IONIC EQUILIBRIUM INVOLVING WEAK ELECTROLYTES by hcj

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```									             IONIC EQUILIBRIUM INVOLVING WEAK ELECTROLYTES
SECTION I

KCl                K+ + Cl-

HCl + H2O                     H3O+ + Cl-

HCl                H+ + Cl-

1.      H2A + H2O                         H3O+ + HA-

2.      HA- + H2O                         H3O + A=

1a.            H2A                        H+ + HA-

2a.            HA-                        H+ + A=

3.             H2A                        2H+ + A=

3a.     H2O + HOAc                                H3O+ + OAc-

4.      K=
H O OAc 
3
                

H 2 O HOAc

5.      K[H2O] =
H O OAc 
3
              

 HOAc
K[H2O] = Ka

6.      Ka =
H O OAc 
3
                

 HOAc

7.      Ka =
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

1.34
Concentration of [H+] which is equal to the concentration of [OAc-] =                        .1 = 1.34  10-3 m.l.
100
1.34
The concentration of HOAc = .1 -           .1 = .099
100

Ka =
134  10 134  10  = 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

NH4OH                 NH4+ + OH-

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.

Kb =
NH OH 
4
        

 NH 4 OH 
x. x
1.8  10-5 =
.5  x

x is very small compared to .5. It can be deleted from the denominator:

x2
1.8  10-5 =
.5
.9  10-5 = x2

9  10-6= x2

9  10 6  x

3  10-3 = x = NH4+ = OH-

concentration of NH 4 + or concentration of OH -
% ionization =                                                           100
concentration of original base

3  10 3
% ionization =              100 = .6%
.5
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.   (a)      [H+] = .05 [Cl-] = .05

(b)      [Na+] = 1.2 [OH-] = 1.2

(c)      [Na+] = .1 M [OAc-] = .1

2.   [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)

PbCl2           Pb+2 + 2Cl-

HCl            H+ + Cl-

If HCl is added, the solubility of PbCl2 will be reduced.

HOAc            H+ + OAc-

Example 1

Calculate the [H+] in a .1 M HOAc which contains .1 M NaOAc (Ka HOAc 1.8  10-5).

HOAc            H+ + OAc-

Solution

HOAc is a weak electrolyte. The Ka expression is given by Equation 7 of Section 1.

 H   OAc  
            
Ki =
 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.

NH3 + H2O               NH4 + + OH-

Kb =
NH  OH 
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           H3O+ + OH-

Reaction 4:           H2O            H+ + OH-

Eq. 8      Ke =
H 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

1  10 14
OH- =
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.

Ka =
H OAc 
           

 HOAc
H+ = x OAc- = x

x. x
1.8  10-5 =
.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                            NH4OH + H+

 NH 4 OH  H  
NH 4   HOH
Eq. 14 Ka =

Ka[HOH] = Kh

 NH 4 OH  H  
NH 
Eq. 15 Kh =

4

Kw
Eq. 16 Kh =
Kb

Eq. 9   [H+][OH-] = Kw

From the hydrolysis of the NH4+
NH4+ + HOH                                   NH4OH + H+

 NH 4 OH  H  
NH 
Eq. 17 Therefore, Kh =

4

Substitute the value of [H+] from Eq. 9 into Eq. 17 and obtain Eq. 18.

 NH 4 OH  Kw
NH  OH 
Eq. 18 Kh =
                 
4

 NH 4 OH                 
1
NH  OH 
but
                         k1
4

Kw
i.e.        = Kh
Kb
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         x. x
5

18  10
.                  .1

10  10 16
 x2
18  10 5
.

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       x. x

4  10 10        .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       2.   [H+] = 1.05  10-5         3.   [H+] = 6  10-10
[H+] = 4.46  10-12            pOH = 9.02                      pH = 9.22
pH = 11.35

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