Practice Test #4

1. The solubility of oxygen gas in water at 25o 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 25oC?

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) ]= KHPA Henry’s Law Therefore, [0.041 g/L][3.0 atm/1.0 atm] = (d) 0.12

(2) Hormone disrupters are believed to bea) responsible for increase in lung cancer among

smokersb) responsible for increasing sperm count among

selected male populationsc) responsible for disruption of the endocrine system in

one way or anotherd) responsible for increase in AIDS

c) responsible for disruption of the endocrine system in one way or another

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.

The amount of oxygen in water, would increase witha) high temperature of water and low atmospheric pressureb) low temperature of water and high atmospheric pressurec) high atmospheric pressure and high water temperatured) not be influenced by water temperature or atmospheric

pressure

(b) Low temperature and high pressure

Which of the following compounds would you expect to see in an anaerobic environment?

(a) H2S (b) SO2 (c) H2SO3 (d) H2SO4

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) H2S.

When considering ground water as the source of drinking water, one must be concerned about

a) depletionb) contamination from leachingc) leakage from underground tanksd) all of the above

(d) All of the above are concerns about using ground water as the source for drinking water.

1. (10 pts) Explain the “oxygen sag” shown in the figure on the right.

The “sag” is caused by twoprocesses occurring simultaneously.The first is a result of a chemicaldischarge that takes the oxygen outof the water by First process is theBOD decay (deoxygenation) which isfirst-order and is shown in curve “e.”.The second process is absorption ofoxygen (reaeration) where oxygen isabsorbed by the moving water.

(10 pts.) Calculate the pH of a 0.15 M solution of acetic acid, CH3COOH [Ka = 1.8 X 10-5]

CH3COOH CH3COO- + 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

Given the following equilibria constants,   PbS Pb2+ + S2- Ksp = 3.0 x10-28

H2S H+ + HS- Ka1 = 9.1 x10-8

HS- H+ + S2- Ka2 = 1.3x10-13

H2O H+ + OH- Kw = 1.0 x10-14

 Calculate the equilibrium constant for the following reaction:   PbS + H2O Pb2+ + HS- + OH-

  K = [Pb2+][HS-][OH-] = [Pb2+][HS-][OH-]{[S2-][H+]/[S2-][H+]} But Ksp= [Pb2+][S2-]; Kw = [OH-][H+]; and 1/Ka2 = [HS-]/[S2-][H+] Therefore: K = KspKw/ Ka2 = 3.0 x10-28 x 1.0 x10-14/ 1.3x10-13

K = 2.3 x10-29