 fi
Hþ þ C3 H5 O2
Equilibrium concentrations:
½Hþ ¼ C3 H5 O2 ¼ 1:4 103 M
½HC3 H5 O2 ¼ 0:15 1:4 103 ¼ 0:15 M neglecting 1:4 103
½Hþ C3 H5 O2
1:4 103 1:4 103
¼ 1:3 105
Ka ¼
¼
0:15
½HC3 H5 O2 Hydrocyanic acid
HCN Ð Hþ þ CN
Equilibrium concentrations:
½Hþ ¼ ½CN ¼ 8:9 106 M
½HCN ¼ 0:20 8:9 106 ¼ 0:20 M neglecting 8:9 106
2
8:9 106
½Hþ ½CN Ka ¼
¼
¼ 4:0 1010
½HCN
0:20
67. Let y ¼ M CaF2 dissolved
Ca2þ þ 2 F
CaF2 ðsÞ fi
y
2y
y ¼ molar solubility
(a)
Ksp ¼ ½Ca2þ ½F 2 ¼ ðyÞð2yÞ2 ¼ 4y3 ¼ 3:9 1011
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
11
3 3:9 10
y¼
¼ 2:1 104 M ðCaF2 dissolvedÞ
4
2:1 104 mol CaF2
1 mol Ca2þ
¼ 2:1 104 M Ca2þ
L
1 mol CaF2
2:1 104 mol CaF2
2 mol F
¼ 4:2 104 M F
L
1 mol CaF2
(b)
2:1 104 mol CaF2
78:08 g
¼ 8:2 103 g CaF2
ð0:500 LÞ
mol
L
68. The molar concentrations of ions, after mixing, are calculated and these concentrations are substituted
into the equilibrium expression. The value obtained is compared to the Ksp of the salt. If the value is
greater than the Ksp, precipitation occurs. If the value is less than the Ksp, no precipitation occurs.
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- Chapter 16 (a)
100: mL 0:010 M Na2 SO4 ! 100: mL 0:010 M SO42
100: mL 0:001 M PbðNO3 Þ2 ! 100: mL 0:001 M Pb2þ
Volume after mixing ¼ 200. mL
Concentrations after mixing :
SO42 ¼ 0:0050 M Pb2þ ¼ 0:0005 M
2þ Pb
SO42 ¼ 5:0 103 5 104 ¼ 3 106
Ksp ¼ 1:3 108 which is less than 3 106 , therefore, precipitation occurs.
(b)
50:0 mL 1:0 104 M AgNO3 ! 50:0 mL 1:0 104 M Agþ
100: mL 1:0 104 M NaCl ! 100: mL 1:0 104 M Cl
Volume after mixing ¼ 150. mL
Concentrations after mixing:
50:0 mL
4
¼ 3:3 105 M Agþ
1:0 10
150: mL
100: mL
4
1:0 10
¼ 6:7 105 M Cl
150: mL
½Agþ ½Cl ¼ 3:3 105 6:7 105 ¼ 2:2 109
Ksp ¼ 1:7 1010 which is less than 2:2 109 , therefore, precipitation occurs.
(c)
Convert g Ca(NO3)2 to g Ca2+
1:0 g CaðNO3 Þ2
1 mol
1 mol Ca2þ
¼ 0:041 M Ca2þ
164:1 g 1 mol CaðNO2 Þ2
0:150 L
250 mL 0:01 M NaOH ! 250 mL 0:01 M OH
Final volume ¼ 4.0 102mL
Concentration after mixing:
150 mL
2þ
ð0:041 M Ca Þ
¼ 0:015 M Ca2þ
4:0 102 mL
250 mL
ð0:01 M OH Þ
¼ 0:0063 M OH
4:0 102 mL
½Ca2þ ½OH 2 ¼ ð0:015Þð0:0063Þ2 ¼ 6:0 107
Ksp ¼ 1:3 106 which is greater than 6:7 107 , therefore, no precipitation occurs.
