Chapter 11 Polyprotic Acids

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Chapter 11 Polyprotic Acids

• Review of Chem 112 and New concepts

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Example H3PO4

• H3PO4 H H2PO4- Ka1 1.1e-2
• H2PO4- H HPO42- Ka2 7.5e-8
• HPO42- H PO43- Ka3 4.8e-13
• For H3PO4 3H PO43-
• Ka1Ka2Ka3 Ka 4.0e-22

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Buffer calculations involving polyprotic acids

• To make exact pH calculations we need to consider
  each equilibrium step. That can get complicated.
  We will instead make a few simplifying
  calculations in this course.
• Example For H3PO4 what are the predominate
  species involved in the buffering at pH 7.40?
• Consider the pKa of each step.
• pKa1 1.96 pKa2 7.12 pKa3 12.32
• remembering the rule that pH buffers work best
  when pKa pH ? 1
• It stands to reason that Ka2 is the equilibrium
  most responsible of pH 7.40 buffering.

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How to make a buffer.

• H2PO4- H HPO42-
• Knowing that H is 10-7.40 4.0e-8 M, what is
  the ratio of HPO42-/H2PO4-
• Therefore a solution of 0.19 Na2HPO4 and 0.10 M
  NaH2PO4 would buffer at pH 7.40.

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Now consider and develop a more rigorous solution
to this buffer problem.

• Now lets consider the relative concentration of
  each form of phosphate at a given pH.
• H3PO4, H2PO4-, HPO42-, PO43-
• We should realize that if we make a 0.10 M H3PO4
  solution the MBE is
• 0.10 M H3PO4 H2PO4- HPO42- PO43-

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For any given concentration it follows that the
MBE will be

• FH3PO4 H3PO4 H2PO4- HPO42- PO43-
• The relative concentrations of each species will
  be
• Fraction of a species is ? species/F

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Relative concentration for each species, ?

• Note that
• 1 ?H3PO4 ?H2PO4- ?HPO4 ?PO4? 1

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Now well look for a relationship between
equation 1 and the MBE.

• This will allow us to calculate concentrations of
  each species.
• Start with..

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Lets find a relationship for each Ka that only
involves 2 variables H and H3PO4, and use
the MBE.
We need to develop eqn 1 and 2 in forms that
only contain H H3PO4
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Lets find a relationship for each Ka that only
involves 2 variables H and H3PO4, and use
the MBE.
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now we can sub eqn 3, 4, 5 into the MBE

• FH3PO4 H3PO4 H2PO4- HPO42- PO43-

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Rearrangement gives
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Example Calculate the concentration of each
species in 0.10 F H3PO4 solution at pH 7.40.
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So what have we discovered?

• The two predominate species of phosphoric acid at
  pH 7.40 are
• HPO42- 0.065M H2PO4- 0.035 M
• The other species are insignificant.
• Note that 0.065/0.035 1.86 1.9

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We can calculate the pH through Ka2 or the
Henderson-Hasselbalch eqn.

• pH pKa log(A-/HA)
• -log(7.5e-8) log(0.065/0.035)
• pH 7.39

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In general for HnA

• D Hn Ka1Hn-1 Ka1Ka2Hn-2 ..
  Ka1Ka2Ka3Kan
• ?HnA Hn/D
• ?Hn-1A Ka1Hn-1/D
• ?Hn-jA Ka1 Ka2.KjHn-j/D
• Remember this formula and practice writing out
  this concept with polyprotic acids

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Example Calculate the relative concentrations
of the various forms of oxalic acid H2C2O4 at pH
2.00

• What is the structure of oxalic acid?
• What are the important equilibria?
• H2C2O4 ? HC2O4- H Ka1 5.6e-2
• HC2O4- ? C2O42- H Ka2 5.42e-5

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Write down the denominator term

• D Hn Ka1Hn-1 Ka1Ka2Hn-2 ..
  Ka1Ka2Ka3Kan
• D H2 Ka1H1 Ka1Ka2
• D (1.0e-2)2 5.6e-21.0e-2 5.6e-25.42e-5
• 6.6e-4

