Homework #7
Chapter 8
Applications of Aqueous Equilibrium
15.
Buffer solution: A solution that resists change in pH when a small amount of acid
or base is added.
Buffer solutions contain a weak acid and its conjugate base or a weak base and its
conjugate acid. For weak acid/conjugate base systems, when a base is added to the
solution it reacts with the weak acid to form more conjugate base. When an acid is
added to the same solution it reacts with conjugate base to form more weak acid. In
both cases the pH remains approximately constant since the number of OH-/H+ ions in
solution is approximately constant. A similar system is set up for weak base/ conjugate
acid buffers. In buffer systems the larger the amount of weak acid/conjugate base or
weak base/conjugate acid the better the buffer. In addition, the best buffers have equal
amounts of weak acid and conjugate base or weak base and conjugate acid.
If the buffer solution is made of NaHCO3 and Na2CO3 then the following equation will
happen in solution
HCO3-(aq) + OH-(aq) ⇌ CO32-(aq) + H2O(l)
CO32-(aq) + H+(aq) ⇌ HCO3-(aq)
16.
Buffer Capacity: An indication of the amount of acid or base that can be added before a
buffer loses its ability to resist the change in pH.
The buffer capacity is greatest when there are equal amount of weak acid and conjugate
base or weak base and conjugate acid.
In addition, as the overall amount of weak acid and conjugate base goes up the
buffering capacity of the solutions go up (same as for weak base and conjugate acid).
Therefore, solution C which has 1.0 M solutions of a weak acid and its conjugate base
will have the greatest buffering capacity.
The Henderson-Hasselbalch equation allows us to calculate the pH of a buffered system.
 A 
pH  pK a  log  HA 


Systems with the greatest buffer capacity will have [A-]=[HA] causing the pH=pKa.
Therefore, when choosing a buffer system you should choose a system that has a pKa
close to what the pH of the overall solution should be.
Since all of the 3 systems have equal amounts of [A-] and [HA] the pH for all the systems
is the same.
1
21.
a)
Major Species: HC3H5O2
No reaction goes to completion
Not a buffer (no conjugate base present)
Use an ICE table to determine pH.
HC3H5O2(aq) ⇌ H+(aq) + C3H5O2-(aq)
HC3H5O2 H+ C3H5O20.100 M
0
0
Initial
-x
+x
+x
Change
0.100-x
x
x
Equilibrium
H C H O  



Ka
3
5
2
HC3 H 5O2 
xx
 1.3  10 5
0.100  x 
Since Ka is small assume 0.100 –x = 0.100
x2
 1.3  10 5
0.100
x  0.0011
Check assumption
0.0011
100%  1.1% Good
0.100
Concentration of H+
H   x  0.0011M

Calculate the pH
 
pH   log H    log0.0011  2.96
b)
+
Major Species: Na and C3H5O2No reaction goes to completion
Not a buffer (no conjugate acid present)
Use an ICE table to determine pH.
Need to determine Kb of:
C3H5O2-(aq) + H2O(l) ⇌ HC3H5O2(aq) + OH-(aq) Kb =?
Kb 
K w 1.0 1014

 7.7 1010
5
K a 1.3 10
Initial
Change
Equilibrium
Kb 
C3H5O20.100 M
-x
0.100-x
 HC3 H 5O2  OH  

HC3H5O2
0
+x
x
OH0
+x
X
xx
 7.7  1010
 0.100  x 
C3 H 5O2 
Since Kb is small assume 0.100 –x = 0.100

x2
 7.7  10 10
0.100
x  8.8  10 6
2
Check assumption
8.8  10 6
 0.0088%
0.100
Good
Concentration of OH-
OH    x  8.8 106 M
Calculate pOH
pOH   log OH     log 8.8 106   5.06
Calculate pH
pH  14  pOH  14  5.06  8.94
c)
For pure water [H+] = [OH-] = 1.0×10-7
 


pH   log H    log 1.0  10 7  7.00
d)
Major Species: HC3H5O2, Na+, and C3H5O2No reaction goes to completion
Buffer
Use Henderson-Hasselbalch equation to solve for pH.
 A 
pH  pK a  log  HA 


0.100
pH   log K a  log  0.100 
pH   log 1.3 105   4.89
You could have also used and ice table to solve for pH.
HC3H5O2(aq) ⇌ H+(aq) + C3H5O2-(aq)
HC3H5O2 H+
C3H5O20.100 M
0
0.100
Initial
-x
+x
+x
Change
0.100-x
x
0.100+x
Equilibrium
H C H O   x0.100  x  1.3 10



Ka
3
5
2
HC3 H 5O2 
0.100  x 
5
Since Ka is small assume 0.100 –x = 0.100
x0.100
 1.3  10 5
0.100
x  1.3  10 5
Check assumption
1.3  10 5
100%  0.013% Good
0.100
Concentration H+
x  1.3  10 5
pH   log H    log 1.3  10 5  4.89
 


