Homework #6
Chapter 7 Homework
Acids and Bases
18.
a)
2H2O(l) ⇌ H3O+(aq) + OH-(aq)
K  H 3O  OH 
Or
H2O(l) ⇌ H+(aq) + OH-(aq)



 
K  H  OH 
b)

HCN(aq) + H2O(l) ⇌ H3O (aq) + CN-(aq)
+
H O CN 
K


3
HCN 
Or
HCN(aq) ⇌ H+(aq) + CN-(aq)
 H   CN  
K
 HCN 
c)
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
NH OH 
K


4
NH 3 
For this question you could have picked any weak acid or base or just used a general
weak acid (HA) and weak base (B)
22.
a)
b)
c)
d)
HClO4
HOCl
H2SO4
H2SO3
strong acid
weak acid
strong acid (1st deprotonation only)
weak acid
25.
The larger the Ka the stronger the acid (table 7.2).
HClO4 > HClO2 > NH4+ > H2O
Strongest acid  Weakest acid
Note: HClO4 is a strong acid, therefore, will have the highest Ka value.
H2O is the conjugate acid of OH-. OH- is a strong base. Strong bases have
the weakest conjugate acids.
(H2O(l) + OH-(aq)  OH-(aq) + H2O(l))
26.
The larger the Ka the stronger the acid (table 7.2), the stronger the acid the weaker the
conjugate base, therefore, the strongest base will have the conjugate acid with the
smallest Ka.
NH3 > ClO2- > H2O > ClO4Note: HClO4 is a strong acid, therefore, will have the highest Ka value.
Acids are classified as strong acids if they have a Ka greater than 1.
H3O+ is one of the weakest strong acids. Therefore, H3O+ (H2O
1
conjugate base) is one of the weakest strong acid and H2O will be
slightly more basic than ClO4-.
27.
The larger the Ka the stronger the acid.
a)
HCl
strong acid and H2O is a weak acid Kw = Ka (1.0×10-14)
b)
HNO2
The Ka of HNO2 (4.0×10-4) > Kw = Ka (1.0×10-14)
c)
HCN
The Ka HCN (6.2×10-10) > Ka of HOC6H5 (1.6×10-10)
28.
The higher the Ka the weaker the conjugate base
a)
Clconjugate base of a strong acid, HCl
H2O
conjugate base of one of the weakest strong acids H3O+
H2O stronger base
b)
H2O
conjugate base of one of the weakest strong acids H3O+
NO2conjugate base of a weak acid
NO2- stronger base
c)
CNconjugate base of the weak acid HCN (Ka = 6.2×10-10)
OC6H5conjugate base of the weak acid HOC6H5 (Ka = 1.6×10-10)
OC6H5 stronger base
29.
a)
b)
c)
CH3CO2H
acid
H2O
base
CH3CO2
conjugate base
H3O+
conjugate acid
Therefore, H2O and CH3CO2- are competing for protons.
CH3CO2- is the stronger base (H3O+ is a strong acid, therefore, forms a weaker
conjugate acid than CH3CO2H)
CH3CO2- is a weak base because it is the conjugate base of a weak acid.
Conjugate bases of weak acid are also weak, therefore, the do no fully dissociate
in solution.
CH3CO2- + H2O(aq)  CH3CO2H(aq) + OH-(aq)
31.
a)
d)
g)
weak acid
strong base
weak acid
b)
e)
h)
strong acid
weak base
strong base
36.
The solution is acidic if [H+]>[OH-] , the solution is basic if [H+]<[OH-], and the solution is
neutral if [H+]=[OH-] .
The relationship between [H+] and [OH-] is:
K w   H   OH    1.0 1014
a)
1.0 10  OH
b)
[H+] = [OH-] neutral solution
8.3 1016 OH    1.0 1014
c)
f)
i)
weak base
weak acid
strong acid (1st deprotonation)
  1.0 1014
OH    1.0 107 M
7



OH   12M
[H+] < [OH-] basic solution

2
c)
12 OH    1.0 1014
OH    8.3 1016 M
[H+] > [OH-] acidic solution
5.4 105  OH    1.0 1014
d)
37.
OH    1.9 1010 M
[H+] > [OH-] acidic solution
If the pH < 7 the solution is acidic, if the pH > 7 the solution is basic, and if the pH = 7 the
solution is neutral.
If the pOH < 7 the solution is basic, if the pOH > 7 the solution is acidic, and if the
pOH = 7 the solution is neutral.
The relationship between pH and [H+] is:
pH   log  H  
The relationship between [H+] and [OH-] is:
a)
Basic
 H    10 pH  107.4  4.0 108 M
 4.0 10  OH
8
b)

