Buffer Calculations for Polyprotic Acids A polyprotic acid

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 D H+ + H 2 PO 4 ka 1 = 1. 1 x 10 -2 This buffer operates in the range: p. H = pka + 1 = 0. 96 – 2. 96

Also, another buffer which is commonly used is the dihydrogen phosphate/hydrogen phosphate buffer. H 2 PO 4 - D H+ + HPO 42 ka 2 = 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 42 - D H+ + PO 43 ka 3 = 4. 8 x 10 -13 This buffer system operates in the p. H range from 11. 3 to 13. 3

The same can be said about carbonic acid/bicarbonate where H 2 CO 3 D H+ + HCO 3 ka 1 = 4. 3 x 10 -7 This buffer operates in the p. H range from 5. 4 to 7. 4; while a more familiar buffer is composed of carbonate and bicarbonate according to the equilibrium: HCO 3 - D H+ + CO 32 ka 2 = 4. 8 x 10 -11 The p. H 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 p. H working ranges. This is true due to the wide variety of their acid dissociation constants.

Example Find the ratio of [H 2 PO 4 -]/[HPO 42 -] if the p. H of the solution containing a mixture of both substances is 7. 4. ka 2 = 7. 5 x 10 -8 Solution The equilibrium equation combining the two species is: H 2 PO 4 - D H+ + HPO 42 ka 2 = 7. 5 x 10 -8 Ka 2 = [H+][HPO 42 -]/[H 2 PO 4 -] [H+] = 10 -7. 4 = 4 x 10 -8 M 7. 5 x 10 -8 = 4 x 10 -8 [HPO 42 -]/[H 2 PO 4 -] = 1. 9

Fractions of Dissociating Species at a Given p. H 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 D H+ + H 2 PO 4 ka 1 = 1. 1 x 10 -2 H 2 PO 4 - D H+ + HPO 42 ka 2 = 7. 5 x 10 -8 HPO 42 - D H+ + PO 43 ka 3 = 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 42 - and PO 43 - where we have, from mass balance: CH 3 PO 4 = [H 3 PO 4] + [H 2 PO 4 -] + [HPO 42 -] + [PO 43 -]

We can write the fractions of each species in solution as a 0 = [H 3 PO 4]/CH 3 PO 4 a 1 = [H 2 PO 4 -]/CH 3 PO 4 a 2 = [HPO 42 -]/CH 3 PO 4 a 3 = [PO 43 -]/CH 3 PO 4 a 0 + a 1 + a 2 + a 3 = 1 ( total value of all fractions sum up to unity).

The value of each fraction depends on p. H of solution. At low p. H dissociation is suppressed and most species will be in the form of H 3 PO 4 while high p. H values will result in greater amounts converted to PO 43 -. Setting up a relation of these species as a function of [H+] is straightforward using the equilibrium constant relations. Let us try finding a 0 where a 0 is a function of undissociated acid. The point is to substitute all fractions by their equivalent as a function of undissociated acid.

Ka 1 = [H 2 PO 4 -][H+]/[H 3 PO 4] Therefore we have [H 2 PO 4 -] = ka 1 [H 3 PO 4]/ [H+] ka 2 = [HPO 42 -][H+]/[H 2 PO-] Multiplying ka 2 time ka 1 and rearranging we get: [HPO 42 -] = ka 1 ka 2 [H 3 PO 4]/[H+]2 ka 3 = [PO 43 -][H+]/[HPO 42 -] Multiplying ka 1 times ka 2 times ka 3 and rearranging we get: [PO 3 -] = ka 1 ka 2 ka 3 [H 3 PO 4]/[H+]3 But we have: CH 3 PO 4 = [H 3 PO 4] + [H 2 PO 4 -] + [HPO 42 -] + [PO 43 -]

Substitution for all species from above gives: CH 3 PO 4 = [H 3 PO 4] + ka 1 [H 3 PO 4]/ [H+] + ka 1 ka 2 [H 3 PO 4]/[H+]2 + ka 1 ka 2 ka 3 [H 3 PO 4]/[H+]3 CH 3 PO 4 = [H 3 PO 4] {1 + ka 1 / [H+] + ka 1 ka 2 /[H+]2 + ka 1 ka 2 ka 3 /[H+]3} [H 3 PO 4]/CH 3 PO 4 = 1/ {1 + ka 1 / [H+] + ka 1 ka 2 /[H+]2 + ka 1 ka 2 ka 3 /[H+]3} ao = [H+]3 / ([H+]3 + ka 1[H+]2 + ka 1 ka 2[H+] + ka 1 ka 2 ka 3) Similar derivations for other fractions results in: a 1 = ka 1[H+]2 / ([H+]3 + ka 1[H+]2 + ka 1 ka 2[H+] + ka 1 ka 2 ka 3) a 2 = ka 1 ka 2 [H+] / ([H+]3 + ka 1[H+]2 + ka 1 ka 2[H+] + ka 1 ka 2 ka 3) a 3 = ka 1 ka 2 ka 3 / ([H+]3 + ka 1[H+]2 + ka 1 ka 2[H+] + ka 1 ka 2 ka 3)

Example Calculate the equilibrium concentrations of the different species in a 0. 10 M phosphoric acid solution at p. H 3. 00. Solution The [H+] = 10 -3. 00 = 1. 0 x 10 -3 M Substitution in the relation for ao gives ao = [H+]3 / ([H+]3 + ka 1[H+]2 + ka 1 ka 2[H+] + ka 1 ka 2 ka 3) ao = (1. 0 x 10 -3)3/{(1. 0 x 10 -3)3 + 1. 1 x 10 -2 (1. 0 x 10 -3)2 + 1. 1 x 10 -2 * 7. 5 x 10 -8 (1. 0 x 10 -3) + 1. 1 x 10 -2 * 7. 5 x 108 * 4. 8 * 10 -13}

ao = 8. 2 x 10 -2 a 0 = [H 3 PO 4]/CH 3 PO 4 8. 2 x 10 -2 = [H 3 PO 4]/0. 10 [H 3 PO 4] = 8. 3 x 10 -3 M Similarly, a 1 = 0. 92, a 1 = [H 2 PO 4 -]/CH 3 PO 4 0. 92 = [H 2 PO 4 -]/0. 10 [H 2 PO 4 -] = 9. 2 x 10 -2 M Other fractions are calculated in the same manner.

p. H 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 p. H for solutions of such salts is straightforward and follows the same scheme described earlier for salts of monoprotic acids.

Example Find the p. H of a 0. 10 M Na 3 PO 4 solution. Solution We have the following equilibrium in water PO 43 - + H 2 O D HPO 42 - + OHThe equilibrium constant which corresponds to this equilibrium is kb where: Kb = kw/ka 3

We used ka 3 since it is the equilibrium constant describing relation between PO 43 - and HPO 42 -. However, in any equilibrium involving salts look at the highest charge on any anion to find which ka to use. Kb = 10 -14/4. 8 x 10 -13 Kb = 0. 020

Kb = 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 p. OH = 1. 44 and p. H = 14 – 1. 44 = 12. 56

2. Protonated Salts These are usually amphoteric salts which react as acids and bases. For example, Na. H 2 PO 4 in water would show the following equilibria: H 2 PO 4 - D H+ + HPO 42 H 2 PO 4 - + H 2 O D OH- + H 3 PO 4 H 2 O D H+ + OH[H+]solution = [H+]H 2 PO 4 - + [H+]H 2 O – [OH-]H 2 PO 4[H+]solution = [HPO 42 -] + [OH-] – [H 3 PO 4]

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