3
22.
a)
Major Species: H+, Cl-, and HC3H5O2
No reaction goes to completion
Not a buffer (no conjugate acid present)
The majority of the H+ ions are coming from the strong acid therefor calculate
pH as you would for a strong acid.
pH   log  H     log  0.020   1.70
We can check this by using the long way
HC3H5O2(aq) ⇌ H+(aq) + C3H5O2-(aq)
HC3H5O2
H+
0.100 M 0.020 M
Initial
-x
+x
Change
0.100-x
0.020+x
Equilibrium
C3H5O20
+x
X
 H   C3 H 5O2   x  0.020  x 
Ka 

 1.3 105
HC
H
O
0.100

x


 3 5 2
Since Ka is small assume 0.023+x = 0.023 and 0.100-x = 0.100
x  0.020 
 1.3 105
 0.100 
x  6.5 105
Check assumptions
6.5 105
100%  0.33%
0.020
6.5 105
100%  0.065%
0.100
Good
Good
Concentration of H+
 H    0.020  x  0.020  6.5 105  0.020M
Calculate the pH
pH   log  H     log  0.020   1.70
Major Species: H+, Cl-, and C3H5O2Yes a reaction goes to completion
H+ + C3H5O2- HC3H4O2(aq)
C3H5O2H+
HC3H5O2
0.100 mol 0.020 mol
0
Initial
0.080
mol
0
0.020
mol
Final
Major Species after reaction: C3H5O2 and HC3H5O2
Buffer
b)
 A 
pH  pK a  log  HA 


pH   log 1.3 105   log
c)

0.080 mol
V
0.020 mol
V
  5.49
Major Species: H+ and ClNo reaction goes to completion
4
Not a buffer
pH   log  H     log  0.020   1.70
d)
Major Species: H+, Cl-, Na+, and C3H5O2Yes a reaction goes to completion
C3H5O2-(aq) + H+(aq)  HC3H5O2(aq)
C3H5O2H+
HC3H5O2
0.100 mol 0.020 mol 0.100 mol
Initial
0.080 mol
0
0.120 mol
Final
Major Species: Cl-, Na+, C3H5O2- and HC3H5O2
Buffer
 A 
pH  pK a  log  HA 


pH   log 1.3 105   log
23.
a)
0.080 mol
V
0.120 mol
V
  4.71
Major Species: HC3H5O2, Na+, and OHYes a reaction goes to completion
HC3H5O2(aq) + OH-(aq)  C3H5O2-(aq) + H2O(l)
HC3H5O2
OHC3H5O20
0.100 mol 0.020 mol
Initial
0.020 mol
0.080 mol
0
Final
+
Major Species: HC3H5O2, Na , and C3H5O2
Buffer
 A 
pH  pK a  log  HA 


pH   log 1.3 105   log
b)


0.020 mol
V
0.080 mol
V
  4.28
Major Species: C3H5O2-, Na+, and OHNo reaction goes to completion.
Not a buffer.
The majority of the OH- ions are from the strong base. Therefore, use the
strong base concentration to calculate pH
pOH   log OH     log  0.020   1.70
pH  pOH  14
pH  14  pOH  14  1.70  12.30
We can check this
C3H5O2-(aq) + H2O(l)⇌ OH-(aq) + HC3H5O2(aq)
Kb  7.7 1010
C3H5O2OHHC3H5O2
0.100 M 0.020 M
0
Initial
-x
+x
+x
Change
0.100-x
0.020+x
X
Equilibrium
5
OH    HC3 H 5O2  x  0.020  x 
Kb 

 7.7 1010

C3 H 5O2 
 0.100  x 
Since K is small assume 0.020+x = 0.020 and 0.100-x = 0.100
x  0.020 
 7.7 1010
 0.100 
x  3.9 1010
Check assumptions
7.7 1010
100%  3.9 106%
0.020
7.7 1010
100%  7.7 107 %
0.100
Good
Good
Concentration of OH[𝑂𝐻 − ] = 0.20 + 𝑥 = 0.020 + 3.9 × 10−10 = 0.020𝑀
Calculate the pH
pOH   log OH     log  0.020   1.70
pH  pOH  14
pH  14  pOH  14  1.70  12.30
c)
Major Species: Na+, and OHNo reaction goes to completion
Not a Buffer
pOH   log OH     log  0.020   1.70
pH  pOH  14
pH  14  pOH  14  1.70  12.30
d)
Major Species: HC3H5O2, C3H5O2-, Na+, and OHYes a reaction goes to completion
HC3H5O2(aq) + OH-(aq)  C3H5O2-(aq) + H2O(l)
HC3H5O2
OHC3H5O20.100 mol 0.020 mol 0.100 mol
Initial
0.120 mol
0.080 mol
0
Final
Major Species: HC3H5O2, C3H5O2-, and Na+
Buffer
 A 
pH  pK a  log  HA 


pH   log 1.3 105   log
25.