  1.0 1014
OH    2.5 107 M
Basic
 H    10 pH  1015.3  5 1016 M
5 10  OH
16
c)
K w   H   OH    1.0 1014

  1.0 1014
OH    20M
Acidic
 H    10 pH  101.0  10M
10  OH    1.0 1014
d)
OH    11015 M
Acidic
 H    10 pH  103.20  6.3 104 M
 2.2 10  OH
4
e)

  1.0 1014
OH    1.6 1011 M
Basic
OH    10 pOH  105.0  1105 M
 H   1106   1.0 1014
f)
 H    1109 M
Acidic
3
OH    10 pOH  109.60  2.5 1010 M
 H   1.5 1010   1.0 1014
 H    4.0 105 M
43.
a)
HNO2 is a weak acid
The major species in solution are HNO2 and H2O
The following equilibrium will occur in solution
HNO2(aq) ⇌ NO2-(aq) + H+(aq)
Ka = 4.0×10-4 (Table 7.2)
+
Calculate the concentration of H ions at equilibrium
HNO2
NO2H+
0.250 M
0
0
Initial
-x
+x
+x
Change
0.250-x
X
X
Equilibrium
 NO2    H  
Ka 
 4.0 10
 HNO2 
4

xx
 0.250  x 
1.0 10 4 4.0 10 4 x  x 2
x 2  4.0 10 4 x  1.0 10
4
0
4
b  b 2  4ac 4.0 10 
x

2a
 4.0 10 
4 2
 4.0 10 4
2
 0.0098 and  0.010
x is 0.0098 M because concentration cannot be negative.
 H    x  0.0098M
Calculate pH
 
pH   log H    log0.0098  2.01
b)
HC2H3O2 is a weak acid
The major species in solution HC2H3O2 and H2O
The following equilibrium will occur in solution
HC2H3O2(aq) ⇌ C2H3O2-(aq) + H+(aq)
Ka = 1.8×10-5 (Table7.2)
Calculate the concentration of H+ ions at equilibrium
HC2H3O2
C2H3O2H+
0.250 M
0
0
Initial
-x
+x
+x
Change
0.250-x
X
X
Equilibrium
C2 H 3O2    H  
Ka 
 1.8 10
 HC2 H3O2 
5

xx
 0.250  x 
Since Ka is small assume 0.250-x = 0.250
1.8  10 5 
xx
0.250
x  0.0021
Check assumption (5% rule)
4
Amount Gained/Lost
0.0021
100% 
100%  0.84%
Original Concentration
0.25
Less than 5%, therefore, good assumption
 H    x  0.0021M
Calculate pH
 
pH   log H    log0.0021  2.68
44.
a)
HOC6H5 is a weak acid therefore
The major species in solution are HOC6H5
The following equilibrium will occur in solution
HOC6H5(aq) ⇌ OC6H5-(aq) + H+(aq)
𝐾𝑎 = 1.6 × 10−10
+
Calculate the concentrations of H at equilirium
HOC6H5
OC6H5H+
0.250 M
0
0
Initial
-x
+x
+x
Change
0.250 – x
x
x
Equilibrium
OC6 H 5   H  
Ka 
 1.6 10
 HOC6 H5 
10

xx
 0.250  x 
Since Ka is small assume 0.250-x = 0.250
xx
1.6  10 10 
 0.250 
x  6.3  106
Check assumption (5% rule)
Amount Gained/Lost
6.3 106
100% 
100%  0.0025%
Original Concentration
0.250
Less than 5%, therefore, good assumption.
 H    x  6.3 106 M
Calculate the concentration of OH- from water equilibrium
  
1.5  10 OH   1.0  10
OH   6.7  10 M
K w  H  OH   1.0  10 14
5


14
10
Calculate pH
pH   log  H     log  6.3 106   5.20
b)
HCN is a weak acid
The major species in solution are HCN and H2O
The following equilibrium will occur in solution
HCN(aq) ⇌ CN-(aq) + H+(aq)
Ka = 6.2×10-10
(Table 7.2)
5
Calculate the concentration of H+ ion at equilibrium
HCN
CNH+
0.250 M
0
0
Initial
-x
+x
+x
Change
0.250-x
x
x
Equilibrium
CN    H  
Ka 
 6.2 10
 HCN 
10

xx
 0.25  x 
Since Ka is small assume 0.250-x = 0.250
6.2  10 10 
xx
0.250
x  1.2  10 5
Check assumption (5% rule)
Amount Gained/Lost
1.2 105
100% 
100%  0.0048%
Original Concentration
0.25
Less than 5%, therefore, good assumption.
 H    x  1.2 105 M
Calculate pH
 


