INDUSTRIAL AND ENGINEERING CHEMISTRY
678
the phases. When 27 per cent of the entering air has been liquefied, for instance, one-half of the original oxygen is present in it, while the other half is in the unliquefied air; similarly the liquid resulting from a 29 per cent liquefaction contains one-half of the argon originally present in the air. Table 111-Partition of Oxygen a n d Argon between Liquid a n d Gaseous Phases in Fractional Liquefaction of Air PERCENT OZIN ENTERING PERCENT ARGONIN ENTBRAIR ING AIR Per cent air UnliqueConUnliqueConcondensed fied densed Total fied densed Total
10.8 12.7 21.0 26.4 37.0 41.5 55.4 61.8 71.4 80.3
16.3 15.0 11.8 10.5
10.2
20.9 20.4 20.7 20.7
6.8 4.3 3.9 2.4 1.7
14.0 16.5 16.7 18.6 (19.5)
20.8 20.8 20.6 21.0 21.2
...
‘s” :); ...
...
...
... ... *..
(0:69)
0.26
0.95
to: 42) (0.39) (0 35 (0:33{ 0.33 (0.36)
0.56 0.62 0.61 0.60 0.59
0.95 0.95 0.97 0.94 0.93 0.95
...
... ... 0.53
Note that the values add up to the normal content to an extent within the experimental error.
The results of Table I1 represent, then, the equilibria a t various temperatures between gaseous and liquid phases, starting from a mixture of constant composition-namely, that of air-78.05 per cent nitrogen, 21.00 per cent oxygen,
Vol. 17, No. 7
and 0.95 per cent argon. Baly’s results show the equilibria resulting on starting with a mixture of variable composition, ranging from 100 per cent nitrogen to 100 per cent oxygen. The present experiments cover only the portion of this range in which the lower limit is given by the composition 21 per cent oxygen in the gaseous phase and 48 per cent oxygen in the liquid and in which the upper limit is the equilibrium between 7 per cent oxygen in the gaseous phase and 21 per cent oxygen in the liquid. The results agree with Baly’s curves for oxygen within this range, and show in addition the corresponding relations for argon. I n Table I1 the argon content is stated also in terms of its relation to the nitrogen and argon residue that would result on removing the oxygen. It is this figure that is of particular interest in the purification of argon mixtures and for oxygen can be removed quite readily, whereas nitrogen offers more difficulty. Evidently, the most favorable condition for obtaining argon from a mixture in nitrogen lies in condensing only a small fraction of the incoming air. To take an extreme case, when 4.5 per cent of the air is liquefied, the liquid contains 47 per cent oxygen and 2.35 per cent argon, which yields after removal of the oxygen 4.5 per cent argon in the nitrogen and argon mixture. This represents an amount equal to 10 per cent of the total argon present in the original air.
An Inexpensive Method for Determining Lead’p2 By Wilfred W. Scott UNIVERSITY OF SOUTHERN
CALIRORNIA, LOS ANOELES,CALIX?.
HE high cost of potassium iodide makes the chromateiodide method for the determination of lead very expensive when used with large classes in quantitative analysis. For this reason the author devised a method using Knop’s reaction with diphenylamine, ferrous sulfate, and chromate solution in connection with the isolation of lead as a chromate salt according to the well-known chromate-iodide method. I n place of potassium iodide being added to the lead chromate solution, the liberated chromic acid is titrated directly by means of ferrous sulfate in presence of diphenylamine. Chromium is reduced from hexavalent to trivalent form giving the ratio 3 Fe = Cr = Pb. The normal equivalent of lead, therefore, is one-third its atomic weight, as in case of the chromate iodide procedure. This method equals the iodide procedure in accuracy, a t about one-eighth the cost. A trial of the method in the hands of students showed that a large amount of hydrochloric acid interferes with the end point to the extent that, instead of producing a deep blue color, the green color deepens to a dark green end pqint, which does not give the contrast desired. Sodium chloride interferes slightly, but to a much less extent. It was thought that the difficulty might be due to the yellow color of iron but this proved to be not so. Later it was discovered that a large excess of hydrochloric acid was responsible and that its effect could be counteracted by addition of an acetate salt. Approximately 1 gram of ammonium acetate per cubic centimeter of free hydrochloric acid is required. In a volume of 200 cc., 50 cc. of free hydrochloric acid (sp. gr. 1.2) required 45 grams of ammonium acetate to give the desired blue end point. I n testing the tolerance of free hydrochloric acid it was found that up to 10 cc. caused no difficulty. In the usual chromate procedure this is more than is required 1 2
Received March 18, 1925. Work done at Colorado School of Mines, Golden, Colo.
in the solution of lead chromate by the hydrochloric acidsodium chloride solution. If a large amount of this solution is used, ammonium acetate must be added to counteract the free hydrochloric acid. The procedure was tested with pure metallic lead and with lead ores. Comparison was made with the chromateiodide method and with the permanganate-oxalate method. The following are typical of results obtained: SAMPLE Pure metallic lead Lead ore
Chromateiodide Per cent
Permanganateoxalate Per cent
Chromateferrous Per cent
99.80 76.58
7k: i 0
99.76 76.95 76.80
PREPARATION OF SAMPLE-The solution of the material, isolation of lead sulfate, and conversion to lead chromate are in accordance with the standard procedure used in the chromate-iodide method. Solution of the lead chromate is effected by adding 50 to 100 cc. of a mixture of hydrochloric acid and saturated salt solution. (1000 cc. saturated salt solution, 120 cc. water, and 100 cc. HC1, sp. gr. 1.20.) TITRATION-The solution is diluted with water to about 150 cc. Ten cubic centimeters of phosphoric-sulfuric acid solution (1:l) and four to six drops of diphenylamine indicator (1 gram salt in 100 cc. HzS04) are added. A 0.1 N solution of ferrous sulfate is now added in excess and the excess determined by titration with 0.1 N potassium dichromate, or potassium permanganate, until the green color changes to blue. If a precipitation of lead occurs the end point will be a navy blue instead of the deep violet blue. Should an excess of hydrochloric acid be present a dark green color will beobtained. Ammonium acetate added to this solution will cause a change from dark green t o blue. A 0.1 N solution is equivalent to approximately 0.0069 gram of lead per cubic centimeter of solution.
