Lecture+3+_+4+homework+solutions

# Lecture+3+_+4+homework+solutions - Solutions for problems...

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Solutions for problems assigned with Lectures 3 and 4 Garrett & Grisham, pg. 45 6. If the internal pH of a muscle cell is 6.8, what is the [HPO 4 2- ]/[H 2 PO 4 - ] ratio in this cell? Answer: The dissociation of phosphoric acid proceeds as follows: + + + + + + 3 4 PO H 2 4 HPO H 4 PO 2 H H 4 PO 3 H the pK a values of 2.15, 7.20 and 12.40 respectively. Now, at pH = 6.8, we expect the first equilibrium to be completely to the right and the last equilibrium to be to the left (i.e., we expect phosphoric acid to be in the doubly or singly protonated forms). From the Henderson Hasselbalch equation we find: pH = pK a + log 10 [A - ] [HA] where [A - ] = [ HPO 4 2 ];[HA ] = [H 2 PO 4 ] log 10 [HPO 4 2 ] [H 2 PO 4 ] = pH pK a [HPO 4 2 ] [H 2 PO 4 ] = 10 (pH pK a ) = 10 (6 .8 7.2) = 0.4 10. Citric acid, a tricarboxylic acid important in intermediary metabolism, can be symbolized as H 3 A. Its dissociation reactions are 6.40 = a pK 3 A H 2 HA 4.76 = a pK 2 HA H A 2 H 3.13 = a pK A 2 H H A 3 H + + + + + + If the total concentration of the acid and its anion forms is 0.02 M, what are the individual concentrations of H 3 A, H 2 A - , HA 2- , and A 3- at pH 5.2? Answer: For citric acid + + + + + + 3 A H 6.40 2 HA H 4.76 A 2 H H 3.13 A 3 H At pH = 5.2 the predominant equilibrium will involve H 2 A and HA 2 -.

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pH = pK a + log 10 [HA 2 ] [H 2 A ] [HA 2 ] [H 2 A ] = 10 (pH pK a ) = 10 (5.2 4.76 ) = 2.754 (1) And, considering the other two equilibria we have [A 3 ] [HA 2 ] = 10 (pH pK a ) = 10 (5.2 6.4 ) = 0.063 (2) and [H 2 A ]
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## This note was uploaded on 01/18/2011 for the course BIS 105 taught by Professor Kennethhilt during the Spring '09 term at UC Davis.

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Lecture+3+_+4+homework+solutions - Solutions for problems...

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