pH   log H    log 1.2  10 5  4.92
45.
HF is a weak acid therefore, the reactions that are occurring in solution are:
HF(aq) ⇌ F-(aq) + H+(aq)
Calcualte the concentrations of HF, F-, and H+ at equilibrium (you will also need to
calculate the consentration of OH- which will be in the soltion from the following
equilibrium H2O(l) ⇌ OH-(aq) + H+(aq))
HF
FH+
0.020 M
0
0
Initial
-x
+x
+x
Change
0.020-x
x
x
Equilibrium
 F    H  
Ka 
 7.2 10
 HF 
4

xx
 0.020  x 
1.44 10 5 7.2 10 4 x  x 2
x 2  7.2 10 4 x  1.44 10
5
0
4
b  b 2  4ac 7.2 10 
x

2a
 7.2 10 
4 2
 5.76 10 5
2
 0.0035 and  0.0042
x is 0.030 M because concentration cannot be negative.
Find the final concentrations
 HF   0.020  x  0.020  0.0035  0.017M
 F     H    x  0.0035M
Calculate the concentration of OH- from water equilibrium
6
  
0.0035OH   1.0  10
OH   2.9  10 M
K w  H  OH   1.0  10 14


14
12
Calculate pH
 
pH   log H    log0.0035  2.46
47.
a)
b)
HA is a weak acid. There are more HA molecules (green and white atoms
together) than H+ ions (single white atom) and A- ions (single green atoms).
This shows that the acid did not completely dissociate making it a weak acid.
Calculate the percentage dissociated
In the picture there are 9 particles that have note dissociated and 1
particle that has
𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑
1
100% =
100% = 10%
𝑡𝑜𝑡𝑎𝑙 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠
1+9
Calculate the equilibrium constant
The reaction of interest is HA(aq) ⇌ H+(aq) + A-(aq)
HA
H+
A0.20 M
0
0
Initial
-x
+x
+x
Change
1
1
9
Equilibrium 0.20− 𝑥=
𝑥=
𝑥=
𝑁𝑎 𝑉
𝑁𝑎 𝑉
𝑁𝑎 𝑉
9
Note from the picture at equilibrium the concentration of HA is 𝑁𝑎 =
𝑉
9
𝑁𝑎 𝑉
Na is
Avogadro’s number and is used to convert the 9 particles to number of moles of
1
HA. The concentration of H+ and A- are 𝑁 𝑉
𝑎
Solve for V
9
0.20 − 𝑥 =
𝑁𝐴 𝑉
1
9
0.20 −
=
𝑁𝑎 𝑉 𝑁𝑎 𝑉
10
50
𝑉=
=
𝑁𝑎 0.20 𝑁𝑎
Solve for K
𝐾=
48.
[𝐻 + ][𝐴− ]
[𝐻𝐴]
=
1
1
50
50
𝑁𝑎 (𝑁 )𝑁𝑎 (𝑁 )
𝑎
𝑎
9
50
𝑁𝑎 (
)
𝑁𝑎
1
= (50)(9) = 0.0022
HC2H2ClO2 is a weak acid, therefore, the equilibrium established is:
HC2H2ClO2(aq) ⇌C2H2ClO2-(aq) + H+(aq)
Find the concentration of H+ at equilirbium
HC2H2ClO2 C2H2ClO2H+
0.10 M
0
0
Initial
-x
x
x
Change
0.10-x
x
x
Equilibrium
7
C2 H 2ClO2    H  
Ka 
 1.35 10
 HC2 H 2ClO2 
3

xx
 0.10  x 
1.4 10 4 1.35 10 3 x  x 2
x 2  1.35 10 3 x  1.4 10
4
0
3
3
4
b  b 2  4ac 1.35 10  1.35 10   5.6 10
x

 0.011 and  0.013
2a
2
2
x is 0.011 M because concentration cannot be negative.
 H    x  0.011M
Calculate pH
 
pH   log H    log0.011  1.96
49.
HIO3 is a weak acid, therefore, the equilibrium established is:
HIO3(aq) ⇌IO3-(aq) + H+(aq)
Find the concentration of H+ at equilibrium
HIO3
IO3H+
0.010 M
0
0
Initial
-x
X
x
Change
0.010-x
X
x
Equilibrium
 IO3   H  
xx
Ka 
 0.17 
 0.010  x 
 HIO3 
0.0017  0.17 x  x 2
x 2  0.17 x  0.0017  0
b  b 2  4ac 0.17 
x

2a
 0.17 
2
 0.0068
2
 0.0094 and  0.18
x is 0.0094 M because concentration cannot be negative.
 H    x  0.0094M
Calculate pH
 
pH   log H    log0.0094  2.03
51.
g
M C6 H5CO2 H  122.13 mol
0.56 g C6 H 5CO2 H
M