0.120 mol
V
0.080 mol
V
  5.06
Major Species: HF, K+, and FNo reaction goes to completion
Buffer
Ka = 7.2×10-4 (Table 7.2)
6
 A 
pH  pK a  log  HA 


M
pH   log  7.2 104   log  1.00
0.60 M   3.36
27.
Major Species: HF, K+, F-, Na+, and OHYes a reaction goes to completion
HF(aq) + OH-(aq)  F-(aq) + H2O(l)
HF
OHF0.60 mol
0.10 mol 1.00 mol
Initial
1.10 mol
0.50 mol
0
Final
+ +
Major Species: HF, K , F , and Na
Buffer
 A 
pH  pK a  log  HA 


pH   log  7.2 104   log
Major Species: HF, K+, F-, H+, and ClYes a reaction goes to completion
F-(aq) + H+(aq)  HF(aq)
FH+
1.00 mol
0.20 mol
Initial
0.80 mol
0
Final
+ +
Major Species: HF, K , F , and Na
Buffer
 A 
pH  pK a  log  HA 


pH   log  7.2 104   log
33.
   3.49
1.10 mol
V
0.50 mol
V
HF
0.60 mol
0.80 mol
   3.14
0.80 mol
V
0.80 mol
V
All of the solutions are buffer solutions because there is a weak base (C5H5N) and its
conjugate acid (C5H5NH+) present, therefore, you can use Henderson-Hasselbalch
equation.
 A 
pH  pK a  log  HA 


 A 
 
 HA
 10 pH  pKa
Need to determine pKa of C5H5NH+
Ka 
Calculate pKa
Kw
Kb
14
10
 1.0
 5.9 106
1.7109


pK a   logK a    log 5.9 10 6  5.23
a)
 A 
 
 HA
 10 pH  pKa  104.505.23  0.19
7
35.
b)
 A 
 
 HA
 10 pH  pKa  105.005.23  0.59
c)
 A 
 
 HA
 10 pH  pKa  105.235.23  1.0
d)
 A 
 
 HA
 10 pH  pKa  105.505.23  1.9
Major Species: Na+, C2H3O2-, H+, and ClYes a reaction will go to completion
H+ + C2H3O2-  HC2H3O2
H+
C2H3O2HC3H5O2
x
1.0 mol
0
Initial
0
1.0 mol-x
X
Final
Major Species: C2H3O2- , HC2H3O2, Na+, and ClBuffer
a)
When the pH = pKa [HC2H3O2] = [C2H3O2-]
 A 
 
 HA
 1  1.0x x
x  0.5M
b)
Since it is a 1.0 L solution 0.5 moles of HCl is needed
Ka = 1.8×10-5
 A 
pH  pK a  log  HA 


 A 
4.20   log 1.8 105   log  HA 


log  HA   0.54


 A 
 A 
 
 HA
 0.29  1.0x x
x  0.78M
Since it is a 1.0 L solution 0.78 moles of HCl is needed.
c)
 A 
pH  pK a  log  HA 


 A 
5.00   log 1.8 105   log  HA 


 A 
log  HA   0.25


 A 
 
 HA
 1.8  1.0x x
x  0.36 M
Since it is a 1.0 L solution 0.36 moles of HCl is needed.
37.
Major Species: Na=, C2H3O2-, and HC2H3O2
No reaction goes to completion
8
Buffer
Ka = 1.8×10-5
 A 
pH  pK a  log  HA 


 A 
  
5.00   log 1.8 105   log  0.200
M 


 A   0.36M
They asked for the mass not the molarity
g
M NaC2 H3O2  82.04 mol
0.500 L C2 H 3O2
38.
a)

0.36 mol C2 H 3O2
1L C2 H 3O2

1mol NaC2 H 3O2
1mol C2 H 3O2

82.04 g NaC2 H 3O2
1mol NaC2 H 3O2
pH   log  5.6 1010   log

Major Species: H+, Cl-, NH3, and NH4+
A reaction will go to completion
NH3(aq) + H+(aq)  NH4+(aq)
NH3
H+
0.125 mol 0.010 mol
Initial
0.115 mol
0
Final
+
Major Species: Cl , NH3, and NH4
Buffer
 A 
pH  pK a  log  HA 


pH   log  5.6 1010   log
40.
2
Major Species: H+, Cl-, NH3, and NH4+
A reaction will go to completion
NH3(aq) + H+(aq)  NH4+(aq)
NH3
H+
Initial 0.0125 mol 0.010 mol
0.0025 mol
0
Final
+
Major Species: Cl , NH3, and NH4
Buffer
 A 
pH  pK a  log  HA 


b)
  15g NaC H O

0.115 mol
V
0.385 mol
V
3
2
NH4+
0.0375 mol
0.0475 mol
0.0025 mol
V
0.0475 mol
V
  7.97
NH4+
0.375 mol
0.385 mol
  8.73
Major Species: HNO3, Na+, and NO2
No reaction goes to completion
Buffer
 A 
pH  pK a  log  HA 


pKa   log  Ka    log  4.0 104   3.40 (Table 7.2)
9
Total volume
x + y = 1.00
x = volume of HNO2
y = volume of NO2Molarity of HNO2 in final solution
M
n x0.50 x0.50


V x  y 1.0 L
Molarity of NO2- in final solution
M
n y 0.50 y 0.50


V x y
1.0 L
Use the Henderson-Hasselbalch equation to find another relationship between x and y
 A 
pH  pK a  log  HA 


