Buffer Calculations for Polyprotic Acids

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).
• H3PO4 D 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
• H2CO3 D 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 HHPO42-/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.
• H3PO4 D 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-

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• 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).

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• 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.

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• 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/H2
• ka3 PO43-H/HPO42-
• Multiplying ka1 times ka2 times ka3 and
  rearranging we get
• PO3- ka1ka2ka3 H3PO4/H3
• But we have
• CH3PO4 H3PO4 H2PO4- HPO42- PO43-

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• Substitution for all species from above gives
• CH3PO4 H3PO4 ka1 H3PO4/ H ka1ka2
  H3PO4/H2 ka1ka2ka3 H3PO4/H3
• CH3PO4 H3PO4 1 ka1 / H ka1ka2 /H2
  ka1ka2ka3 /H3
• H3PO4/CH3PO4 1/ 1 ka1 / H ka1ka2
  /H2 ka1ka2ka3 /H3
• ao H3 / (H3 ka1H2 ka1ka2H
  ka1ka2ka3)
• Similar derivations for other fractions results
  in
• a1 ka1H2 / (H3 ka1H2 ka1ka2H
  ka1ka2ka3)
• a2 ka1ka2 H / (H3 ka1H2 ka1ka2H
  ka1ka2ka3)
• a3 ka1ka2ka3 / (H3 ka1H2 ka1ka2H
  ka1ka2ka3)

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• 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 H3 / (H3 ka1H2 ka1ka2H
  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

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• 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

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• 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 gtgt 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

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• 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-
• Hsolution HH2PO4- HH2O OH-H2PO4-
• Hsolution HPO42- OH- H3PO4

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• 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 ka1ka2H2PO4-)/(ka1
  H2PO4-1/2
• At high salt concentration and low ka1 this
  relation may be approximated to
• H ka1ka21/2
• Where the pH will be independent on salt
  concentration but only on the equilibrium
  constants.