1mol C6 H 5CO2 H
122.13 g C6 H 5CO2 H
  0.0046mol C H CO H
6
5
2
n 0.0046mol

 0.0046M
V
1.0 L
C6H5CO2H is a weak acid, therefore, the equilibrium established is:
H2O(l) ⇌ OH-(aq) + H+(aq)
C6H5CO2H(aq) ⇌C6H5CO2-(aq) + H+(aq)
8
Calcualte the equilirium concnetrations
C6H5CO2H
C6H5CO20.0046 M
0
Initial
-x
+x
Change
0.0046-x
X
Equilibrium
C6 H5CO 2    H  
Ka 
 6.4 10
C6 H5CO2 H 
5

H+
0
+x
X
xx
 0.0046  x 
2.9 10 7 6.4 10 5 x  x 2
x 2  6.4 10 5 x  2.9 10
7
0
5
b  b 2  4ac 6.4 10 
x

2a
 6.4 10 
5 2
 1.2 10 6
2
 5.2 104 and  5.8 104
x is 5.2×10-4 M because concentration cannot be negative.
Find the final concentrations
C6 H5CO2 H   0.0046  x  0.0046  5.2 104  0.0041M
C6 H5CO2    H   x  5.2 104 M
Calculate the concentration of OH- from water equilibrium
K w  H  OH   1.0  10 14
  
5.2  10 OH   1.0  10
OH   1.9  10 M
4


14
11
Calculate pH
 


pH   log H    log 5.2  10 4  3.28
52.
C6H5CO2H is a solid before it is put into water. Therefore, two equilibriums are
established.
C6H5CO2H(s) ⇌ C6H5CO2H(aq)
C6H5CO2H(aq) ⇌ C6H5CO2-(aq) + H+(aq)
They want to know how much C6H5CO2H will dissolve. This will be the amount of
C6H5CO2H that starts in the second equilibrium which we will denote as s.
C6H5CO2H
C6H5CO2H+
S
0
0
Initial
-x
+x
+x
Change
s-x
x
x
Equilibrium
The pH of the solution will give us the equilibrium concentration of the H+ ions
 
pH   log H 
H   10

 pH
Initial
Change
Equilibrium
 10 2.80  0.0016M
C6H5CO2H
s
-0.0016
s-0.0016
C6H5CO20
+0.0016
0.0016
H+
0
+0.0016
0.0016
9
C6 H 5CO2    H  
 0.0016  0.0016
Ka 
 6.4 105 
S  0.0016
C6 H 5CO2 H 
S  0.042M
Convert to grams per 100 ml
0.042𝑚𝑜𝑙
1𝐿
122.13𝑔
𝑔
𝑆=(
)(
)(
) = 0.0051
1𝐿
1000𝑚𝐿
1𝑚𝑜𝑙
𝑚𝐿
Therefore the water solubility per 100 ml is 0.51 g per 100 ml
55.
a)
The first solution contains 0.10 M HCl (strong acid) and 0.10 M HOCl (weak
acid). Since there is both a strong and weak acid with equal concentrations of
each the number of H+ ions from the weak acid will be minimal therefore the pH
will be completely from the strong acid (𝑝𝐻 = − log[𝐻 + ] = − log 0.10 = 1.00
This can be proven using an ICE table
HOCl(aq) ⇌ H+(aq) + OCl-(aq) 𝐾𝑎 = 3.5 × 20−8
HOCl
OClH+
0.10 M
0
0.10
Initial
-x
+x
+x
Change
0.10-x
x
0.10+x
Equilibrium
OCl    H  
x  0.10  x 
Ka 
 3.5 10 8 
 0.10  x 
 HOCl 
Since Ka is small assume 0.250-x = 0.250
x  0.10 
3.5  10 8 
 0.10 
x  3.5  108
Check assumption (5% rule)
Amount Gained/Lost
3.5 108
100% 
100%  3.5 105%
Original Concentration
0.10
Less than 5%, therefore, good assumption.
 H    0.10  x  0.10  3.5 108  0.10M
Calculate pH
pH   log  H     log  0.10   1.00
b)
The second solution contains 0.050 M HNO3 (strong acid) and 0.50 M HC2H3O2
(weak acid). Since there is both a strong and weak acid and even though the
concentration of the weak acid is 10x less than the concentration of the strong
acid the number of H+ ions from the weak acid will be minimal therefore the pH
will be completely from the strong acid (𝑝𝐻 = − log[𝐻 + ] = − log 0.050 =
1.30
This can be proven using an ICE table
HC2H3O2(aq) ⇌ H+(aq) + C2H3O2-(aq)
𝐾𝑎 = 1.8 × 20−5
HC2H3O2
C2H3O2H+
0.50 M
0
0.050
Initial
-x
+x
+x
Change
0.50-x
X
0.050+x
Equilibrium
10
C2 H 3O2    H  
x  0.050  x 
Ka  
 1.8 10 5 
 0.50  x 
 HC2 H3O2 
Since Ka is small assume 0.50-x = 0.50 and 0.05+x=0.05
x  0.05 
1.8  10 5 
 0.50 
x  1.8  104
Check assumption (5% rule) Note:0.05<0.50 therefore is the assumption
works for 0.05-x then it will work form 0.50-x
Amount Gained/Lost
1.8 104
100% 
100%  0.36%
Original Concentration
0.05
Less than 5%, therefore, good assumption.
 H    0.050  x  0.050  1.8 104  0.050
Calculate pH
pH   log  H     log  0.50   1.30
59.
HOCN is a weak acid, therefore, the following equilibrium is established:
HOCN(aq) ⇌ OCN-(aq) + H+(aq)
HOCN
OCNH+
1.00×10-2 M
0
0
Initial
-x
+x
+x
Change
-2
1.00×10 -x
x
x
Equilibrium
The pH of the solution at equilibrium will allow us to calculate the concentration of H+
ions at equilibrium =x
 
