# ap acid chapters 15 & 16 practice problems

```15.116 First we must calculate the molarity of the trifluoromethane sulfonic acid. (Molar mass  150.1 g/mol)
1 mol
150.1 g
 0.0164 M
0.250 L
0.616 g 
Molarity 
Since trifluoromethane sulfonic acid is a strong acid and is 100% ionized, the [H ] is 0.0165 M.
pH  log(0.0164)  1.79
15.126 0.100 M Na2CO3  0.200 M Na  0.100 M CO32
First stage:
Initial (M):
Change (M):
Equilibrium (M):
CO32 (aq)  H2O(l )
0.100
x
0.100  x
K1 
K1 
HCO3 (aq)  OH(aq)
0
0
x
x
x
x
Kw
1.0  1014

 2.1  104
K2
4.8  1011
[HCO3 ][OH ]
[CO32 ]
2.1  104 
x2
x2

0.100  x
0.100
x  4.6  103 M  [HCO3 ]  [OH]
Second stage:
HCO3 (aq)  H2O(l)
Initial (M):
Change (M):
Equilibrium (M):
H2CO3(aq)  OH(aq)
4.6  103
y
0
y
4.6  103
y
(4.6  103)  y
y
(4.6  103)  y
K2 
[H2CO3 ][OH ]
[HCO3 ]
2.4  108 
y[(4.6  103 )  y]
(4.6 103 )  y

( y)(4.6  103 )
(4.6  103 )
y  2.4  108 M
At equilibrium:
[Na]  0.200 M
[CO23 ] = 0.100 M  4.6  103 M = 0.095 M
[HCO 3 ]  (4.6  103) M  (2.4  108) M  4.6  103 M
[H2CO3]  2.4  108 M
[OH]  (4.6  103) M  (2.4  108] M  4.6  103 M
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
[H  ] 
1.0  1014
4.6  10
3
543
 2.2  1012 M
15.129 When the pH is 10.00, the pOH is 4.00 and the concentration of hydroxide ion is 1.0  104 M. The
concentration of HCN must be the same. (Why?) If the concentration of NaCN is x, the table looks like:
Initial (M):
Change (M):
Equilibrium (M):
Kb 
CN(aq)  H2O(l)
x
1.0  104
HCN(aq)
0
1.0  104
(x  1.0  104)

(1.0  104)
OH(aq)
0
1.0  104
(1.0  104)
[HCN][OH  ]
[CN  ]
2.0  105 
(1.0  104 )2
( x  1.0  104 )
x  6.0  104 M  [CN]0
Amount of NaCN  250 mL 
16.14
6.0  104 mol NaCN 49.01 g NaCN

 7.4  103 g NaCN
1000 mL
1 mol NaCN
Step 1: Write the equilibrium that occurs between H2 PO4 and HPO24 . Set up a table relating the initial
concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations.
Initial (M):
Change (M):
Equilibrium (M):
H2 PO4 (aq)
0.15
x
0.15  x
H(aq)  HPO24 (aq)
0
0.10
x
x
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 ][HPO24 ]
[H2 PO4 ]
You can look up the Ka value for dihydrogen phosphate in Table 15.5 of your text.
6.2  108 
( x)(0.10  x)
(0.15  x)
6.2  108 
( 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.
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CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
pH  log[H]  log(9.3  108)  7.03
16.15
Using the HendersonHasselbalch equation:
pH  pKa  log
[CH3COO ]
[CH3COOH]
4.50  4.74  log
[CH3COO ]
[CH3COOH]
Thus,
[CH3COO ]
 0.58
[CH3COOH]
16.18
As calculated in Problem 16.12, the pH of this buffer system is equal to pKa.
pH  pKa  log(1.8  105)  4.74
(a)
The added NaOH will react completely with the acid component of the buffer, CH 3COOH. NaOH
ionizes completely; therefore, 0.080 mol of OH are added to the buffer.
Step 1: The neutralization reaction is:
Initial (mol):
Change (mol):
Final (mol):
CH3COOH(aq)  OH(aq) 
 CH3COO(aq)  H2O(l)
1.00
0.080
1.00
0.080
0.080
0.080
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):
0.92
Change (M):
x
Equilibrium (M):
0.92  x
H(aq)  CH3COO(aq)
0
1.08
x
x
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.
pH  log[H]  log(1.5  105)  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, CH 3COO. HCl ionizes
completely; therefore, 0.12 mol of H ion are added to the buffer
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
545
Step 1: The neutralization reaction is:
Initial (mol):
Change (mol):
Final (mol):
CH3COO(aq)  H(aq) 
 CH3COOH(aq)
1.00
0.12
1.00
0.12
0.12
0.12
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.
H(aq)  CH3COO(aq)
0
0.88
x
x
x
0.88  x
CH3COOH(aq)
Initial (M):
1.12
Change (M):
x
Equilibrium (M):
1.12  x
Write the Ka expression, then solve for x.
Ka 
[H  ][CH3COO ]
[CH3COOH]
1.8  105 
( x)(0.88  x)
x(0.88)

(1.12  x)
1.12
x  [H]  2.3  105 M
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CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
Step 3: Having solved for the [H], calculate the pH of the solution.
pH  log[H]  log(2.3  105)  4.64
The pH of the buffer decreased from 4.74 to 4.64 upon addition of 0.12 mol of strong acid.
16.29
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:
mol NaOH  12.5 mL H 2SO4 
concentration of NaOH 
16.53
(a)
0.500 mol H2SO4 2 mol NaOH

 0.0125 mol NaOH
1000 mL soln
1 mol H 2SO4
0.0125 mol NaOH
50.0  103 L soln
 0.25 M
The solubility equilibrium is given by the equation
AgI(s)
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  ] 
Ksp

[Ag ]
(b)

8.3  1017
9.1  10
9
 9.1  109 M
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)
Al3(aq)  3OH(aq)
Ksp  [Al3][OH]3
Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum
ion is:
Ksp
1.8  1033
[Al 3 ] 

 7.4  108 M
 3
9 3
[OH ]
(2.9  10 )
What is the pH of this solution? Will the aluminum concentration change if the pH is altered?
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
16.54
547
Strategy: In each part, we can calculate the number of moles of compound dissolved in one liter of solution
(the molar solubility). Then, from the molar solubility, s, we can determine Ksp.
Solution:
(a)
7.3  102 g SrF2
1 mol SrF2

 5.8  104 mol/L = s
1 L soln
125.6 g SrF2
Consider the dissociation of SrF2 in water. Let s be the molar solubility of SrF2.
SrF2(s)
Initial (M):
Change (M):
Equilibrium (M):
s
Sr2(aq)  2F(aq)
0
0
s
2s
s
2s
Ksp  [Sr2][F]2  (s)(2s)2  4s3
The molar solubility (s) was calculated above. Substitute into the equilibrium constant expression to
solve for Ksp.
Ksp  [Sr2][F]2  4s3  4(5.8  104)3  7.8  1010
(b)
6.7  103 g Ag3PO4
1 mol Ag3PO4

 1.6  105 mol/L = s
1 L soln
418.7 g Ag3PO4
(b) is solved in a similar manner to (a)
The equilibrium equation is:
Ag3PO4(s)
Initial (M):
Change (M):
Equilibrium (M):
s
3Ag(aq)  PO34 (aq)
0
0
3s
s
3s
s
Ksp  [Ag]3 [PO34 ]  (3s)3(s)  27s4  27(1.6  105)4  1.8  1018
```