3.55  3.40  log
 
0.50 y
1.0 L
0.50 x
1.0 L
 y
0.15  log  
x
y  1.4 x
The total volume of the solution is 1 L (x+y=1). Use this equation to solve for x and y.
y  1.4 x
x  y  1.00
x  1.4 x  2.4 x  1.00
x  0.42 L  VHNO2
y  .58L  VNO
2
41.
a)
Major Species: H2PO4- and HPO42No reaction goes to completion.
Buffer


pK a   logK a    log 6.2  10 8  7.21
A
pH  pK a  log  HA 



 HPO4 2 
7.15  7.21  log  H PO   
 2 4 
b)
 HPO4 2 


 H 2 PO4 


 0.87
 H 2 PO4 


 HPO4 2 



1
 1.1
0.87
The pH of intracellular fluid is ~7.15 (part b). The pKa of H3PO4 is 2.12. The best
buffers have equal amounts of A- and HA, therefore, the pH = pKa. In order to
10
42.
a)
have a pH of 7.15 with a solution containing H3PO4/H2PO4- there would have to
be much more H2PO4- in solution that H3PO4 making it an ineffective buffer.
Major Species: H2CO3 and HCO3No reaction goes to completion
Buffer
pKa   log  Ka    log  4.3 107   6.37
 A 
pH  pK a  log  HA 


 HCO3 
 HCO3 


7.40  6.37  log   H 2CO3    6.37  log  0.0012
M 




 HCO3   0.013M
44.
a)
b)
c)
d)
45.
a)
b)
Major Species: K+, OH-, CH3NH3+, and ClYes a reaction goes to completion
OH-(aq) + CH3NH3+(aq)  H2O(l) + CH3NH2(aq)
OHCH3NH3+
CH3NH2
0
0.1 mol
0.1 mol
Initial
0.1
mol
0
0
Final
+
Major Species: K , CH3NH2, and Cl
Not a buffer
Major Species: K+, OH-, and CH3NH2
No reaction goes to completion
Not a buffer
Major Species: K+, OH-, CH3NH3+, and ClYes a reaction goes to completion
OH-(aq) + CH3NH3+(aq)  H2O(l) + CH3NH2(aq)
OHCH3NH3+
CH3NH2
0
0.2 mol
0.1 mol
Initial
0.1
mol
0.1 mol
0
Final
+
Major Species: K , OH , CH3NH2, and Cl
Not a buffer
Major Species: K+, OH-, CH3NH3+, and ClYes a reaction goes to completion
OH-(aq) + CH3NH3+(aq)  H2O(l) + CH3NH2(aq)
OHCH3NH3+
CH3NH2
0
0.1 mol
0.2 mol
Initial
0.1
mol
0
0.1 mol
Final
+
Major Species: K , CH3NH2, Cl , and CH3NH2
Buffer
Major Species: H+, NO3-, Na+, and NO3No reaction goes to completion
Not a buffer
Major Species: H+, NO3-, HF
No reaction goes to completion
11
Not a buffer
c)
d)
46.
Major Species: H+, NO3-, Na+, and FYes a reaction goes to completion
H+(aq) + F-(aq)  HF(aq)
H+
F0.2 mol
0.4 mol
Initial
0
0.2 mol
Final
+
Major Species: NO3 , Na , and F
Buffer
Major Species: H+, NO3-, Na+, and OHYes a reaction goes to completion
H+(aq) + OH-(aq)  H2O(aq)
H+
OH0.2 mol
0.4 mol
Initial
0
0.2 mol
Final
+
Major Species: NO3 , Na , and OHNot a Buffer
Major Species: Na+, F-, H+, and ClA reaction goes to completion
H+ + F-  HF
H+
F0.0025 mol
0.0100 mol
Initial
0
0.0075 mol
Final
Major Species: Na+, F-, Cl-, and HF
Buffer
Ka = 7.2×10-4 (Table 7.2)
 A 
pH  pK a  log  HA 


