IONIC EQUILIBRIUM INVOLVING WEAK ELECTROLYTES
SECTION I
K+ + Cl-
KCl
HCl + H2O
H3O+ + Cl-
H+ + Cl-
HCl
1.
H2A + H2O
H3O+ + HA-
2.
HA- + H2O
H3O + A=
1a.
H2A
H+ + HA-
2a.
HA-
H+ + A=
3.
H2A
2H+ + A=
H3O+ + OAc-
3a.
H2O + HOAc
4.
H O OAc 
K=
5.
H O OAc 
K[H O] =


3
 H 2 O HOAc


3
 HOAc
2
K[H2O] = Ka
H O OAc 

6.
Ka =
7.
Ka =

3
 HOAc
H OAc 


 HOAc
Example 1
Calculate the Ka for HOAc if .1 M solution of HOAc is 1.34% ionizable.
Solution
HOAc
H+ + OAc-
Since HOAc is 1.34% ionizable
Concentration of [H+] which is equal to the concentration of [OAc-] =
134
.
 .1 = 1.34  10-3 m.l.
100
The concentration of HOAc = .1 -
134
.
 .1 = .099
100
.  10 134
.  10 
134
Ka =
= 1.8  10
3
3
-5
.099
Example 2
Kb for NH4OH is 1.8  10-5. Calculate the concentration of NH4+ and the OH- and the % ionization if
.5 molar NH4OH solution.
Solution
NH4+ + OH-
NH4OH
The concentration of NH4+ = the concentration of OH-. Let “x” be the concentration of NH4+. Then
“x” is the concentration of OH- and the concentration of NH4OH is .5 - x.
NH OH 
Kb =


4
 NH 4 OH
1.8  10-5 =
x. x
.5  x
x is very small compared to .5. It can be deleted from the denominator:
1.8  10-5 =
x2
.5
.9  10-5 = x2
9  10-6= x2
9  10 6  x
3  10-3 = x = NH4+ = OH% ionization =
concentration of NH 4 + or concentration of OH concentration of original base
% ionization =
3  10 3
 100 = .6%
.5
 100
SECTION II
1.
Calculate the concentration of each ion in the following solutions.
(a)
.05 M HCl
(b)
1.2 M NaOH
(c)
.1 M NaOAc
2.
Calculate the H+ and the % ionization in a .1 M HCN (Ka HCN 4  10-10).
3.
.1 M NH4OH solution is 1.34% ionizable. Calculate the Kb for NH4OH.
4.
Calculate the H+ and the % ionization in a 1 M HOAc (Ka 1.8  10-5).
ANSWERS FOR SECTION II:
1.
2.
(a)
[H+] = .05 [Cl-] = .05
(b)
[Na+] = 1.2 [OH-] = 1.2
(c)
[Na+] = .1 M [OAc-] = .1
[H+] = 6.3  10-6
% ionization = 6.3  10-3%
3.
1.85  10-5
4.
[H+] = 4.24  10-3
% ionization = .424%
SECTION III
(Common Ion)
Pb+2 + 2Cl-
PbCl2
H+ + Cl-
HCl
If HCl is added, the solubility of PbCl2 will be reduced.
H+ + OAc-
HOAc
Example 1
Calculate the [H+] in a .1 M HOAc which contains .1 M NaOAc (Ka HOAc 1.8  10-5).
H+ + OAc-
HOAc
Solution
HOAc is a weak electrolyte. The Ka expression is given by Equation 7 of Section 1.
Ki =
 H   OAc  
 HOAc 
Assume the concentration of H+ = x. The concentration of OAc- from OHAc = x. NaOAc is a
strong electrolyte; thus the concentration of the OAc- = .1, and the total concentration of OAc- is .1
+ x.
1.8  10-5 =
 x .1  x
.1
The value of x compared to .1 is very small. It can be neglected. Hence:
1.8  10-5 =
 x .1
.1
H+ = x = 1.8  10-5
Reaction 1:
HOAc
H+ + OAc-
Reaction 2:
NaOAc
Na+ + OAc-
H+ + OH
H2O
Example 2
Calculate the [OH-] in a .2 M NH3 solution which also contains .5 M NH4Cl. Kb NH3 is
1.8  10-5.
Solution
NH4OH is a weak electrolyte.
NH4 + + OH-
NH3 + H2O
NH  OH 
Kb =