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Now for each component

• ?HnA Hn/D H2/D
• ?H2A H2/D (1.0e-2)2/6.6e-4 0.15
• ?Hn-1A Ka1Hn-1/D Ka1H1/D
• ?HA- Ka1H/D (1.0e-25.6e-2)/6.6e-4 0.85
• ?Hn-jA Ka1 Ka2.KjHn-j/D Ka1 Ka2/D
• ?A Ka1 Ka2/D 5.6e-25.42e-5/6.6e-4 0.0046

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We can always check our answer

• 1 ?H2A ?HA- ?A
• 0.15 0.85 0.0046 1.00

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amphoteric acids and bases

• The salts of H2PO4-, HPO42- can either be
  acidic (proton donor) or basic (proton acceptor)
• Is a solution of 0.10 M NaH2PO4 acidic or basic?
• Two possible Reactions
• H2PO4- H HPO42- Ka2 6.32e-8
• H2PO4- H2O H3PO4 HO-
• Kb Kw/Ka1 1e-14/7.11e-3 1.41e-12
• We can see that Ka2 gtgt Kb therefore its acidic.

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How can we calculate pH of an amphoteric species?

• We will find two relationships of which you will
  be required to know the simpler one.
• You will not be expected to derive these
  relationships.
• In order to describe H we need to develop a
  PBE. We will only do a PBE this one time, it is
  necessary for the derivation of an formula that
  we use often.

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PBE is keeping track of where protons come from

• Example
• H3A we have the following
• H3A H2A- H Overall rxns.
• H2A- HA2- H H3A HA- 2H 1
• HA2- A3- H H3A A3- 3H 2
• H2O H OH-

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The PBE for this example is

• Note that 2 protons came off of H3A to make HA2-
  hence the multiplier of 2. Three for A3-.

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PBE For NaH2PO4

• H2PO4- H HPO42- Ka2
• H2PO4- H2O H3PO4 HO- Kb
• H2O H OH-
• PBE H HPO42- - H3PO4 OH-
• Note its - H3PO4 because that step consumes a
  proton by making OH-.

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We now seek an eqn that expresses only H and
H2PO4-

• Now for substitutions

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H3PO4 HPO42- expressed only in terms of
H2PO4- H

• H3PO4 H2PO4-H/Ka1 1
• And we find
• HPO42- Ka2H2PO4-/H 2

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PBE H HPO42- - H3PO4 OH-
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• This is eqn 11-13/14 in Harris

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This eqn allows us to calculate pH from the
intermediate form of a polyprotic acid.

• For 0.10 M NaH2PO4
• H2PO4- H2O H3PO4 HO-
• Kb Kw/Ka1 1e-14/7.11e-3 1.41e-12
• H2PO4- ? HPO42- H Ka2 6.32e-8
• Ka1 7.11e-3 Ka2 6.32e-8

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Now for the H
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Example Calculate the pH of 0.050 M H2SO3

• H2SO3 H HSO3- Ka 1.23e-2
• 0.050 -x x x
• x2/(0.050 x) 1.23e-2
• use quadratic formula
• x 1.94e-2 pH 1.71

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Example Calculate the pH of 0.050 M NaHSO3
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Simplifications
Remember this equation
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Example what is the pH of 0.10 M NaH2PO4 ?

• Ka1 7.11e-3 pKa1 2.148
• Ka2 6.32e-8 pKa2 7.199
• pH ½ (2.148 7.199) 4.67
• compare with 4.69 (slide 32)

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0.050 M NaHSO3

• Ka1 1.23e-2 pKa1 1.910
• Ka2 6.6e-8 pKa2 7.18
• pH ½ (1.910 7.18) 4.54
• compare to 4.59 (slide 34)

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pH buffer system of human blood

• Human blood is at pH 7.2 0.1
• H2PO4-(aq) ? H(aq) HPO42-(aq) pKa 7.199
• H2CO3(aq) ? H(aq) HCO3-(aq) pKa 6.352
• http//scifun.chem.wisc.edu/chemweek/BioBuff/BioBu
  ffers.html
• http//www.chemistry.wustl.edu/edudev/LabTutorial
  s/Buffer/Buffer.html