Ch. 11Polyprotic AcidBase Equilibria

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Ch. 11-Polyprotic Acid-Base Equilibria
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Proteins as Polyprotic Acids and Bases
Amino acids
Peptides/proteins
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Amino Acids

• Amino acid building blocks of proteins have the
  general structure
• R is a different side chain for each amino acid
• At low pH, both the ammonium and carboxyl groups
  are protonated. At high pH, neither is.
• Were going to look at leucine as an example

A zwitterion is a molecule with both positive and
negative charges
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Amino Acids
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Leucine

• The side group on leucine is an isobutyl group
• Equilibrium constants are for these reactions
• Well now calculate pH for solutions of these

H2L
HL
L-
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H2L (Acid form)

• Whats the pH of a 0.0500 M solution of H2L
• Because Ka14.69x10-3 and Ka21.79x10-10, were
  going to assume that H2L acts as a monoprotic
  acid
• We can check our assumption by calculating L-

H2L (0.050-x)
HL (x)
H (x)
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L- (Base Form)

• Whats the pH of a 0.0500 M solution of L-?
• Again, because Kb2ltltKb1were going to assume that
  H2L acts as a monoprotic acid
• We can check our assumption by calculating H2L

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HL (Middle Form)

• What is the pH of a solution of 0.050 M HL?
• HL is amphiprotic, meaning it can both donate and
  accept a proton
• Wed expect this solution to be slightly acidic
  because Ka1 is larger than Kb2, but we have to
  use the systematic treatment to determine pH

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Example-Leucine (HL)

• Reactions
• Charge balance
• Equilibrium constants

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Example-Leucine (cont)

• If we assume that neither reaction proceeds very
  far because both Ks are small, then we can put
  the Formal concentration into HL

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Example-Leucine (cont)

• If HLF0.050 M
• We can again check our assumptions
• The assumptions were valid, both of these are
  much lower concentration than HL (0.050 M)

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Simplified Equation

• Some conditions usually exist to simplify the
  general equation
• If Ka2(F)gtgtKw, than the 2nd term in the numerator
  can be dropped and if Ka1ltltF, the 1st term in the
  denominator can also be neglected
• After canceling F and doing some algebra
• This eqn usually works for the intermediate form
  of a diprotic acid Here, it gives pH6.04
  instead of 6.06

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Example-KHP

• Potassium hydrogen phthalate is a salt of the
  intermediate form of phthalic acid. Calculate
  the pH of 0.10 and 0.010 M KHP solution

H2P (Phthalic acid)
HP- (monohydrogen phthalate)
P2- (phthalate)
(For any molarity)
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Summary of Solving Diprotics

• Solving for H2A
• Treat as monoprotic with KaKa1
• Solving for HA-
• Use the approximation HA-F
• Solving A2-
• Treat A2- as monobasic with KbKb1

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Diprotic Buffers

• A buffer from a diprotic (or polyprotic) acid is
  treated the same as for a monoprotic
• For H2A, we can write two Henderson-Hasselbalch
  equations, both of which must be true.

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A Buffer System

• What is the pH of a solution prepared by
  dissolving 1.0 g KHP and 1.20 g Na2P in 50.0 mL
  water?
• We know A2- and HA-, so we use pK2
• Since volume cancels anyway, we just used moles

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Polyprotic Acids and Bases

• Polyprotics essentially can be treated similar to
  diprotics
• H3A is treated as a monprotic with KaK1
• H2A- is treated as an intermediate form
• HA2- is treated as an intermediate form
• A3- is treated as monobasic with KbKb1

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Graphic of Treating Triprotics