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Exact pH calculation

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Exact pH calculation

The paradigm given in Weak Acids and Bases deals with the equilibria of solutions containing one weak acid or one weak base. In most cases, the amount of H+ from the autoionization of water is negligible. For very dilute solutions, the amount of H+ ions from the autoionization of water must also be taken into account. Thus, a strategy is given here to deal with these systems.

When two or more acids are present in a solution, the concentration of H+ (or pH) of the solution depends on the concentrations of the acids and their acidic constants Ka. The hydrogen ion is produced by the ionization of all acids, but the ionizations of the acids are governed by their equilibrium constants, Ka's.

Similarly, the concentration of OH- ions in a solution containing two or more weak bases depends on the concentrations and Kb values of the bases.

For simplicity, we consider two acids in this module, but the strategies used to discuss equilibria of two acids apply equally well to that of two bases.

Two Equilibria with a Common Ion H+

If the pH is between 6 and 8, the contribution due to autoionization of water to [H+] should also be considered. When autoionization of water is considered, the method is called the exact pH calculation or the exact treatment. This method is illustrated in this module. The need for the eaxct treatment has been pointed out in the discussion of Example 2 in Weak Acids and Bases.

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

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

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


 [H+] = (x+y),
[A-] = x,
[OH-] = y,
and the two equilibrium constants are (x + y) x Ka = --------- - - (1) C - x and Kw = (x + y) y, - - - (2)
Although you may use the method of successive approximation, but the formula to calculate the pH can be derived.

Solving for x from (2) gives

  x = ---- - y

and substituting this expression into in (1) results in

        (x+y) (Kw/y - y)
  Ka = -------------------
         C - Kw/y + y

Rearrange this equation to give:

  [H+] = (x+y)

        C - Kw/y + y
     = ------------- Ka
          Kw/y - y

Note that Kw/y = [H+], and y = Kw/[H+].
Thus, we get:

           C - [H+] + Kw/[H+]
 [H+] =  ----------------------  Ka
             [H+] - Kw/[H+]

To simplify the formula, let

   h = [H+]

Then, the formula becomes 

        C - h + Kw/h
  h = --------------- Ka
         h - Kw/h

Now, consider the approximations we can make when we use this formula due to conditions.

Case 1

If h > 1E-6, then Kw/h < 1E-8, small indeed compared to h and C.


        C - h
 h = --------------- Ka

h2 + Ka h - C Ka = 0 and -Ka + (Ka2 + 4 C Ka)(1/2) h = ----------------------- 2

Case 2

If h << C, then C - h => C
 h = --------------- Ka

 h = (C Ka)1/2

The treatment presented as Case 3 at the beginning of this tutorial is more general, and may be applied to problems involving two or more weak acids in one solution.

Example 1

Calculate the [H+], [Ac-], and [Cc-] when the solution contains 0.200 M HAc (Ka = 1.8E-5), and 0.100 M HCc (the acidity constant Kc = 1.4E-3). (HAc is acetic acid whereas HCc is Chloroacetic acid).

Assume x and y to be the concentrations of Ac- and Cc-, respectively, and write the concentrations below the equations:

     HAc   =  H+ + Ac- 
   0.200-x    x     x    

     HCc   =  H+ + Cc-
   0.100-y    y     y   

   [H+]  =  (x + y)

Thus, you have

     (x + y) x
    -----------  = 1.8E-5  - - - (1)
     0.200 - x

     (x + y) y
    -----------  = 1.4E-3  - - - (2)
     0.100 - y
Solving for x and y from (1) and (2) may seem difficult, but you can often make some assumptions to simplify the solution procedure. Since HAc is a weaker acid than is HCc, you expect x << y. Further, y << 0.100. Therefore, x + y => y and 0.100 - y => 0.100. Equation 2 becomes:
     (    y) y
    -----------  = 1.4E-3  - - - (2')

which leads to

     y  = ((1.4E-3)(0.100))1/2
        =  0.012

Substituting y in (1) results in

     (x + 0.012) x
    ---------------  = 1.8E-5  - - - (1')
       0.200 - x
This equation is easily solved, but you may further assume that 0.200 - x => 0.200, since x << 0.200. Thus,
          -0.012 + (1.44E-4 + 1.44E-5)1/2
   x  =  -----------------------------

      = 2.9E-4  -  -  -  - Small indeed compared to 0.200
You had a value of 0.012 for y by neglecting the value of x in (2). You can now recalculate the value for y by substituting values for x and y in (2).
     (2.9E-4 + y) y
    ----------------  = 1.4E-3  - - - (2")
     0.100 - 0.012

Solving for y in the above equation gives

    y  =  0.011.
You have improved the y value from 0.012 to 0.011. Substituting the new value for y in a successive approximation to recalculate the value for x improves its value from 2.9E-4 to a new value of 3.2E-4. Use your calculator to obtain these values.

Further refinement does not lead to any significant changes for x or y. Thus, your mission is accomplished.

You should write down these calculation on your note pad, since reading alone does not lead to thorough understanding.

Example 2

A weak acid HA has a Ka value of 4.0E-11. What are the pH and the equilibrium concentration of A- in a solution of 0.0010 M HA?

For the solution of this problem, two methods are given here. If you like the x and y representation, you may use method (a).

Method (a) The two equilibrium equations are:

      HA      =   H+  +  A-;
   0.0010-x       x      x   

      H2O     =   H+  +  OH- 
                  y      y    

      [H+]  =  (x+y)

      (x+y) x
      --------  =  4.0E-11  - - (3); 

         (x+y) y = 1e-14  - - - (4)

Assume y << x, and x << 0.0010, then you have

      (x  ) x
      --------  =  4.0E-11  - - - (3')

      x = ((0.0010)(4.0e-11))1/2
        = 2.0E-7
Substituting 2.0E-7 for x in 4 and solve the quadratic equation for y gives,
      (2.0E-7+y) y = 1E--14
      y  =  4.1E-8
Substituting 4.1E-8 in (3), but still approximate 0.0010-x by 0.0010
      (x+4.1E-8) x
      --------------  =  4.0E-11 - - - (3'')
Solving this quadratic equation for a positive root results in
      x = 1.8E-7 M   <----- Recall x = [A-]

     [H+] = x + y
            = (1.8 + 0.41)1E-7
            = 2.2E-7
     pH  =  6.65
The next method uses the formula derived earlier.

Method (b)Using the formula from the exact treatment, and using 2E-7 for all the [H+] values on the right hand side, you obtain a new value of [H+] on the left hand side,

          C - [H+] + Kw/[H+]
 [H+] = ----------------------  Ka
            [H+] - Kw/[H+]

        = 2.24E-7

     pH = 6.65.
The new [H+] enables you to recalculate [A-] from the formula:
     (2.24E-7) [A-] = C Ka

     [A-] = (0.0010) (4.0E-11) / (2.24E-7)
            = 1.8E-7


Often, you may be attempted to use the approximation method:

  x = (C Ka)1/2
    = 2.0E-7 M A-, or H+;  pH = 6.70
and obtained a pH of 6.70.

The value 6.70 is greater than 6.65 by less than 1 %. However, when an approximation is made, you have no confidence in the calculated pH of 6.70.

Calculating pH by Approximations

Water is both an acid and a base due to the autoionization, H2O = H+ + OH-

However, the amount of H+ ions from water may be very small compared to the amount from an acid if the concentration of the acid is high.

When calculating [H+] in an acidic solution, approximation method or using the quadratic formula has been discussed in the modules on weak acids.

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