1. Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can form a buffer when combined with its conjugate base (dihydrogen phosphate). H 3 PO 4  H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 This buffer operates in the range: pH = pk a + 1 = 0.96 – 2.96

2. Also, another buffer which is commonly used is the dihydrogen phosphate/hydrogen phosphate buffer. H 2 PO 4 -  H + + HPO 4 2- k a2 = 7.5 x 10 -8 This buffer operates in the range from 6.1 to 8.1 A third buffer can be prepared by mixing hydrogen phosphate with orthophosphate as the following equilibrium suggests: HPO 4 2-  H + + PO 4 3- k a3 = 4.8 x 10 -13 This buffer system operates in the pH range from 11.3 to 13.3

3. The same can be said about carbonic acid/bicarbonate where H 2 CO 3  H + + HCO 3 - k a1 = 4.3 x 10 -7 This buffer operates in the pH range from 5.4 to 7.4; while a more familiar buffer is composed of carbonate and bicarbonate according to the equilibrium: HCO 3 -  H + + CO 3 2- k a2 = 4.8 x 10 -11 The pH range of the buffer is 9.3 to 11.3. Polyprotic acids and their salts are handy materials which can be used to prepare buffer solutions of desired pH working ranges. This is true due to the wide variety of their acid dissociation constants.

4. Example Find the ratio of [H 2 PO 4 - ]/[HPO 4 2- ] if the pH of the solution containing a mixture of both substances is 7.4. k a2 = 7.5x10 -8 Solution The equilibrium equation combining the two species is: H 2 PO 4 -  H + + HPO 4 2- k a2 = 7.5 x 10 -8 K a2 = [H + ][HPO 4 2- ]/[H 2 PO 4 - ] [H + ] = 10 -7.4 = 4x10 -8 M 7.5x10 -8 = 4x10 -8 [HPO 4 2- ]/[H 2 PO 4 - ] [HPO 4 2- ]/[H 2 PO 4 - ] = 1.9

5. Fractions of Dissociating Species at a Given pH Consider the situation where, for example, 0.1 mol of H 3 PO 4 is dissolved in 1 L of solution. H 3 PO 4  H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 H 2 PO 4 -  H + + HPO 4 2- k a2 = 7.5 x 10 -8 HPO 4 2-  H + + PO 4 3- k a3 = 4.8 x 10 -13 Some of the acid will remain undissociated (H 3 PO 4 ), some will be converted to H 2 PO 4 -, HPO 4 2- and PO 4 3- where we have, from mass balance: C H3PO4 = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ]

6. We can write the fractions of each species in solution as  0 = [H 3 PO 4 ]/C H3PO4  1 = [H 2 PO 4 - ]/C H3PO4  2 = [HPO 4 2- ]/C H3PO4  3 = [PO 4 3- ]/C H3PO4  0 +  1 +  2 +  3 = 1 ( total value of all fractions sum up to unity).

7. The value of each fraction depends on pH of solution. At low pH dissociation is suppressed and most species will be in the form of H 3 PO 4 while high pH values will result in greater amounts converted to PO 4 3-. Setting up a relation of these species as a function of [H + ] is straightforward using the equilibrium constant relations. Let us try finding  0 where  0 is a function of undissociated acid. The point is to substitute all fractions by their equivalent as a function of undissociated acid.

8. K a1 = [H 2 PO 4 - ][H + ]/[H 3 PO 4 ] Therefore we have [H 2 PO 4 - ] = k a1 [H 3 PO 4 ]/ [H + ] k a2 = [HPO 4 2- ][H + ]/[H 2 PO - ] Multiplying k a2 time k a1 and rearranging we get: [HPO 4 2- ] = k a1 k a2 [H 3 PO 4 ]/[H + ] 2 k a3 = [PO 4 3- ][H + ]/[HPO 4 2- ] Multiplying k a1 times k a2 times k a3 and rearranging we get: [PO 3- ] = k a1 k a2 k a3 [H 3 PO 4 ]/[H + ] 3 But we have: C H3PO4 = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ]