pH   log H 
H   10

 pH
 10 2.77  0.0017
Updating the ICE table gives
HOCN
1.00×10-2 M
-0.0017
0.0083
OCN0
+0.0017
0.0017
H+
0
+0.0017
0.0017
Initial
Change
Equilibrium
0.00170.0017  3.5  10 4
OCN  H 
Ka 

0.0083
HOCN 

63.
 
The larger the Kb the stronger the base. Therefore the strongest base will have the
largest Kb.
NH3 > C5H5N > H2O > NO3Strongest base  Weakest base
Note: HNO3 is a strong acid. The stronger the acid the weaker the
conjugate base. Therefore, NO3- is a very weak base.
Note: H2O is the conjugate base of (H3O+). H3O+ is one of the weakest strong
acids. Therefore, H2O is only slightly more basic than NO3-
11
64.
The smaller the Kb the stronger the conjugate acid.
HNO3 > C5H5NH+ > NH4+ > H2O
Strongest acid  Weakest acid
Note: HNO3 is a strong acid.
Note: C5H5N and NH3 are both weak bases, therefore, will have weak conjugate
acids. The Kb of NH3 > kb of C5H5N resulting in C5H5NH+ being a stronger
acid than NH4+.
Note: H2O is the conjugate acid of OH-. OH- is a strong base. Strong bases have
the weakest conjugate acids.
(H2O(l) + OH-(aq)  OH-(aq) + H2O(l))
68.
a)
Ca(OH)2 is a strong base (fully dissociates)
OH    0.00040M Ca(OH ) 2

2 mol Ca ( OH )2
1mol Ca ( OH )2
  0.00080M OH

pOH   log OH     log  0.00080   3.10
pH  pOH  14
pH  14  pOH  14  3.10  10.90
b)
KOH is a strong base (fully dissociates)
g
M KOH  56.11 mol
25 g KOH

1mol KOH
56.11 g KOH

1mol OH 
1mol KOH
  0.45mol OH

n 0.45mol
OH    
 0.45M
V
1L
pOH   log OH     log  0.45   0.35
pH  pOH  14
pH  14  pOH  14  0.35  13.65
c)
NaOH is a strong base (fully dissociates)
g
M Na 5OH  40.00 mol
150.0 g NaOH

1mol NaOH
40.00 g NaOH

1mol OH 
1mol NaOH
  3.750mol OH

n 3.750mol
OH    
 3.750M
V
1L
pOH   log OH     log  3.750   0.5740
pH  pOH  14
pH  14  pOH  14  0.5740  14.5740
12
69.
Ba(OH)2 is a strong base (fully dissociates)
Calculate the concentration of OH-
14  pH  pOH
pOH  14  pH  14  10.50  3.50

