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Chemistry LibreTexts

Solutions

1. Calculate the equilibrium concentrations and pH for a 0.20 M proprionic acid solution.
 
 
 

Complete the equilibrium and express the equilibrium concentrations in terms of a single unknown 'x'.

 
HC3H5O2 (aq)
H+(aq)
+ C3H5O2- (aq)
initial concentration
0.2 M
 
0
0
change
- x
 
+ x
+ x
equilibrium concentration
0.2 - x
 
x
x

 

Write the equilibrium expression (Ka) and express the Ka in terms of the equilibrium concentrations.
 

  [H+][C3H5O2-]  
(x)(x)
 
Ka
_______________
 =
___________
= 1.3 x 10-5
     [HC3H5O2]  
(0.2 - x)
 (from table)

 

Solve for x; since the initial concentration is greater than 0.1 and the Ka less than 10-4, we can drop the '- x' term.
 

x2
 
solving for 'x' gives
 
_______
= 1.3 x 10-5
 
x = 1.61 x 10-3
0.2
     

 

Solve for the equilibrium concentrations by substituting the value of 'x' into the equilibrium concentrations: [H+] = [C3H5O2-] = x, [HC3H5O2] = 0.2 - x

[H+] = [C3H5O2-] = 1.61 x 10-3 M, [HC3H5O2] = 1.98 x 10-1 M;

pH = - log [H+] = - log(1.61 x 10-3)

pH = 2.79

2. Calculate the equilibrium concentrations and pH for a 0.20 M carbonic acid solution.
 Compare the answer to problem 1. Note which has the higher Ka and lower pH.
This is done the same way as problem 1, using a Ka value of 4.2 x 10-7
 

 
H2CO3 (aq)
H+(aq)
+  HCO3- (aq)
initial concentration
0.2 M
 
0
0
change
- x
 
+ x
+ x
equilibrium concentration
0.2 - x
 
x
x


 

  [H+][HCO3-]  
(x)(x)
 
Ka
_______________
 =
___________
= 4.2 x 10-7
     [H2CO3]  
(0.2 - x)
 (from table)


 

x2
 
solving for 'x' gives
 
_______
= 4.2 x 10-7
 
x = 2.90 x 10-4
0.2
     

 

[H+] = [HCO3-] = 2.90 x 10-4 M, [H2CO3] = 2.0 x 10-1 M; pH = 3.54

The higher the Ka, the lower the pH for the same concentration of weak acid.  The higher Ka value tells us we have more dissociation of the weak acid, giving a greater [H+].

3. Calculate the Ka for a 0.3 M solution of HA (weak acid) if the pH = 3.65
First calculate the [H+] concentration from the pH
[H+] = 10-pH; [H+] = 10-3.65; [H+] = 2.24 x 10-4

since 1:1 mole ratio

[H+] = [A-]; [HA] = 0.3 - [H+]

Substitute into the Ka expression

  [H+][A-]  
(2.24 x 10-4)(2.24 x 10-4)
Ka
________
 =
________________________
     [HA]  
(0.3 - 2.24 x 10-4)

Ka= 1.67 x 10-7

 

4. Calculate the pH for a 0.2 M pyridine solution.
 

 
C5H5N (aq)
OH-(aq)
+  C5H5NH+ (aq)
initial concentration
0.2 M
 
0
0
change
- x
 
+ x
+ x
equilibrium concentration
0.2 - x
 
x
x


 

  [OH-][C5H5NH+]  
(x)(x)
 
Kb
________________
 =
___________
= 2.0 x 10-9
 

      [C5H5N]

 
(0.2 - x)
 (from table)

 

Solve for 'x'

x2
 
solving for 'x' gives
 
_______
= 2.0 x 10-9
 
x = 2.0 x 10-5
0.2
     

 

[OH-] = [C5H5NH+] = 2.0 x 10-5 M, [C5H5N] = 2.0 x 10-1 M

Using the following equations:  pOH = - log [OH-];   pH + pOH = 14

pOH = 4.70, pH = 9.30

5. Calculate the pH of a 0.34 M HCl  solution.

pH = 0.47,  strong acid, 100 % ionization; 1:1 mole ratio, [HCl] = [H+]