4
NH 3 
NH4Cl
NH4+ + Cl-
Assume the concentration of OH- = x.
Concentration of NH4+ from NH4OH is also x. But in addition to NH4+ from NH4OH, we have
NH4 from NH4Cl which is .5 M. The total NH4+ = .5 + x.
+
Substitute the values obtained from the ionization expression for NH4OH
1.8  10-5 =
. 5  x  x
. 2
x is very small compared to .5; thus it can be neglected.
1.8  10-5 =
. 5  x
.2
.36  10-5 = .5x
.72  10-5 = x
7.2  10-6 = x
x = 7.2  10-6
Section IV
1.
Calculate the concentration of [H+] in a .1 M HCN solution which also contains .2 M NaCN
(Ka HCN 4  10-10).
2.
Calculate the [H+] in a .2 M HOAc solution which also contains .2 M NaOAc
(Ka HOAc 1.8  10-5).
3.
How many m/l of hydrogen ions are contained in a .1 M HOAc solution which contains .2 Mole of
NaOAc? (Ka HOAc 1.8  10-5 ).
ANSWERS TO SECTION IV:
1.
[H+] = 2  10-10 mole/liter
2.
[H+] = 1.8  10-5 mole/liter
3.
[H+] = 9  10-6 mole/liter
Section V
(Ionization of Water)
Reaction 3:
H2O + H2O
Reaction 4:
H2O
Ke =
H+ + OH-
 H OH 

Eq. 8
H3O+ + OH-

 H 2 O
Ke[H2O] = [H+][OH-]
Eq. 9
Kw = [H+][OH-]
Kw is the ion product constant for water. Kw is a constant at 25C and has a value of 1  10-14.
Eq. 10 [H+][OH-] = 1 10-14
It follows that in an aqueous solution at 25C.
Eq. 11 pH + pOH = 14
In pure water the [H+] = [OH-] = 1 10-7.
For example, calculate the [OH-] in a .01 molar HCl.
Solution
[H+][OH-] = 1 10-14
[1  10-2][OH-] = 1  10-14
OH- =
1  10 14
1  10 2
= 1  10-12
Section VI
(pH)
Eq. 12 pH = -log[H+]
Eq. 13 pOH = -log[OH-]
Example 1
Calculate the pH of a solution in which [H+] is 1  10-3 m/l.
Solution
Substitute the value given in Equation 12.
pH = -log[H+]
= -log[1  10-3]
= -[log 1 + log 10-3]
= -[-3 log 10]
= -[-3] = 3
Example 2
Calculate the pH of .0055 molar HCl.
Solution
HCl is completely ionzied; thus the H+ is .0055 or 5.5  10-3. Substitute the value in Equation 12.
pH = -log[5.5  10-3
= [log 5.5 + log 10-3]
= -[.74 - 3]
= -[-2.26]
= 2.26
If the pOH is required, remember that:
pH + pOH = 14
pOH = 14 - 2.26
pOH = 11.74
Example 3
Calculate the pH of .1 M HOAc. Ka 1.8 10-5.
Solution
HOAc is a weak electrolyte. Ki expression is given by Equation 7.
H OAc 
Ka =