69. With a known Ba2þ concentration, the SO42 concentration can be calculated using the Ksp value.
BaSO4 ðsÞ fi
Ba2þ þ SO42
Ksp ¼ ½Ba2þ SO42 ¼ 1:5 109
Ba2þ ¼ 0:050 M
Ksp
1:5 109
¼
¼ 3:0 108 M SO42 in solution
(a) SO42 ¼
½Ba2þ 0:050
M SO42 ¼ M BaSO4 in solution
3:0 108 mol BaSO4
233:4 g
(b)
ð0:100 LÞ
mol
L
¼ 7:0 107 g BaSO4 remain in solution
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- Chapter 16 70. If Pb2þ ½Cl 2 exceeds the Ksp , precipitation will occur.
Ksp ¼ Pb2þ ½Cl 2 ¼ 2:0 105
0:050 M PbðNO3 Þ2 ! 0:050 M Pb2þ
0:010 M NaCl ! 0:010 M Cl
ð0:050Þð0:010Þ2 ¼ 5:0 106
2þ 2
Pb ½Cl is smaller than the Ksp value. Therefore, no precipitate of PbCl2 will form.
71. ½Ba2þ SO2
½Sr2 SO42 ¼ 3:5 107
¼ 1:5 109
4
Both cations are present in equal concentrations ð0:10 M Þ. Therefore, as SO42 is added, the Ksp of
BaSO4 will be exceeded before that of SrSO4 . BaSO4 precipitates first.
72. 2 SO2 ðgÞ þ O2 ðgÞ fi
2
Keq ¼
½SO3 ½SO2 2 ½O2 ¼
2 SO3 ðgÞ
ð11:0Þ2
¼ 1:1 104
2
3
ð4:20Þ 0:60 10
1 mol
73. ð0:048 g BaF2 Þ
¼ 2:7 104 mol BaF2
175:3 g
2:7 104 mol
¼ 1:8 102 M BaF2 dissolved
0:015 L
2 F
1:8 102
2 1:8 102
ðmolar concentrationÞ
2
Ksp ¼ ½Ba2þ ½F 2 ¼ 1:8 102 3:6 102 ¼ 2:3 105
BaF2 ðsÞ fi
Ba2þ
74. N2 þ 3 H2 fi
þ
2 NH3
2
Keq ¼
4:0 ¼
½NH3 ½N2 ½H2 3
¼ 4:0
y2
3
ð2:0Þð2:0Þ
y ¼ 8:0 M ¼ ½NH3 Let y ¼ ½NH3 y2 ¼ 64
y¼
pffiffiffiffiffi
64
75. Total volume of mixture ¼ 40:0 mL ð0:0400 LÞ
Ksp ¼ ½Sr2þ SO42 ¼ 7:6 107
3
1:0
10
M
ð0:0250 LÞ
½Sr2þ ¼
¼ 6:3 104 M
0:0400 L
2 2:0 103 M ð0:0150 LÞ
¼ 7:5 104 M
¼
SO4
0:0400 L
½Sr2þ SO42 ¼ 6:3 104 7:5 104 ¼ 4:7 107
4:7 107 < 7:6 107 no precipitation should occur:
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- Chapter 16 76. First change g Hg2 I2 ! mol Hg2 I2
3:04 107 g Hg2 I2
1 mol
¼ 4:64 1010 M Hg2 I2
L
655:0 g
Hg2 I2 fi
þ
Hg22þ
ðmolar solubilityÞ
2 I
4:64 1010 M
2 4:64 1010 M
2
Ksp ¼ Hg22þ ½I 2 ¼ 4:64 1010 9:28 1010 ¼ 4:00 1028
77. 3 O2 ðgÞ þ heat Ð 2 O3 ðgÞ
Three ways to increase ozone
(a) increase heat
(b) increase amount of O2
(c) increase pressure
(d) remove O3 as it is made
H2 OðgÞ
78. H2 Oðl Þ fi
Conditions on the second day
(a) the temperature could have been cooler
(b) the humidity in the air could have been higher
(c) the air pressure could have been greater
79. COðgÞ þ H2 OðgÞ Ð CO2 ðgÞ þ H2 ðgÞ
(c) is the correct answer
Keq ¼
½CO2 ½H2 ¼1
½CO½H2 O
With equal concentrations of products and reactants, the Keq value will equal 1.