pH   log  7.2 104   log

HF
0
0.2 mol
HF
0
0.0025 mol
0.0075 mol
V
0.0025 mol
V
  3.62
47.
A buffer has the greatest buffer capacity when [HA] = [A-]. When this happens the pH =
pKa. Therefore, the best acid would have a pKa of 7.00. The acid that has the closest pKa
value is HOCl which has a pKa of 7.46. In order to make a 1 L solution you would mix
equal amounts of HOCl with NaOCl (or another salt containing OCl-).
52.
a)
The blue line is the weak acid and the red line is the weak acid. You can tell the
difference between the two plots because of the following 3 reasons. 1) The
equivalence point of a weak acids pH curve is at a pH great than 7 and the
equivalence point of a strong acid pH curve is at 7. 2) At the start of the pH
curve of a weak acid titration the pH changes quickly and then levels off. This
does not happen in a strong acid titration curve. 3) Since both acids have the
same initial concentration the starting point of the strong acid should be at a
lower pH than the weak acid.
12
b)
c)
d)
53.
a)
b)
c)
d)
e)
A buffer solution is a solution that resist change in pH. In the weak acid/strong
base titration, as the strong base is added to the weak acid resulting in the
formation of the conjugate base of the weak acid which will be a weak base.
When there is equal amount of the weak acid and weak base this is called the
half equivalence point. At this point if an acid is added the weak base can react
with the acid to absorb the added H+ in the solution resulting in approximately
the same pH. If a base is added the weak acid can react with the base to absorb
the added OH- in the solution resulting in approximately the same pH.
Therefore, the middle of the buffer region sits at the half equivalence point.
Although the pH appears to be stable at the half equivalence point for the
strong acid/strong base titration this system is not considered a buffer. When
strong base is added to the acid system the H+ reacts with the OH- to form
water. However, water is neutral therefore, if an acid was added to the system
there would be nothing to react with it causing the pH to go down dramatically.
The statement is true. Since both acid have the same initial amount and initial
concentrations the amount of base added to get to the equivalence point will be
the same for both systems. The definition of the equivalence point is the point
at which enough titrant has been added to fully react the analyte. Since the
moles of acid is the same for both the strong and weak acid the number of
moles of base needed to fully react each system is the same.
The statement is false. When a strong acid reacts with a strong base a neutral
salt and water is formed. This results in a pH of 7.00 at the equivalence point.
When a weak acid reacts with a strong base the conjugate base of the weak acid
is formed. At the equivalence point all of the weak acid will be converted into
the conjugate base of the acid resulting in a pH that is greater than 7.00 at the
equivalence point.
The acid is a weak acid. You can tell this because at the beginning of the
titration (beaker 1) all of the hydrogen atoms (blue atoms) are attached to their
counter ions (green atoms). As strong base is added, hydrogen atoms are
removed from their counter ions forming water and leaving behind the
conjugate base (or counter ion) of the acid. If the system was a strong acid
solution add you would see only hydrogen ions (blue atoms) in the solution at
the beginning of the titration. As strong base is added, the hydrogen atoms
should disappear because they are reacting with OH- to form water which is not
shown in the picture.
c(start of titration) ae (half equivalence point)b (equivalence
point)d(end of titration)
The pH=pKa for beaker e. This is the half equivalence point or the point where
half of the weak acid (green atom bonded to blue atom) is converted to the
conjugate base (green atom) of the weak acid.
Beaker b represents the equivalence point. The equivalence point occurs when
all of the weak acid (green atom bonded to blue atom) is converted to the
conjugate base (green atom) of the weak acid.
You would not need to know the Ka value of the weak acid to determine the pH
of the system for beaker d. At this point all of the weak acid has been converted
into the conjugate base of the weak acid. In addition, more strong base has
been added. In a system with a strong base and a weak base the strong base
13
54.
determines the pH of the system due to the dramatically larger number of OHions from the strong base.
a)
The equivalence point occurs when equal mole amounts of weak acid and
strong base have been added to the system. For this plot it occurs after ~22 mL
of base are added. The pH at the equivalence point will be greater than 7.
b) & c) The maximum buffering occurs when the [HA] = [A-] or pH = pKa. This will be in
the middle of the flat region (~12).
d)
The pH only depends on [HA] on the far left of the plot because no [A-] is
present.
e)
The pH only depends on [A-] at the equivalence point, when ~22 mL of base are
added
f)
The pH depends only on the amount of excess base on the far right of the plot.
55.
For weak base titrated with a strong acid the equivalence point should occur at a pH
lower than 7 due to the weak conjugate acid that is generated.
59.
a)
b)
c)
f (The pH initially increases more rapidly for weak acids than it does for strong
acids. In, addition the weaker the acid the stronger the conjugate base
therefore the weakest acid should have the highest pH at the equivalence point)
a. If you did not know the initial concentration of the acids and wanted to know
if the acid was strong or weak you could look at the pH at the equivalence point.
If the pH at the equivalence point is 7.0 then it is a strong acid if the pH at the
equivalence point is greater than 7.0 it is a weak acid.
d (The pKa of the system is 6.0, therefore, when [HA] = [A-] the pH of the system
should be 6.0. This is referred to as the ½ equivalence point because ½ the
amount of base has been added that would be needed to get to the equivalence
point. For this graph the ½ equivalence point occurs after ~25 mL of NaOH has
been added. The line that has a pH closest to 6 at the ½ equivalence point is d.
14
61.
This is a strong acid strong base titration
Calculate the initial moles of HClO4
0.0400L HClO4
a)

0.200 mol HClO4
1L HClO4

1mol H 
1mol HClO4
  0.00800mol H

Major Species: H+, and ClO4No reaction goes to completion
Not a buffer
 0.00800mol 
pH   log  H     log 
  0.699
 0.0400 L 
b)
Major Species: H+, ClO4-, K+, and OHA reaction goes to completion
H+(aq) + OH-(aq)  H2O(l)
H+
OH0.00800 mol 0.00100 mol
Initial
0.00700 mol
0
Final
+
+
Major Species: H , ClO4 , and K
Not a buffer
0.00700mol


pH   log  H     log 
  0.854
 0.0400 L  0.0100 L 
c)
Major Species: H+, ClO4-, K+, and OHA reaction goes to completion
H+(aq) + OH-(aq)  H2O(l)
H+
0.00800 mol
Initial (mol)
0.00400 mol
Final (mol)
Major Species: H+, ClO4-, and K+
Not a buffer
OH0.00400 mol
0
0.00400mol


