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Buffer Calculations for Polyprotic Acids

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 - k a1 = 1.1 x 10 -2

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Buffer Calculations for Polyprotic Acids

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  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). H3PO4D H+ + H2PO4- ka1 = 1.1 x 10-2 This buffer operates in the range: pH = pka+ 1 = 0.96 – 2.96

  2. Also, another buffer which is commonly used is the dihydrogen phosphate/hydrogen phosphate buffer. H2PO4-D H+ + HPO42- ka2 = 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: HPO42- D H+ + PO43- ka3 = 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 H2CO3D H+ + HCO3- ka1 = 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: HCO3-D H+ + CO32- ka2 = 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 [H2PO4-]/[HPO42-] if the pH of the solution containing a mixture of both substances is 7.4. ka2 = 7.5x10-8 Solution The equilibrium equation combining the two species is: H2PO4-D H+ + HPO42- ka2 = 7.5 x 10-8 Ka2 = [H+][HPO42-]/[H2PO4-] [H+] = 10-7.4 = 4x10-8 M 7.5x10-8 = 4x10-8 [HPO42-]/[H2PO4-] [HPO42-]/[H2PO4-] = 1.9

  5. Fractions of Dissociating Species at a Given pH Consider the situation where, for example, 0.1 mol of H3PO4 is dissolved in 1 L of solution. H3PO4D H+ + H2PO4- ka1 = 1.1 x 10-2 H2PO4-D H+ + HPO42- ka2 = 7.5 x 10-8 HPO42- D H+ + PO43- ka3 = 4.8 x 10-13 Some of the acid will remain undissociated (H3PO4), some will be converted to H2PO4-, HPO42- and PO43- where we have, from mass balance: CH3PO4 = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]

  6. We can write the fractions of each species in solution as a0 = [H3PO4]/CH3PO4 a1 = [H2PO4-]/CH3PO4 a2 = [HPO42-]/CH3PO4 a3 = [PO43-]/CH3PO4 a0 + a1 + a2 + a3 = 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 H3PO4 while high pH values will result in greater amounts converted to PO43-. Setting up a relation of these species as a function of [H+] is straightforward using the equilibrium constant relations. Let us try finding a0 where a0 is a function of undissociated acid. The point is to substitute all fractions by their equivalent as a function of undissociated acid.

  8. Ka1 = [H2PO4-][H+]/[H3PO4] Therefore we have [H2PO4-] = ka1 [H3PO4]/ [H+] ka2 = [HPO42-][H+]/[H2PO-] Multiplying ka2 time ka1 and rearranging we get: [HPO42-] = ka1ka2 [H3PO4]/[H+]2 ka3 = [PO43-][H+]/[HPO42-] Multiplying ka1 times ka2 times ka3 and rearranging we get: [PO3-] = ka1ka2ka3 [H3PO4]/[H+]3 But we have: CH3PO4 = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]

  9. Substitution for all species from above gives: CH3PO4 = [H3PO4] + ka1 [H3PO4]/ [H+] + ka1ka2 [H3PO4]/[H+]2 + ka1ka2ka3 [H3PO4]/[H+]3 CH3PO4 = [H3PO4] {1 + ka1 / [H+] + ka1ka2 /[H+]2 + ka1ka2ka3 /[H+]3} [H3PO4]/CH3PO4 = 1/ {1 + ka1 / [H+] + ka1ka2 /[H+]2 + ka1ka2ka3 /[H+]3} ao = [H+]3 / ([H+]3 + ka1[H+]2 + ka1ka2[H+] + ka1ka2ka3) Similar derivations for other fractions results in: a1 = ka1[H+]2 / ([H+]3 + ka1[H+]2 + ka1ka2[H+] + ka1ka2ka3) a2 = ka1ka2 [H+] / ([H+]3 + ka1[H+]2 + ka1ka2[H+] + ka1ka2ka3) a3 = ka1ka2ka3 / ([H+]3 + ka1[H+]2 + ka1ka2[H+] + ka1ka2ka3)

  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 ao gives ao = [H+]3 / ([H+]3 + ka1[H+]2 + ka1ka2[H+] + ka1ka2ka3) ao = (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. ao = 8.2x10-2 a0 = [H3PO4]/CH3PO4 8.2x10-2 = [H3PO4]/0.10 [H3PO4] = 8.3x10-3 M Similarly, a1 = 0.92, a1 = [H2PO4-]/CH3PO4 0.92 = [H2PO4-]/0.10 [H2PO4-] = 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: Na3PO4 and Na2CO3. 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 Na3PO4 solution. Solution We have the following equilibrium in water PO43- + H2O D HPO42- + OH- The equilibrium constant which corresponds to this equilibrium is kb where: Kb = kw/ka3

  14. We used ka3 since it is the equilibrium constant describing relation between PO43- and HPO42-. However, in any equilibrium involving salts look at the highest charge on any anion to find which ka to use. Kb = 10-14/4.8x10-13 Kb = 0.020

  15. Kb = x * x/0.10 – x Assume 0.10 >> x 0.02 = x2/0.10 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, NaH2PO4 in water would show the following equilibria: H2PO4-D H+ + HPO42- H2PO4- + H2O D OH- + H3PO4 H2O D H+ + OH- [H+]solution = [H+]H2PO4- + [H+]H2O– [OH-]H2PO4- [H+]solution = [HPO42-] + [OH-] – [H3PO4]

  17. Now make all terms as functions in either H+ or H2PO4-, then we have: [H+] = {ka2 [H2PO4-]/[H+]} + kw/[H+] –{[H2PO4-][H+]/ka1} Rearrangement gives [H+] = {(ka1kw + ka1ka2[H2PO4-])/(ka1 + [H2PO4‑]}1/2 At high salt concentration and low ka1 this relation may be approximated to: [H+] = {ka1ka2}1/2 Where; the pH will be independent on salt concentration but only on the equilibrium constants.

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