 HOAc
H+ = x OAc- = x
1.8  10-5 =
x. x
.1
1.8  10-6 = x2
1.34  10-3 = x
therefore,
H+ = 1.34  10-3
therefore,
pH = -log[1.34  10-3]
= -[.11 - 3]
= -[-2.89]
= 2.89
IMPORTANT POINTS TO REMEMBER
1.
The pH of a neutral solution is 7; at this point the [H+] = [OH-] = 1  10-7 m/l.
2.
If the pH of solution is less than 7, then the solution is acidic and the [H +] is larger than 1  10-7.
3.
If the pH of a solution is more than 7, then the solution is basic and the [H +] is less than 1  10-7.
Section VIII
1.
Calculate the pH of .1 M HCl.
2.
Calculate the pH of .1 M HOAc (Ka HOAc 1.8  10-5)
3.
Calculate the pH, pOH of .0005 M HCl.
4.
Calculate the pH of .05 M HCN (Ka 4  10-10)
5.
Calculate the pOH of .5 M HOA (Ka HOAc 1.8  10-5)
ANSWERS TO SECTION VIII:
1.
pH = 1
2.
pH = 2.89
3.
pH = 3.3
pH = 10.7
4.
pH = 5.35
5.
pOH = 11.48
Section IX
(Hydrolysis)
Reaction 5:
NH4Cl + H2O
NH4OH + HCl
Reaction 6:
NaOAc + H2O
NaOH + HOAc
Reaction 7:
NH4+ + HOH
Eq. 14 Ka =
NH4OH + H+
 NH 4 OH  H  
NH 4   HOH
Ka[HOH] = Kh
Eq. 15 Kh =
 NH 4 OH  H  
NH 

4
Kw
Kb
Eq. 16 Kh =
Eq. 9
[H+][OH-] = Kw
From the hydrolysis of the NH4+
NH4+ + HOH
Eq. 17 Therefore, Kh =
NH4OH + H+
 NH 4 OH  H  
NH 

4
Substitute the value of [H+] from Eq. 9 into Eq. 17 and obtain Eq. 18.
Eq. 18 Kh =
 NH 4 OH  Kw
NH  OH 


4
but
 NH 4 OH
NH  OH 

4
i.e.
Kw
= Kh
Kb


1
k1
Example 1
Calculate the pH of a solution which is .1 M NaOAc. (Ka HOAc 1.8  10-5)
Solution
OAc- + HOH
HOAc + OH-
The hydrolysis of OAc- indicates an increase in the OH-; hence, the solution should be basic.
Assume the concentration of HOAc = x. Then the concentration of OH- = x.

Kw  HOAc OH
Kh =

Ka
OAc 

1  10 14
18
.  10
5
10  10 16
18
.  10 5




x. x
.1
 x2
5.5  10-11 = x2
55  10-12 = x
7.5  10-6 = x
pOH = -log[OH-]
= -log[7.5  10-6]
= [.88 - 6]
= 5.12
therefore, pH = 14 - 5.12 = 8.88
Example 2
Calculate the pH of a .1 M NaCN solution. Ka HCN 4  10-10.
Solution
CN- + HOH
HCN + OH-
From the reaction between CN- and water, it can be seen that there is an increase in the OH-, and
therefore the solution will be basic. Let x be the concentration of HCN; then the concentration of OH - is x.
That is, [HCN] = [OH-] = x.

Kw  HCN OH
Eq. 19 Kh =

Ka
CN 




10  10 15
4  10
10

x. x
.1
2.5  10-6 = x2
1.58  10-3 = x
pOH = -log[1.58  10-3]
= [.2 - 3]
= 2.8
pH = 14 - 2.8 = 11.2
Section X
1.
Calculate the [OH-], [H+], and the pH in .2 M (NaCN). (Ka HCN = 4  10-10)
2.
Calculate the [H+] and the pOH in .2 M NH4Cl. (Kb NH4OH = 1.8  10-5)
3.
Calculate the [H+] and the pH of .5 M NaOAc. (Ka HOAc 1.8  10-5)
ANSWERS FOR SECTION X:
1.
[OH-] = 2.24  10-3
[H+] = 4.46  10-12
pH = 11.35
2.
[H+] = 1.05  10-5
pOH = 9.02
3.
[H+] = 6  10-10
pH = 9.22