pOH   log OH 
OH   10

 pOH

 10 3.50  3.2  10  4
Calculate the concentration of Ba(OH)2
3.2 104 M OH 

1mol Ba OH 2
2 mol OH 
  1.6 10
4
M Ba  OH 2
73.
The presence of nitrogen most often results in a basic molecule. Nitrogen has an
unpaired electron, therefore, bonds easily to H+ ions.
93.
The larger the pH the more basic a solution is
a)
HI strong acid
HF weak acid
NaF (weak base)
NaF(s)  Na+(aq) + F-(aq)
Na+ + H2O
NaOH + H+ (NaOH is a strong base)
Na+ is neutral
F- + H2O
HF + OHF- is basic
NaI (neutral)
NaI(s)  Na+(aq) + I-(aq)
Na+ + H2O
NaOH + H+ (NaOH is a strong base)
Na+ is neutral
I- + H2O
HI + OH- (HI is a strong acid)
I- is neutral
HI < HF < NaI < NaF
 Increasing pH
b)
NH4Br (weak acid)
NH4Br(s)  NH4+(aq) + Br-(aq)
NH4+ + H2O
NH3 + H3O+
NH4+ is acidic
Br- + H2O
HBr + OH- (HBr is a strong acid)
Br- is neutral
HBr strong acid
KBr (neutral)
KBr(s)  K+(aq) + Br-(aq)
K+ + H2O
KOH + H+ (KOH is a strong base)
K+ is neutral
Br- + H2O
HBr + OH- (HBr is a strong acid)
Br- is neutral
NH3 weak base
13
14
c)
HBr < NH4Br < KBr < NH3
 Increasing pH
C6N5NH3NO3 (weak acid)
C6N5NH3NO3(s)  C6N5NH3+(aq) + NO3-(aq)
C6N5NH3+ + H2O
C6N5NH2 + H3O+
C6N5NH3+ is acidic
NO3- + H2O
HNO3 + OH- (HNO3 is a strong acid)
NO3- is neutral
NaNO3 (neutral)
NaNO3(s)  Na+(aq) + NO3-(aq)
Na+ + H2O
NaOH + H+ (NaOH is a strong base)
Na+ is neutral
NO3- + H2O
HNO3 + OH- (HNO3 is a strong acid)
NO3- is neutral
NaOH strong base
HOC6H5 weak acid (Ka = 1.6×10-10)
KOC6H5 (weak base)
KOC6H5(s)  K+(aq) + OC6H5-(aq)
K+ + H2O
KOH + H+ (KOH is a strong base)
K+ is neutral
OC6H5- + H2O
HOC6H5 + OHOC6H5- is basic
C6H5NH2 weak base
HNO3 Strong acid
There are 2 weak acid C6N5NH3NO3 (C6H5NH3+) and HOC6H5
Compare the Ka values to determine the weaker acid
HOC6H5
Ka = 1.6×10-10
+
The Ka value of C6H5NH3 is not given therefore, it must be calculated
C6H5NH2(aq)
Kb = 3.8×10-10
Ka 
K w 1.0 1014

 2.6 105
Kb 3.8 1010
Therefore, C6N5NH2NO3 is a stronger acid than HOC6H5
There are 2 weak bases C6N5NH2 and KOC6H5 (OC6H5-)
Compare the Kb values to determine the weaker base
C6N5NH2
Kb = 3.8×10-10
The Kb value of OC6H5- is not given therefore, it must be calculated
HOC6H5(aq)
Ka = 1.6×10-10
Kb 
K w 1.0 1014

 6.2 105
10
Kb 1.6 10
Therefore, KOC6H5 is a stronger base than C6N5NH2
HNO3< C6N5NH3NO3 < HOC6H5 < NaNO3< C6H5NH2 < KOC6H5 < NaOH
15
95.
See chart for problem 85.
a)
Sr(NO3)2 (salt formed form a strong acid and strong base)
Sr(NO3)2
Sr(NO3)2(s)  Sr2+(aq) + 2NO3-(aq)
Sr2+ + 2H2O
Sr(OH)2 + 2H+ (Sr(OH)2 is a strong base)
Sr2+ is neutral
NO3- + H2O
HNO3 + OH- (HNO3 is a strong acid)
NO3- is neutral
Therefore, Sr(NO3)2 is neutral
b)
C2H5NH3CN
C2H5NH3CN(s)  C2H5NH3+(aq) + CN-(aq)
C2H5NH3+ + H2O
C2H5NH2 + H3O+
C2H5NH3+ is acidic
CN- + H2O
HCN + OHCN- is basic
Therefore, need to compare Ka and Kb values of reactions
We are not given the Ka of C2H5NH3+, therefore, we need to calculate it
C2H5NH2(aq)
Kb = 5.6×10-4
Ka 
K w 1.0 1014

 1.8 1011
4
Kb 5.6 10
We are not given the Kb of CN-, therefore, we need to calculate it
HCN(aq)
Ka = 6.2×10-10
Kb 
c)
K w 1.0 1014

