Quiz 1, BICH 410 (MWF3-3:50), Friday, Sept. 11, 1998

Clearly circle the letter corresponding to your answer for each problem. There are five problems, each of which counts 4 points. No partial credit given for multiple-choice questions.
gas constant R 8.315 J/mole-K°


1. The pKa's for the ionization of phosphoric acid (H3PO4) are 2.14, 6.86, and 12.4. If 1 liter of a 0.1 M solution of Na3PO4 is mixed with 0.1 liter of a 1 M solution of Na2HPO4, what is the pH of the resulting solution?

Answer: E); Phosphoric acid contains three dissociable protons with three different pKa's. According to the problem, there are (1 Liter)(0.1 moles/Liter) = 0.1 moles of PO4-3 and (0.1 Liter)(1 mole/Liter) = 0.1 moles of HPO4-2. When you have an equal amount of the weak acid (HPO4-2) and the conjugate base (PO4-3), then the pH of the solution is the pKa for that dissociation, which is 12.4.


2. Which of the following statements is true about the dielectric constant?

Answer: E)


3. If an endothermic reaction results in a decrease in entropy, what would you predict about the free energy change for this reaction?

Answer: C); deltaG = deltaH - TdeltaS; deltaH is greater than zero for an endothermic reaction, and deltaS is less than zero for a decrease; hence, a positive number minus a negative number must be a positive number.


4. In the pathway for the degradation of glucose is the following reaction:
1,3-bisphosphoglycerate + ADP <-> 3-phosphoglycerate + ATP
Given the information below, calculate the standard free energy change for the above reaction.
1,3-bisphosphoglycerate <-> 3-phosphoglycerate + Pi; std. free energy change = -49 kJ/mole
ATP <-> ADP + Pi; std. free energy change = -31 kJ/mole
glucose + Pi <-> glucose-6-phosphate; std. free energy change = +14 kJ/mole

Answer: A); Sum the top reaction plus the inversion of the second reaction (ATP synthesis); hence, standard free energy change for the overall reaction is -49 kJ/mole + -(-31 kJ/mole) = -18 kJ/mole

5. What is the approximate pH of a 0.01 M solution of a weak acid, HA, that dissociates with a Ka = 10-4 M?

Answer: B); dissociation reaction for weak acid HA <-> H+ + A-
[H+] = [A-]; also, since very little of HA dissociates, [HA] = (approximately) [HA]o, the starting conc. of HA
Ka = [H+][A-] / [HA] = [H+][H+] / [HA]o
So, [H+] = square root of (Ka)([HA]o)
pH = -log[H+] = -log(square root of (Ka)([HA]o)) = -log(sq.root of (10-4)(10-2)) = -log(10-3) = 3