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How to Calculate the Fraction of EDTA

Ethylenediaminetetraacetic acid (EDTA) is a weak acid although it can also act as a weak base. It can pick up either one or two protons as a base or donate up to four protons as an acid. Consequently, it has no fewer than six different forms you could find in solution, depending on the pH. The most important of these -- the one that is actually useful -- is the completely deprotonated form, EDTA-4. You can use the dissociation constants and the pH to calculate the fraction of EDTA-4 in solution.

Instructions

- 1
  Write down the equation you'll use to solve this problem:
  
  ( K1 K2 K3 K4 K5 K6) ) / ( [H+]^6 + K1 [H+]^5 + K1 K2 [H+]^4 + K1 K2 K3 [H+]^3 + K1 K2 K3 K4 [H+]^2 + K1 K2 K3 K4 K5 [H+] + K1 K2 K3 K4 K5 K6 )
  
  This equation may look complicated, but it will look simpler once you break it down into smaller pieces.
- 2
  Start by calculating the numerator. The numerator is just the product of the six acid dissociation constants for EDTA. These acid dissociation constants are as follows:
  
  K1 = 1
  
  K2 = 0.0316
  
  K3 = 0.01
  
  K4 = 2.04 x 10^-3
  
  K5 = 7.41 x 10^-7
  
  K6 = 4.27 x 10^-11
  
  If you multiply all six of these numbers together, you get 2.04 x 10^-23.
- 3
  Convert the pH to [H+], the hydrogen ion concentration. Remember that [H+] is just equal to 10^-pH. If the pH is 8, for example, the hydrogen ion concentration is 10^-8 = 1 x 10^-8.
- 4
  Calculate the first four terms of the denominator, which are as follows:
  
  [H+]^6 + K1 [H+]^5 + K1 K2 [H+]^4 + K1 K2 K3 [H+]^3
  
  In the example, [H+] = 1 x 10^-8, so once you substitute this number in for [H+] and raise it to each power, you have the following:
  
  1 x 10^-48 + K1 (1 x 10^-40) + K1 K2 (1 x 10^-32) + K1 K2 K3 (1 x 10^-24)
  
  Now multiply the last three terms in this expression by the appropriate K values. This gives you the following:
  
  1 x 10^-48 + (1) (1 x 10^-40) + (1) (0.0316) (1 x 10^-32) + (1) (0.0316) (0.01) (1 x 10^-24) =
  
  1 x 10^-48 + (1 x 10^-40) + (3.16 x 10^-34) + (3.16 x 10^-28) = 3.16 x 10^-28
- 5
  Calculate the last three terms of the denominator:
  
  K1 K2 K3 K4 [H+]^2 + K1 K2 K3 K4 K5 [H+] + K1 K2 K3 K4 K5 K6
  
  K1 = 1
  
  K2 = 0.0316
  
  K3 = 0.01
  
  K4 = 2.04 x 10^-3
  
  K5 = 7.41 x 10^-7
  
  K6 = 4.27 x 10^-11
  
  Start by substituting in the [H+] value you calculated and your K values to give the following:
  
  (1) (0.0316) (0.01) (2.04 x 10^-3) (1x10^-8)^2 + (1) (0.0316) (0.01) (2.04 x 10^-3) (7.41 x 10^-7) (1x10^-8) + (1) (0.0316) (0.01) (2.04 x 10^-3) (7.41 x 10^-7) (4.27 x 10^-11)
  
  =
  
  6.45 x 10^-15 + 4.78 x 10^-21 + 2.04 x 10^-23
  
  =
  
  6.45 x 10^-15
- 6
  Add your result from Step 5 and your result from Step 4 together. In the example, this will give you the following:
  
  6.45 x 10^-15 + 3.16 x 10^-28 = 6.45 x 10^-15
  
  In this case, the second result is so much larger than the first that adding the first to it doesn't really change it at all.
- 7
  Divide the numerator (your result from Step 2) by the denominator (your result from Step 6) to obtain the following:
  
  2.04 x 10^-23 / 6.45 x 10^-15 = 3.16 x 10^-9
  
  This is the fraction of unbound EDTA that is completely deprotonated. As you can see, at pH 8 it's very small, and lowering the pH would make it smaller still. At higher pH values, however, it will approach 1, because the last term of the denominator in the equation will not change, while the first six terms of the denominator will become smaller as pH increases.

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