pH   log  H     log 
  1.301
 0.0400 L  0.0400 L 
d)
e)
Major Species: H+, ClO4-, K+, and OHA reaction goes to completion
H+(aq) + OH-(aq)  H2O(l)
H+
OH0.00800 mol 0.00800 mol
Initial (mol)
0
0
Final (mol)
+
Major Species: ClO4 , and K
Not a buffer
Since the moles of OH- equal the moles of H+ the solution will be neutral 7.00.
Another way to state this is that we are at the equivalence point.
Major Species: H+, ClO4-, K+, and OHA reaction goes to completion
H+(aq) + OH-(aq)  H2O(l)
H+
OH0.00800 mol 0.01000 mol
Initial (mol)
15
Final (mol)
Major Species:
Not a buffer
ClO4-,
0
+
0.00200 mol
-
K , and OH
0.0020mol


pOH   log OH     log  0.014
  1.85
0.0400 L  0.0800 L 

pH  14  pOH  14  1.85  12.15
62.
Ba(OH)2 is a strong base and HCl is a strong acid
Calculate the moles of Ba(OH)2 that you start with
0.0800L Ba  OH 2
a)

0.100 mol Ba OH 2
1L Ba OH 2

2 mol OH 
1mol Ba  OH 2
  0.0160mol OH

Major Species: Ba2+, and OHNo reaction goes to completion
Not a buffer
 .0160mol 
pOH   log OH     log 
  0.699
 0.0800 L 
pH  14  pOH  14  0.699  13.301
b)
Major Species: Ba2+, OH-, H+, and ClYes a reaction goes to completion
H+(aq) + OH-(aq)  H2O(l)
OH0.0160 mol
Initial (mol)
0.0080 mol
Final (mol)
2+
Major Species: Ba , OH , and ClNot a buffer
H+
0.00800 mol
0
0.0080mol


pOH   log OH     log 
  1.10
 0.0800 L  0.0200 L 
pH  14  pOH  14  1.10  12.90
c)
Major Species: Ba2+, OH-, H+, and ClYes a reaction goes to completion
H+(aq) + OH-(aq)  H2O(l)
OH0.0160 mol
Initial (mol)
0.0040 mol
Final (mol)
2+
Major Species: Ba , OH , and ClNot a buffer
H+
0.0120 mol
0
0.0040mol


pOH   log OH     log 
  1.44
 0.0800 L  0.0300 L 
pH  14  pOH  14  1.44  12.56
d)
Major Species: Ba2+, OH-, H+, and ClYes a reaction goes to completion
H+(aq) + OH-(aq)  H2O(l)
16
e)
OHH+
0.0160 mol 0.0160 mol
Initial (mol)
0
0
Final (mol)
2+
Major Species: Ba , and Cl
Not a buffer
Since the moles of OH- equal the moles of H+ the solution will be neutral 7.00.
Another way to state this is that we are at the equivalence point.
Major Species: Ba2+, OH-, H+, and ClYes a reaction goes to completion
H+(aq) + OH-(aq)  H2O(l)
OHH+
0.0160 mol 0.0320 mol
Initial (mol)
0
0.0160 mol
Final (mol)
Major Species: Ba2+, H+, and ClNot a buffer
0.0160mol


pH   log  H     log 
  1.000
 0.0800 L  0.0800 L 
63.
This is a weak acid/strong base titration
a)
Major Species: HC2H3O2
No reaction goes to completion
Not a buffer
HC2H3O2(aq) ⇌ H+(aq) + C2H3O2-(aq)
HC2H3O2 H+ C2H3O20.200 M
0
0
Initial
-x
+x
+x
Change
0.200-x
x
x
Equilibrium
Ka 
xx
 1.8  10 5
(0.200  x)
Since Ka is small assume 0.200-x = 0.200
xx
 1.8  10 5
(0.200)
x  0.0019
Check assumption
0.0019
100%  0.95% Good
0.200
Calculate pH
H   x  0.0019
pH   logH    log0.0019  2.72


b)
Major Species: HC2H3O2, K+, and OHA reaction goes to completion
HC2H3O2(aq) + OH-(aq)  H2O(l) + C2H3O2-(aq)
17
HC2H3O2
OH0.0200 mol 0.00500 mol
Initial (mol)
0.0150 mol
0
Final (mol)
+
Major Species: HC2H3O2, K , and C2H3O2
Buffer
 A 
pH  pK a  log  HA    log 1.8 105   log


c)

