JOHN J. ALEXANDER University of Cincinnati Cincinnati, 45221
Stoichiometric Calculations Gordon A. P a r k e r University of Toledo Toledo, Ohio 43606
= 0.360 mg 0% I-'
dissolved oxygen
General Chemistry-Pressure Measurement
This question, designed for a sophomore or junior level quantitative analysis course, illustrates calculations based upon oxidation-reduction reactions as applied to a common environmental problem and requires cognitive skill a t the application level. Question Determination of elemental oxygen in natural waters has taken on increased importance with the emphasis on pollution studies. Dissolved oxygen (D.O.) content is a criterion for ascertaining the ability of a stream to support life. The Wiukler test for oxygen (Winkler, L. W., Chem. Ber., 21,2843 (1888)) is the recognized procedure for D.O. determination. The Winkler method uses an alkaline iodide solution added to an aqueous sample containing dissolved oxygen. A manganese(I1) solution is then added and the freshly precipitated manganese(I1) hydroxide formed reacts with the oxygen present. T h e manganese(I1) is oxidized to a higher valence state. Upon acidification the oxidized manganese reacts with the iodide present oxidizing it to free iodine. Iodine, equivalent to the amount of oxygen present, is then determined by titration with standardized sodium thiosulfate. 0 2 + 4 M n 2 + + 2 H 2 0 = 4 M n 3 ++4OH2 Mn3+ 2 I- = 2 Mn2++It
+
I2 + 2 s20a2= 2 I-
+ sp0ez-
60.0 ml of lake water was made basic, excess iodide added followed by Mn(I1) solution. All additions occurred in an air tight container. 27.4 ml of 1.00 X M NazS208 was required after acidification to titrate the liberated iodine to the starch endpoint. A correction of 0.005 mg oxygen per liter is subtracted from the experimental value to correct for dissolved oxygen in the added reagents. Calculate the mg of oxygen per liter present in the lake water. Answer mmales S203- = (27.4 ml) X (1.M)X 10-'mmoles ml-') = 2.74 X low3mmoles Sz08 1 mmole b
- 2 mmoles 5 ~ 0 3 ~ -
Thus 1.37 X
Applying the correction 0.365 mg 0 2 I-' - O.M)5mg Fz 1-' in the lake water
mmoles I2were present 11~-2~n~+-%o~
M a r g a r e t J. Steffel Ohio State University Marion Campus Marion, Ohio 43302 Understandine of (1)the o ~ e r a t i o nof barometers and (2) vapor pressure a;e te&d hy ;his freshman level question:~t reauires a ~-~ l i c a t i of o nknowledge about v a.w r Dressure to the . experimental task of measuring pressure with a barometer.
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Question In theory, any liquid can he used in a barometer. However, it is practical t o use a liquid only if it has a high density and a low vaDor oressure. At 24%. the densitv of mercurv is 14 iorr, At 2 4 " ~ the , g/ml a n i itsLvaporpressure is 2 X density of chloroform is 1.4 g/ml and its vapor pressure is 180 torr. (a) What is the height of a mercury column that can he supported by standard atmospheric pressure? tb) Explam what effect mercury v n p r in qudihr~umwith llquid will haw on the hereht of the liquid column in a mercury harameter. (c) Determine the height of a chloroform column that can he supported by standard atmospheric pressure. (d) Explain what effect chloroform vapor in equilibrium with liquid would have on the height of the chloroform column. (e) Determine the actual height of the liquid column in a chloroform barometer at 24'C by taking into account your answer to (dl. Answer 760 mm The vaoor oressure of m e r w is extremelv low at 24T. The verv slight pressure of mercury vapor above th; liquid column will ndt have a detectable effect on the height of the column. (c) Chloroform is 1/10 as dense as mercury. Consequently, the chloroform column that can be supported by standard atmospheric pressure is ten times as tall as the mercury column that can be supported by the same pressure: 7600 mm. (d) The chloroform vapor in the space above the liquid column will ~ u s downward b on the liauid decreasina the heieht of the column ihat would be supported by the atmospiere if t h k is no opposing pressure inside the barometer tube. (e) The vapor pressure of chloroform is 180 tom, or 180 mm of Hg. A pressure of 180 mm of Hg is the equivalent of 1800 mm of chloroform. Consequently, the chloroform vapor above the liquid column will depress the liquid column by 1800mm. Its height will be (7600 - 1800)or 5800 mm. (a) (b) . .
and mmoles 0 2 were in the sample 0.685 X mmoles 0 2 ) X (32.0 mg mmole-') (0.686 X mg 0 2 ml-' = 60.0 ml
102 1 Jowml of Chemical Education
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