9. Substitution for all species from above gives: C H3PO4 = [H 3 PO 4 ] + k a1 [H 3 PO 4 ]/ [H + ] + k a1 k a2 [H 3 PO 4 ]/[H + ] 2 + k a1 k a2 k a3 [H 3 PO 4 ]/[H + ] 3 C H3PO4 = [H 3 PO 4 ] {1 + k a1 / [H + ] + k a1 k a2 /[H + ] 2 + k a1 k a2 k a3 /[H + ] 3 } [H 3 PO 4 ]/C H3PO4 = 1/ {1 + k a1 / [H + ] + k a1 k a2 /[H + ] 2 + k a1 k a2 k a3 /[H + ] 3 }  o = [H + ] 3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) Similar derivations for other fractions results in:  1 = k a1     / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 )  2 = k a1 k a2 [H + ] / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 )  3 = k a1 k a2 k a3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 )

10. Example Calculate the equilibrium concentrations of the different species in a 0.10 M phosphoric acid solution at pH 3.00. Solution The [H + ] = 10 -3.00 = 1.0x10 -3 M Substitution in the relation for  o gives  o = [H + ] 3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 )  o = (1.0x10 -3 ) 3 /{(1.0x10 -3 ) 3 + 1.1x10 -2 (1.0x10 -3 ) 2 + 1.1x10 -2 * 7.5x10 -8 (1.0x10 -3 ) + 1.1x10 -2 * 7.5x10 - 8 * 4.8 * 10 -13 }

11.  o = 8.2x10 -2  0 = [H 3 PO 4 ]/C H3PO4 8.2x10 -2 = [H 3 PO 4 ]/0.10 [H 3 PO 4 ] = 8.3x10 -3 M Similarly,  1 = 0.92,  1 = [H 2 PO 4 - ]/C H3PO4 0.92 = [H 2 PO 4 - ]/0.10 [H 2 PO 4 - ] = 9.2x10 -2 M Other fractions are calculated in the same manner.

12. pH Calculations for Salts of Polyprotic Acids Two types of salts exist for polyprotic acids. These include: 1. Unprotonated salts These are salts which are proton free which means they are not associated with any protons. Examples are: Na 3 PO 4 and Na 2 CO 3. Calculation of pH for solutions of such salts is straightforward and follows the same scheme described earlier for salts of monoprotic acids.

13. Example Find the pH of a 0.10 M Na 3 PO 4 solution. Solution We have the following equilibrium in water PO 4 3- + H 2 O  HPO 4 2- + OH - The equilibrium constant which corresponds to this equilibrium is k b where: K b = k w /k a3

14. We used k a3 since it is the equilibrium constant describing relation between PO 4 3- and HPO 4 2-. However, in any equilibrium involving salts look at the highest charge on any anion to find which k a to use. K b = 10 -14 /4.8x10 -13 K b = 0.020

15. K b = x * x/0.10 – x Assume 0.10 >> x 0.02 = x 2 /0.10 x = 0.045 Relative error = (0.045/0.10) x 100 = 45% Therefore, assumption is invalid and we have to use the quadratic equation. If we solve the quadratic equation we get: X = 0.036 Therefore, [OH - ] = 0.036 M pOH = 1.44 and pH = 14 – 1.44 = 12.56

16. 2. Protonated Salts These are usually amphoteric salts which react as acids and bases. For example, NaH 2 PO 4 in water would show the following equilibria: H 2 PO 4 -  H + + HPO 4 2- H 2 PO 4 - + H 2 O  OH - + H 3 PO 4 H 2 O  H + + OH - [H + ] solution = [H + ] H2PO4 - + [H + ] H2O – [OH - ] H2PO4 - [H + ] solution = [HPO 4 2- ] + [OH - ] – [H 3 PO 4 ]

17. Now make all terms as functions in either H + or H 2 PO 4 -, then we have: [H + ] = {k a2 [H 2 PO 4 - ]/[H + ]} + k w /[H + ] –{[H 2 PO 4 - ][H + ]/k a1 } Rearrangement gives [H + ] = {(k a1 k w + k a1 k a2 [H 2 PO 4 - ])/(k a1 + [H 2 PO 4 ‑ ]} 1/2 At high salt concentration and low k a1 this relation may be approximated to: [H + ] = {k a1 k a2 } 1/2 Where; the pH will be independent on salt concentration but only on the equilibrium constants.