C2H3O20
0.00500 mol
0.00500 mol
V
0.0150 mol
V
Major Species: HC2H3O2, K+, and OHA reaction goes to completion
HC2H3O2(aq) + OH-(aq)  H2O(l) + C2H3O2-(aq)
HC2H3O2
OH0.0200 mol 0.01000 mol
Initial (mol)
0.0100 mol
0
Final (mol)
+
Major Species: HC2H3O2, K , and C2H3O2
Buffer
  4.27
C2H3O20
0.01000 mol
 A 
pH  pK a  log  HA    log 1.8 105   log


d)
Major Species: HC2H3O2, K+, and OHA reaction goes to completion
HC2H3O2(aq) + OH-(aq)  H2O(l) + C2H3O2-(aq)
HC2H3O2
OH0.0200 mol
0.01500 mol
Initial (mol)
0.0050 mol
0
Final (mol)
+
Major Species: HC2H3O2, K , and C2H3O2
Buffer
Major Species: HC2H3O2, K+, and OHA reaction goes to completion
HC2H3O2
OH0.0200 mol 0.02000 mol
Initial (mol)
0
0
Final (mol)
+
Major Species: C2H3O2 , and K
Not a buffer
C2H3O2-(aq) + H2O(l) ⇌ HC2H3O2(aq) + OH-(aq)
C2H3O2HC2H3O2 OH0.02000 mol
0
0
Initial (mol)
0.0667 M
0
0
Initial (M)
Change
-x
+x
+x
Equilibrium
0.0667-x
x
x
Calculate Kb
Kb 
0.01000 mol
V
0.01000 mol
V
  4.74
C2H3O20
0.01500 mol
 A 
pH  pK a  log  HA    log 1.8 105   log


e)


0.01500 mol
V
0.0050 mol
V
  5.22
C2H3O20
0.02000 mol
Kb
K w 1.0 1014

 5.6 1010
K a 1.8 105
18
Kb 
xx
 5.6  10 10
0.0667  x 
Since Kb is very small assume 0.0600-x = 0.0600
xx
 5.6  10 10
0.0667
x  6.1  10 6
Check Assumption
6.1  10 6
100%  0.0091% Good
0.0667
Calculate pH
[OH  ]  x  6.1106 M
pOH   log OH     log  6.1106   5.21
pH  14  pOH  14  5.21  8.79
f)
Major Species: HC2H3O2, K+, and OHA reaction goes to completion
HC2H3O2
OH0.0200 mol 0.02500 mol
Initial (mol)
0
0.00500
Final (mol)
+
Major Species: C2H3O2 , K , and OH
The majority of the OH- will come from OH-
C2H3O20
0.02000 mol
 0.00500mol 
pOH   log OH     log 
  1.84
 0.100 L  0.2500 L 
pH  14  pOH  14  1.84  12.16
64.
This is a weak base/strong acid titration
a)
Major Species: H2NNH2
No reaction goes to completion.
Not a buffer
H2NNH2(aq) + H2O(l) ⇌ H2NNH3+(aq) + OH-(aq)
H2NNH2
H2NNH3+
OH0.100 M
0
0
Initial
-x
+x
+x
Change
0.100-x
X
X
Equilibrium
Kb 
xx
 3.0 106
(0.100  x)
Since Ka is small assume 0.100-x = 0.100
xx
 3.0 106
(0.100)
x  5.5 104
19
Check assumption
5.5 104
100%  0.55% Good
0.100
Calculate pH
OH    x  5.5 104
pOH   log OH     log  5.5 104   3.26
pH  14  pOH  14  3.26  10.74
b)
Major Species: H2NNH2, H+, and NO3A reaction goes to completion
H2NNH2(aq) + H+(aq)  H2NNH3+(aq)
H2NNH2
H+
0.0100 mol 0.00400 mol
Initial (mol)
0.0060 mol
0
Final (mol)
Major Species: H2NNH2, NO3-, and H2NNH3+
Buffer
Determine the Ka
H2NNH3+
0
0.00400 mol
1.0 1014
 3.3 109
3.0 106
 A 
pH  pK a  log  HA    log  3.3 109   log


Ka 
c)
Kw
Kb

Major Species: H2NNH2, H+, and NO3A reaction goes to completion
H2NNH2(aq) + H+(aq)  H2NNH3+(aq)
H2NNH2
H+
0.0100 mol 0.00500 mol
Initial (mol)
0.0050 mol
0
Final (mol)
+
Major Species: H2NNH2, NO3 , and H2NNH3
Buffer

H2NNH3+
0
0.00500 mol
 A 
pH  pK a  log  HA    log  3.3 109   log


d)
Major Species: H2NNH2, H+, and NO3A reaction goes to completion
H2NNH2(aq) + H+(aq)  H2NNH3+(aq)
H2NNH2
H+
0.0100 mol 0.00800 mol
Initial (mol)
0.0020 mol
0
Final (mol)
+
Major Species: H2NNH2, NO3 , and H2NNH3
Buffer

0.0500 mol
V
0.0500 mol
V
  8.48
H2NNH3+
0
0.00800 mol
 A 
pH  pK a  log  HA    log  3.3 109   log


e)
  8.66
0.0060 mol
V
0.00400 mol
V

0.0020 mol
V
0.00800 mol
V
  7.88
Major Species: H2NNH2, H+, and NO3A reaction goes to completion
H2NNH2(aq) + H+(aq)  H2NNH3+(aq)
20
H2NNH2
H+
0.0100 mol 0.0100 mol
Initial (mol)
0
0
Final (mol)
Major Species: H2NNH3+, and NO3Not a buffer
H2NNH3+(aq) ⇌ H2NNH2(aq) + H+(aq)
H2NNH3+
H2NNH2 H+
0.0100 mol
0
0
Initial (mol)
0.0667
0
0
Initial (M)
-x
+x
+x
Change
0.0667-x
x
x
Equilibrium
Ka 
H2NNH3+
0
0.0100 mol
xx
 3.3  109
0.0667  x 
Since Ka is small assume 0.0667-x = 0.0667
xx
 3.3  10 9
(0.0667)
x  1.5  10 5
Check assumption
1.5  105
100%  0.022% Good
0.0667
Calculate pH
H   x  1.5 10
pH   logH    log 1.5  10   4.82

