CHEMISTRY 59-320
ANALYTICAL CHEMISTRY
Fall - 2010
Lecture 15
Chapter 10 (continued)
Summary of diprotic acid
calculations
• A diprotic acid H2A exists as H2A, HA- and A2- in
solution and thus acid may be prepared with any
of the above substance.
• Case 1: The acid solution is prepared with H2A.
• Case 2: The acid solution is prepared with HA-.
• Case 3: The solution is prepared with A2-.
10-2 Diprotic buffers
• It acts in the same way as a buffer prepared from
a monoprotic acid, except here we have two
pairs of acid and conjugate base, i.e. H2A/HAand HA-/A2-.
• Depending on the starting material, one can use
one of the following two equations to calculate
pH of the buffer solution
[ HA ]
pH  pK1  log
[ H 2 A]
[ A2  ]
pH  pK2  log
[ HA ]
Diprotic buffer calculation
• 10-12: How many milliliters of 0.202 M NaOH should be added to
25.0 ml of 0.0233 M salicylic acid (2-hydroxybenzoic acid) to adjust
the pH to 3.50?
• Solution: From Appendix G, pK1= 2.972; pK2 = 13.7
the addition of strong base produced equivalent number of moles of
HA-.
OH- + H2A ↔ HA- + H2O
[ HA ]
pH  pK1  log
[ H 2 A]
Assuming x ml of NaOH is required; [HA-] = x*0.202/(0.025+ x)
[H2A] = (0.0233*0.025 – x*0.202)/(0.025+x)
3.5 = 2.972 + log([HA-]/[H2A])
10-3 Polyprotic acids and bases
• The treatment of ployprotic acids (HnA) is the same as
diprotic acid.
• Treating HnA as a monoprotic acid and An- as
monobasic.
• Other intermediates can be dealt with the following
equations for the calculation of pH.
• For Hn-1A-
•
For Hn-2A2-
Example of polyprotic acid calculation
• Problem 10-17: (a) calculate the quotient
[H3PO4]/[H2PO4-] in 0.0500 M KH2PO4. (b) find the same
quotient for 0.0500 M K2HPO4.
• Solution: From Appendix G, find pK1 = 2.148, pK2 =
7.198, and pK3 = 12.375.
H3PO4 ↔ H+ + H2PO4(K1)
H2PO4- ↔ H+ + HPO42(K2)
HPO42- ↔ H+ + PO43(K3)
(a) using equation 10-13 to calculate [H+], in which F =
0.0500M. The calculation yields [H+] = 1.988 x 10-5 M.
then, employing the definition k1 = [H+][H2PO4-]/[H3PO4]
to calculate the ratio of [H3PO4]/[H2PO4-]
(b) is the same as (a), but uses equation 10-14.
10-4 Which is the principal species
• For monoprotic acid: Following Henderson-Hasselbalch
equation pH = pKa + log([A-]/[HA]), when the pH > pKa,
the basic species is the dominant one, whereas at pH <
pKa, the acidic species is the dominant one.
• For polyprotic acid: the reasoning is the same as
monoprotic acid, except here there are multiple pKa
values to separate different dominant species.
Example of calculating pH and
dominant species of polyprotic acid
• Problem 10-23: The diprotic acid H2A has pK1 = 4.00 and pK2 =
8.00. (a) At what pH is [H2A] = [HA-]? (b) at what pH is [HA-] = [A2-]?
(c) Which is the principle species at pH 2.00: H2A, HA-, A2-? (d)
Which is the principle species at pH = 6.00?
• Solution:
(a) using Henderson-Hasselbalch equation, when [H2A] =
[HA-], pH = pKa1, i.e. pH = 4.00.
(b) the same as (a), when [H2A] = [HA-], pH = pKa2, pH =
8.00.
(c) Since pH = 2.00 < pKa1, the H2A should be the dominant
species.
(d) since pH > pKa1 & pH < pKa2, HA- is the dominant one.
10-5 Fractional composition
equations
• In addition to estimating the dominant species
based on pH, one can calculate the fraction of
each species.
• The fraction of molecules in monoprotic acid
Fractional composition of Diprotic
system
Example of calculating pH and
dominant species of polyprotic acid
Example: calculate the pH and composition of individual
solutions of (a) 0.050 0 M H2L+, (b) 0.050 0 M HL, and (c)
0.050 0 ML−.
The Acidic Form, H2L+ - diprotic acid
Leucine hydrochloride contains the protonated
species H2L+
HL is an even
weaker acid,
because
K2 = 1.80 × 10−10.
10-6 Isoelectric and isoionic pH
• Isoionic pH is the pH of the pure, neutral,
polyprotic acid.
• Isoelectric pH is the pH at which average charge
of the polyorotic acid is 0.
• For a diprotic amino acid, the isoelectric
pH is halfway between the two pKa
values.
• For a diprotic amino acid, the isoionic pH
is given by equation 10-22, where [H2A-] is
not exactly equal to [A-]
Applications of isoelectric pH in biochemistry:
Isoelectric focusing