CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010

CHEMISTRY 59-320CHEMISTRY 59-320ANALYTICAL CHEMISTRYANALYTICAL CHEMISTRY

Fall - 2010Fall - 2010

Lecture 15Chapter 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/HA- and HA-/A2-.

• Depending on the starting material, one can use one of the following two equations to calculate pH of the buffer solution

][][log

][][log

2

2

21

HAApKpH

AHHApKpH

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

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])

][][log

21 AH

HApKpH

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+ + HPO4

2- (K2)

HPO42- ↔ H+ + PO4

3- (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

The Acidic Form, H2L+ - diprotic 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−.

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