Example 1

In a solution of acetic acid, the equilibrium concentrations are found to be [CH3COOH] = 0.100; [CH3COO-] = 0.0042. Evaluate the pH of this solution and the equilibrium constant of ionization of acetic acid.

Solution
From the ionization of acetic acid,

CH3COOH = CH3COO- + H+
0.100             0.0042     0.0042

we conclude that
[H+] = [CH3COO-]
      = 0.0042.
Thus, pH = -log0.0042 = 2.376.

The equilibrium constant of ionzation,

        (0.0042)2
K = ------------- = 1.78x10-4
          0.100
 

Discussion
The equilibrium constant of an acid is represented by Ka; and similar to the pH scale, a pKa scale is defined by

 

pKa = - log Ka

and for acetic acid, pKa = 4.75. Note that Ka = 10-pKa

The pH and pKa of Weak acid

There are many weak acids, which do not completely dissociate in aqueous solution. As a general discussion of weak acids, let HA represent a typical weak acid. Then its ionization can be written as:

HA = H+ + A-

In a solution whose label concentration is C (= [HA] + [A-]), let us assume that x is the concentration that has undergone ionization. Thus, at equilibrium, the concentrations are

[HA] = C - x
[H+] = [A-] = x

Make sure you understand why they are so, because you will have to setup these relationship in your problem solving. In summary, we formulate them as

HA = H+ + A-
C                   - initial concentration, assume x M ionized
C-x    x       x - equilibrium concentration

            x2
Ka = -------
          C - x
 

 

Ka = 10-pKa

pKa = - log Ka
 

 

The pKa values of many weak acids are listed in table form in handbooks, and some of these values are given in the Handbook of CAcT, pKa of Acids.

Example 2

The pKa of acetic acid is 4.75. Find the pH of acetic acid solutions of labeled concentrations of 1.0 M, 0.010 M, and 0.00010 M.

Solution
Assume the label concentration as C and x mole ionized, then the ionization and the equilibrium concentrations can be represented by the formulation below.

CH3COOH = CH3COO- + H+
C - x                     x             x

            x2
Ka = -------
          C - x
The equation is then
x2 + Ka x = C Ka = 0
 

The solution of x is then 

        - Ka + (Ka2 + 4 C Ka)(1/2)
  x  =  ---------------------------
                    2
Recall that Ka - 1.78e-5, the values of x for various
C are given below:
C = 1.0 0.010 0.00010 M
x = 0.0042 0.00041 0.0000342 M
pH = 2.38 3.39 4.47

Discussion
In the above calculations, the following cases may be considered: 

  1. If x is small (< 1% of) compared to C, then C-x is
approximately C. Thus,

    x  = (Ka * C)(1/2)

    Note that you are comparing x with C here. If C >
100*Ka, the above method gives satisfactory results.

  2. If x is not small by comparison with C, or C is not large
in comparison to Ka, then the equation takes this form:

    x2 + Ka x - C Ka = 0,

    and the solution for x, which must not be negative, has been
given above.

  3. Both cases 1 and 2 neglect the contribution of [H+] from the ionization
of water. However, if the pH calculated from cases 1 and 2 falls in the
range between 6 and 7, the concentration from self-ionization of water
cannot be neglected.

    When the contribution of pH due to self-ionization of water cannot be neglected, there are two equilibria to be considered. 

        HA  =  H+  +  A-
   C-x     x     x

    H2O = H+ + OH- 55.6 y y <- - - ([H2O] = 55.6)

    Thus, [H+] = (x+y), [A-] = x, [OH-] = y,

    and the two equilibria are

    (x+y) x Ka = --------- ........ (1) C - x

    and Kw = (x+y) y, ........... (2) (Kw = 1E-14) 

     
    There are two unknown quantities, x and y in two equations, and (1) may
be rearranged to give

     

    x2  + (y + Ka) x - C Ka = 0
     
             -(y+Ka) + ((y+Ka)2 + 4 C Ka)(1/2)
   x  =  -------------------------------------
                        2

     

    4. One of the many methods to find a suitable solution for this problem is to use iterations, or successive approximations.

     

    1. Assume that  y = 1E-7

       

    2. Calculate an x value using the quadratic form

       

                    -(y+Ka) + ((y+Ka)2 + 4 CKa)(1/2)
     x  =  -------------------------------------
                           2 

    3. Calculate a new y value (yn) from the x just obtained using

       

      yn = 1E-14/(x+y)

       

    4. Replace y in step (2) by yn, and recalculate x.

       

    5. Repeat steps (2) and (3) until the new values and the
    old values differ insignificantly.

The above procedure is actually a general method that always gives a
satisfactory solution. This technique have to be used to calculate the pH
of dilute weak acid solutions. Further discussion is given in the
Exact Calculation of pH.

Calculate pOH of Basic Solutions

The discussion on weak acids provide a paradigm for the discussion of weak bases. For weak base B, the ionization is

B- + H2O = HB + OH-
and
        [HB] [OH-]
Kb = -------------
            [B-]

The pOH can be calculated for a basic solution if Kb is given. In this case, the discussion is similar and parallel to that given above for the calculation of pH of weak acids when Ka is know.

Confidence Building Questions