 1.6 105
10
K a 6.2 10
Since Ka < Kb C2H5NH3CN is basic
C5H5NHF
C5H5NHF(s)  C5H5NH+(aq) + F-(aq)
C5H5NH+ + H2O
C5H5N + H3O+
C5H5NH+ is acidic
F- + H2O
HF + OHF- is basic
Therefore, need to compare Ka and Kb values of reactions
We are not given the Ka of C5H5NH+, therefore, we need to calculate it
C2H5N(aq)
Kb = 1.7×10-9
Ka 
K w 1.0 1014

 5.9 106
9
Kb 1.7 10
We are not given the Kb of F-, therefore, we need to calculate it
HF(aq)
Ka = 7.2×10-4
Kb 
K w 1.0 1014

 1.4 1011
4
K a 7.2 10
Since Ka > Kb C5H5NHF is acidic
16
d)
NH4C2H3O2
NH4C2H3O2(s)  NH4+(aq) + C2H3O2-(aq)
NH4+ + H2O
NH3 + H3O+
NH4+ is acidic
C2H3O2- + H2O
HC2H3O2 + OHC2H3O2- is basic
Therefore, need to compare Ka and Kb values of reactions
NH4+ Ka = 5.6×10-10
We are not given the Kb of C2H3O2- therefore we need to calculate it
HC2H3O2(aq)
Ka = 1.8×10-5
Kb 
e)
K w 1.0 1014

 5.6 1010
K a 1.8 105
Ka = Kb NH4C2H3O2 is neutral
NaHCO3
NaHCO3(s)  Na+(aq) + HCO3-(aq)
Na+ + H2O
Na(OH) + H+ (NaOH is a strong base)
Na+ is neutral
HCO3- + H2O
H2CO3 + OHHCO3- + H2O
CO32- + H3O+
Since H2CO3 is a polyprotic acid must look at Ka and Kb of the two possible
reactions.
HCO3Ka = 4.8×10-11
We are not given the Kb of HCO3 therefore we need to calculate it
H2CO3(aq)
Ka = 4.3×10-7
K w 1.0 1014
Kb 

 2.3 108
7
K a 4.3 10
Since Ka < Kb HCO3- is basic
Therefore, NaHCO3 is basic
97.
a)
KCl KNO2
KCl(s)  K+(aq) + Cl-(aq)
K+ + H2O
KOH + H+ (KOH is a strong base)
K+ is neutral
Cl- + H2O
HCl + OH- (HCl is a strong acid)
Cl- is neutral
Therefore, KCl is neutral
Therefore, the solution will be neutral pH=7.00. For neutral solutions the
concentration of H+ and OH- equals 1.0×10-7 M.
17
b)
KF
KF(s)  K+(aq) + F-(aq)
K+ + H2O
KOH + H+ (KOH is a strong base)
K+ is neutral
F- + H2O
HF + OHF- is basic
Therefore, KF is basic
Interested in the following reaction
F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq)
[𝐻𝐹][𝑂𝐻 − ]
𝐾𝑏 =
[𝐹 − ]
Problem we do not know Kb
HF(aq)
Ka=7.2×10-4
𝐾𝑏 =
𝐾𝑤
𝐾𝑎
=
1.0×10−14
7.2×10−4
= 1.4 × 10−11
Calculate the concentration of OH- ions
F1.0 M
Initial
-x
Change
1.0-x
Equilibrium
OH    HF 
xx
Kb 

 1.4 1011

1.0  x
 F 
OH0
+x
x
HF
0
+x
x
Since Kb is small assume that 1.0-x = 1.0
xx
1.0
6
x  3.7 10 M
1.4 1011 
Check assumption (5% rule)
Amount Gained/Lost
3.7 106
100% 
100%  3.7 104%
Original Concentration
1.0
Less than 5%, therefore, good assumption.
Calculate OH- concentration
OH    x  3.7 106 M
Calculate H+ concentration
[𝐻 + ][𝑂𝐻− ] = 1.0 × 10−14
1.0 × 10−14
[𝐻 + ] =
= 2.7 × 10−9
3.7 × 10−6
Calculate pH
pH   log  H     log 2.7 109 M  8.57


18
101.
a)
KNO2
KNO2(s)  K+(aq) + NO2-(aq)
K+ + H2O
KOH + H+ (KOH is a strong base)
K+ is neutral
NO2- + H2O
HNO2 + OHNO2- is basic
Therefore, KNO2 is basic
Interested in the following reaction
NO2-(aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq)
Kb

HNO2 OH  

NO 

2
Problem we don’t know Kb
Know
HNO2(aq)
Kb 
Ka = 4.0×10-4
14
K w 1.0 10

 2.5 1011
4
K a 4.0 10
Calculate the concentration of OH- ions
NO20.12 M
Initial
-x
Change
0.12-x
Equilibrium
Kb 
 HNO2  OH  
 NO 

