1. Practice Test #4

2. 1. The solubility of oxygen gas in water at 25 o and 1.0 atm pressure of oxygen is 0.041 g/L. What is the solubility (in g/L) of oxygen in water at 3.0 atm and 25 o C? a. 0.041 b. 0.014 c. 0.31 d. 0.12 Remember that the solubility is directly proportional to the pressure of the gas at a specific temperature. [A(g) ]= K H P A Henry’s Law Therefore, [0.041 g/L][3.0 atm/1.0 atm] = (d) 0.12

3. (2) Hormone disrupters are believed to be a) responsible for increase in lung cancer among smokers b) responsible for increasing sperm count among selected male populations c) responsible for disruption of the endocrine system in one way or another d) responsible for increase in AIDS c) responsible for disruption of the endocrine system in one way or another

4. When evaluating the toxicity of a compound, (a) it is impossible to extrapolate the dose-response curve with absolute certainty. (b) one must know the KOW and half-life (c) one must understand the chemical structure of the compound. (d) toxicity is directly correlated with water solubility of the compound (a) it is impossible to extrapolate the dose-response curve with absolute certainty.

5. The amount of oxygen in water, would increase with a) high temperature of water and low atmospheric pressure b) low temperature of water and high atmospheric pressure c) high atmospheric pressure and high water temperature d) not be influenced by water temperature or atmospheric pressure (b) Low temperature and high pressure

6. Which of the following compounds would you expect to see in an anaerobic environment? (a) H 2 S(b) SO 2 (c) H 2 SO 3 (d) H 2 SO 4 When you consider the oxidation numbers of the sulfurs in the compound, you arrive at -2, +4, +4, and +6. Since the question specifies an anaerobic environment, or reductive environment, the answer would be (a) H 2 S.

7. When considering ground water as the source of drinking water, one must be concerned about a) depletion b) contamination from leaching c) leakage from underground tanks d) all of the above (d) All of the above are concerns about using ground water as the source for drinking water.

8. 1. (10 pts) Explain the “oxygen sag” shown in the figure on the right. The “sag” is caused by two processes occurring simultaneously. The first is a result of a chemical discharge that takes the oxygen out of the water by First process is the BOD decay (deoxygenation) which is first-order and is shown in curve “e.”. The second process is absorption of oxygen (reaeration) where oxygen is absorbed by the moving water.

9. (10 pts.) Calculate the pH of a 0.15 M solution of acetic acid, CH 3 COOH [Ka = 1.8 X 10 -5 ] CH 3 COOH  CH 3 COO - + H + 0.15 -- --- 0.15-x x x 1.8 X 10 -5 = [x][x]/[0.15-x] = [x][x]/0.15 [x] 2 = 2.7x10 -6 [x] = 1.64x10 -3 pH = 2.78

10. Given the following equilibria constants, PbS  Pb 2+ + S 2- K sp = 3.0 x10 -28 H 2 S  H + + HS - Ka 1 = 9.1 x10 -8 HS -  H + + S 2- K a2 = 1.3x10 -13 H 2 O  H + + OH - K w = 1.0 x10 -14 Calculate the equilibrium constant for the following reaction: PbS + H 2 O  Pb 2+ + HS - + OH - K = [Pb 2+ ][HS - ][OH - ] = [Pb 2+ ][HS - ][OH - ]{[S 2- ][H + ]/[S 2- ][H + ]} But K sp = [Pb 2+ ][S 2- ]; K w = [OH - ][H + ]; and 1/K a2 = [HS - ]/[S 2- ][H + ] Therefore: K = K sp K w / K a2 = 3.0 x10 -28 x 1.0 x10 -14 / 1.3x10 -13 K = 2.3 x10 -29