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Questions [and] Answers. J. A. Campbell. J. Chem. Educ. , 1977, 54 (7), p 437. DOI: 10.1021/ed054p437. Publication Date: July 1977. Cite this:J. Chem...
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Questions 6309. Br- is toxic a t 10 mM whereas the normal [CI-] is 100 mM. T h e usual treatment for poisoning due t o bromide ion is a dosage of 5-10 g (twice normal intake) of sodium or ammonium chloride per day. Discuss the probable reason for both t h e toxicity, and the effectiveness of the antidote. Suggest a n analogous treatment for poisoning by thallous ion, TI+. 6310. Alcohol, CzHsOH, is rapidly absorbed from the stomach and is then metabolized by a mechanism zero order in alcohol. T h e rate varies with the individual, b u t averages about 10 ml of C2HSOH per hour. T h e net reaction is known t o he

globin containing 4 g of iron. How many grams of cyanide could be complexed if half t h e hemoglobin were oxidized? How much sodium nitrite would be required? The net reaction is known t o be

A.

H3CCHzOH

+ NADC

-

NADH

+ HBCCHO+ H+

T h e catalyzing enzyme is alcohol dehydrogenase. Suggest some interpretations of the zero order in C2H50H. 6311. T h e tolerable dose for cyanide ion increased three-fold bv intravenous administration of sodium thiosulfate, fi\r:-fol(.l by sodium nitrite, and 18-fold hy a comtination of the two. T h e mechanism involves the oxidation of hemoglobin to methemoglot~inin the presence of nitrite. This frees the Iron in the heme so it coordinares the cvani(le. The thiosulfate then overcomes the normal body deficiency in sulfur and favors the CN- t o CNS- reaction catalyzed by sulfurtransferase. A normal adult has about 1000 g of blood hemo-

6312 Anesthetic gases, X , dissolve from t h e air, a, into the blood, b, according t o Henry's Law: S = [Xb]/[XJ. Assuming two gases were effective anesthetics a t the same [XJ, which would be a more rapid anesthetic, t h e one with the larger or with t h e smaller solubility. 6313. Most insect pheromones have 8-18 carbon atoms and a molecular weight of 100-300 so a s t o be volatile yet unique. Some 100 are now known b u t many more must exist if an appreciable fraction of the lo6 insect species use them for communication a s seems likelv. Some oheromones are known to be effective at a received rate of about 1moleculelsec cm3 of air. Make some reasonable assumntions and calculate the rate a t which a female might emit a sex pheromone in order t o attract males from a s far a s 2 km. 6314. If a toothpick is laid across a n ice cube and table salt is then shaken onto the cube and pick, the cube freezes to the toothpick and can be lifted by it. Why?

Answers A309. CI- and Brr have similar chemistries,but the larger Bralmost certainly cannot cross cell membranes at the same rate as CI- so physiological imbalances result. The slow onset of Brr toxicity is consistent with this possibility. Raising the general halide level leads to increased excretion of halides, primarily in the urine. Doubling the rate of halide excretion doubles the rate of bromide excretion and more quickly drops the dose below the toxic level. Administration of KC1 is the standard antidote to thallous poisoning. Do you see why? A310. Since the rate is zero order in C2HsOH it must be determined either by a limiting amount of NADt or of alcohol dehydrogenase, or both. Thus the rate-determining step could be between the enzyme-alcohol complex and the NADt and could be zero order in CzH50Heither because the enzyme or the NAD+ was the limiting reactant. The Kd,, of the enzyme-alcoholcomplex is known to be 2 X 10W Ethanol is intoxicating at 2-8 X 10+M so the ratio of enzyme to enzyme-alcohol complex varies from 1 to 0.2. Thus, the enzyme is not the limiting reagent, but the NADf is. As a matter of fact, increasing [NADt], ashy adding glucose, increases the rate. [See Westerfeld, Arner. J. Clin. Nutr., 9 426 (1961)l. A31 1. Assume a 1 to 1 Fe-CN complex. This will contain 56 g Fe per 26 g CN or, far 4g Fe half liberated, about 1 g of CN can be complexed. This is well above the lethal dose for an adult. 2 g of free Fe will require the oxidation of 2/56 or about 0.04 moles of Fe, hence 0.04 moles of hemoglobin requiring 0.02 moles of NO2from the net equation. 0.02 moles Na NO2 is 0.02 X (23 14 32) = 1.4 e of NaNO?. This could be dissolved in.. sav. .,100 cm30f H.0 to eive a 0.2 M solution. Formation of the Fe-CN comnlex wmld

+ +

-

~

so the antidote needs to be administered quickly and skillfully.

A312. The more soluble gas will require a higher blood eoncentration to reach its effective concentration. Thus, more of it must cross from the air into the blood. The rate of diffusion is dependent primarily on the partial pressure of the gas in the lungs which we shall assume is constant for all anesthetics. If so, the more soluble the anesthetic is in the blood the longer it will take for it to reach its effective concentration in the blood and the slower the anesthetic action. A313. Assume air movement of 4 kmhr and an angular conical dispersion of loometers at 2 km distance. The pheromone in the lower half of the cone is assumed adsorbed on vegetation. Then at 2 km there is an effective circular front of ndV4 10' m2 = 10s cm2,each cubic centimeter of which contains 1 molecule of pheromone. The front moves at a velocity of 4kmhr = 4 X 105crnjhr = 102cm/sec.In effect,108cm2X 102cm/sec= 1O10em3Isecmave past the horizontal plane 2 km from the source and carry 10'0 moleculeslsec of pheromone. Thus the female must exude 10'0 molecules per second as a lower limit. Since the wind velocity is 4 kmlhr she must continue for at least a couple of hours in order toallow time for the pheromone to reach the male (minimum 0.5 hr) and for him to react and reach her (perhaps an hour of upwind flying). Thus, she must release at least loL0(molecules/sec)x 2 hr X 3600 (seelhr) = 1014molecules. If the molecular weight is 200, she must emit 10" X 20016 X 1Ui3 = 3 X 1 0 F gor 30 ngof pheromone over a two hour period. Note this is probably a lower limit at this range. A314. The added salt falls on the ice around the edge of the toothpick, but not on the ice under the toothpick. As the ice melts where salt has fallen, heat of fusion is required and the local temperature falls. This lowers the temperature of the ice, and its water film, under the toothpick. The water film freezes "gluing" ice and toothpick together.

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Volume 54, Number 7, July 1977 / 437