2

HNO2
0
+x
X
OH0
+x
x
xx
 2.5 1011
0.12  x
Since Kb is small assume that 0.12-x = 0.12
xx
0.12
6
x  1.7 10 M
2.5 1011 
Check assumption (5% rule)
Amount Gained/Lost
1.7 106
100% 
100%  0.0014%
Original Concentration
0.12
Less than 5%, therefore, good assumption.
Calculate OH+ concentration
OH    x  1.7 106 M
Calculate pH
pOH   log OH     log 1.7 106   5.77
pH  pOH  14
pH  14  pOH  14  5.77  8.23
19
b)
NaOCl
NaOCl(s)  Na+(aq) + OCl-(aq)
Na+ + H2O
NaOH + H+ (NaOH is a strong base)
Na+ is neutral
OCl- + H2O
HOCl + OHOCl- is basic
Therefore, NaOCl is basic
Interested in the following reaction
OCl-(aq) + H2O(l) ⇌ HOCl(aq) + OH-(aq)
Kb 
 HOCl  OH  
OCl  
Problem we don’t know Kb
HOCl(aq)
Kb 
1.0 1014
 2.9 107
8
3.5 10
Calculate the concentration of OH- ions
OCl0.45 M
Initial
-x
Change
0.45-x
Equilibrium
Kb 
Ka = 3.5×10-8
 HOCl  OH  
OCl 


HOCl
0
+x
X
OH0
+x
x
xx
 2.9 107
0.45  x
Since Kb is small assume that 0.45-x = 0.45
xx
0.45
4
x  3.6 10 M
2.9 107 
Check assumption (5% rule)
Amount Gained/Lost
3.6 104
100% 
100%  0.08%
Original Concentration
0.45
Less than 5%, therefore, good assumption.
Calculate OH+ concentration
OH    x  3.6 104 M
Calculate pH
pOH   log OH     log  3.6 104   3.44
pH  pOH  14
pH  14  pOH  14  3.44  10.56
20
c)
NH4OCl4 (a salt from a weak base and a strong acid)
NH4OCl4(s)  NH4+(aq) + OCl4-(aq)
NH4+ + H2O
NH3 + H3O+
NH4+ is acidic
OCl4- + H2O
HOCl4 + OH- (HOCl4 strong acid)
OCl4- is neutral
Therefore, NH4OCl4 is acidic
Interested in the following reaction
NH4+(aq) ⇌ NH3(aq) + H+(aq)

K w  NH 3   H 
Ka 

 5.6 1010

Kb
 NH 4 
Calculate the concentration of H+ ions
NH4+
0.40 M
Initial
-x
Change
0.40-x
Equilibrium
Ka 
 NH3   H  
 NH 

4

NH3
0
+x
x
H+
0
+x
X
xx
 5.6 1010
0.40  x
Since Kb is small assume that 0.40-x = 0.40
xx
0.40
5
x  1.5 10 M
5.6 1010 
Check assumption (5% rule)
Amount Gained/Lost
1.5 105
100% 
100%  0.0038%
Original Concentration
0.40
Less than 5%, therefore, good assumption.
Calculate H+ concentration
 H    x  1.4 105 M
Calculate pH
pH   log  H     log 1.5 105   4.82
121.
Because the solution has a pH of 4.6 we know that the solution is acidic, therefore,
NaOH, NH3, and NaCN cannot be the solute.
Because the bulb is bright it means that there are a lot of ions in solution. Therefore, a
strong electrolyte is need. Only HCl, NaOH, and NH4Cl are strong electrolytes.
NaOH was already eliminated because it was basic.
HCl could light the bulb bright because it is a strong acid but the pH of a 1.0 M solution
of HCl is
 
pH   log H    log1.0  0
21
Confirm the solution is made from NH4Cl by compare actual to calculated Ka values
Determine the Ka of the weak acid.
HA(aq) ⇌ A-(aq) + H+(aq)
Ka = unknown
 A   H  
Ka 
 HA
Initial
Change
Equilibrium
HA
1.0 M
-x
1.0-x
A0
+x
x
H+
0
+x
x
Determine x
𝑥 = [𝐻 + ]
𝑝𝐻 = − log[𝐻 + ]
4.6 = − log[𝐻 + ]
[𝐻 + ] = 2.5 × 10−10
 A   H    2.5 105  2.5 105 
Ka 

 6.3 1010
5
1.0  2.5 10
 HA
This Ka of NH4+ is 5.6×10-10 which is close to the calculated value. Therefore the solution
must contain NH4Cl
22