80.
(a) Keq ¼
(b) Keq ¼
81.
½O3 2
(c) Keq ¼ ½CO2 ðgÞ
½O2 3
½H2 Oðl Þ
½H2 OðgÞ
½Hþ (d) Keq ¼ 2
Bi3þ ½H2 S3
6
2A
þ
B
fi
1:0 M
1:0 M
1:0 2ð0:30Þ
1:0 0:30
0:4 M
0:7 M
½C
0:30
Keq ¼
¼
¼3
½A2 ½B ð0:4Þ2 ð0:7Þ
C
0
0:30
0:30 M
Initial conditions
Equilibrium concentrations
82. Since the second reaction is the reverse of the first, the Keq value of the second reaction will be the
reciprocal of the Keq value of the first reaction.
½I2 ½Cl2 ¼ 2:2 103 ðfirst reactionÞ
Keq ¼
½ICl2
Keq ¼
½ICl2
½I2 ½Cl2 Keq ¼
1
¼ 450
2:2 103
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Page 224
- Chapter 16 83. HNO2 ðaqÞ Ð Hþ ðaqÞ þ NO2 ðaqÞ
OH reacts with Hþ and equilibrium shifts to the right.
(c)
After an initial increase, ½OH will be neutralized and equilibrium shifts to the right.
½Hþ will be reduced (reacts with OH ). Equilibrium shifts to the right.
NO2 increases as equilibrium shifts to the right.
(d)
½HNO2 decreases and equilibrium shifts to the right.
(a)
(b)
84. CaSO4 ðsÞ fi
Ca2þ ðaqÞ þ SO42 ðaqÞ
Ksp ¼ ½Ca2þ SO42 ¼ 2:0 104
Let x ¼ moles CaSO4 that dissolve per L ¼ ½Ca2þ ¼ SO42
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðxÞðxÞ ¼ 2:0 104
x ¼ 2:0 104
x ¼ 0:014 M CaSO4
M ! moles ! grams
0:014 mol CaSO4
136:2 g
¼ 1:1 g CaSO4
ð0:600 LÞ
mol
L
85. PbF2 ðsÞ fi
Pb2þ þ 2 F
change g PbF2 ! mol PbF2
0:098 g PbF2
1 mol
¼ 1:0 103 mol=L ¼ 1:0 103 M PbF2
245:2 g
0:400 L
Ksp ¼ Pb2þ ðF Þ2
2
Pb ¼ 1:0 103 ; ½F ¼ 2 1:0 103 ¼ 2:0 103
2
Ksp ¼ 1:0 103 2:0 103 ¼ 4:0 109
86. Treat this is an equilibrium where W ¼ whole nuts, S ¼ shell halves, and K ¼ kernels
Ð
2S þ K
W
144
0
0
amount before creacking
144 x
2x
x
144 x þ 2x þ x ¼ 194 total pieces
144 þ 2x ¼ 194; 2x ¼ 50
x ¼ 25 kernels; 50 shell halves;
2
Keq ¼
x ¼ number of kernels after creacking
119 whole nuts left
2
ð2xÞ ðxÞ ð50Þ ð25Þ
¼
¼ 5:3 102
144 x
119
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Page 225
- Chapter 16 87.
SO2 ðgÞ
0:50 M
þ NO2 ðgÞ
0:50 M
Ð
SO3 ðgÞ þ NOðgÞ
0
0
0:50 x
0:50 x
x
½SO3 ½NO
x2
¼
Keq ¼
¼ 81
½SO2 ½NO2 ð0:50 xÞ2
x
Take the square root of both sides
x
¼ 9:0
x ¼ 0:45 M
0:50 x
½SO3 ¼ ½NO ¼ 0:45 M
½SO2 ¼ ½NO2 ¼ 0:05 M
- 225 -
Initial conditions
Equilibrium concentrations