5

f)
5
Major Species: H2NNH2, H+, and NO3A reaction goes to completion
H2NNH2(aq) + H+(aq)  H2NNH3+(aq)
H2NNH2
H+
H2NNH3+
0
0.0100 mol 0.0200 mol
Initial (mol)
0
0.0100 mol 0.0100 mol
Final (mol)
Major Species: H2NNH3+, H+, and NO3Not a buffer
The H+ concentration will be mainly from the excess strong acid.
 0.0100mol 
pH   log  H     log 
  1.301
 0.100 L  0.100 L 
95.
a)
CaC2O4(s) ⇌ Ca2+(aq) + C2O42-(aq)


K sp  Ca 2  C2O4
Initial
Change
Equilibrium
2

Ca2+
0
+x
x
C2O420
+x
x
* x is the solubility
21
The concentrations of Ca2+ and C2O42- must be in M.
g
M CaC2O4  128.10 mol

6.1103 g CaC2O4
1L CaC2O4

1mol CaC2O4
128.10 g CaC2O4
  4.8 10
5
M
K sp  Ca 2  C2O4 2   xx  x 2   4.8 105   2.3 109
2
b)
BiI3(s) ⇌ Bi3+(aq) + 3I-(aq)
  
K sp  Bi 3 I 
3
Bi3+
0
+x
x
Initial
Change
Equilibrium
I0
+3x
3x
* x is the solubility
  


K sp  Bi 3 I   x3x   27 x 4  27 1.32  105  8.20  1019
3
100.
a)
3
4
Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)
Ag+
SO420
0
Initial
+2x
+x
Change
2x
x
Equilibrium

K sp  Ag 
 SO   1.2  10
2
2
4
5
2 x 2 x  1.2  10 5
x  0.014M
b)
Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)
Ag+
SO420.10 M
0
Initial
+2x
+x
Change
0.10+2x
x
Equilibrium
Assume that x is small due to the small equilibrium constant
2
K sp   Ag    SO42   1.2 105
 0.10  2 x 
2
x  1.2 105
Assume that 0.10 +2x =0.10
 0.10 
2
x  1.2 105
x  0.0012M
Check assumption
2  0.0012 
0.10
100%  2.4% Good
22
c)
Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)
Ag+
SO420
0.20 M
Initial
+2x
+x
Change
2x
0.20+x
Equilibrium
Assume that x is small due to the small equilibrium constant
2
K sp   Ag    SO42   1.2 105
 2 x   0.20  x   1.2 105
2
Assume that 0.20 +x =0.20
 2x 
2
0.20  1.2 105
x  0.0039M
Check assumption
0.0039
100%  1.95% Good
0.20
101.
a)
Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq)


K sp  Fe 3 OH 

3
This problem is slightly different because there is some OH- in water. From the
Kw we know that the initial [OH-]=1.0×10-7.
Fe3+
OH0
1×10-7
Initial
+x
+3x
Change
x
1×10-7+3x
Equilibrium
* x is the solubility
Since Ksp is small assume that 1×10-7+3x = 1×10-7



3


3
K sp  Fe 3 OH   x 1  107  4  1038
x  4  1017 M
Check assumption


3 4  1017
100%  1  10 7% Good
7
1  10
b)
Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq)


K sp  Fe 3 OH 

3
Calculate the concentration of OH- ions when the pH is 5.0
 
pH   log H 
H   10  10  1 10
H OH   1 10
OH   1H10   1110
 1  10
10

 pH



 5.0
5
14
14

14
9
5
23
Fe3+
OH0
1×10-9
Initial
+x
+3x
Change
-9*
x
1×10
Equilibrium
* x is the solubility
* The question told you to assume that the pH is constant



3


3
K sp  Fe 3 OH   x 1  109  4  1038
x  4  1011 M
c)
Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq)


K sp  Fe 3 OH 

3
Calculate the concentration of OH- ions when the pH is 11.0
 
pH   log H 
H   10  10  1 10
H OH   1 10
OH   1H10   1110
 0.001
10

 pH


11.0
11
14
14

14

11
Fe3+
OHInitial
0
0.001
Gain/Lose
+x
+3x
Equilibrium
x
0.001*
* x is the solubility
* The question told you to assume that the pH is constant
Since Ksp is small assume that 0.001 + 3x = 0.001



K sp  Fe 3 OH   x0.001  4  1038
3
3
x  4  10 29 M
118.
Mn2+(aq) + C2O42-(aq) ⇌ MnC2O4(aq)
MnC2O4(aq) + C2O42-(aq) ⇌ Mn(C2O4)22-(aq)
Mn2+(aq) +2C2O42-(aq) ⇌ Mn(C2O4)22-(aq)



K1
K2
K  K1K2
K  K1K2  7.9  10 7.9  10  6.2 10
3
1
5
24