T H E J O U R N A L O F I N D U S T R I A L A-VD E N G I N E E R I N G C H E M I S T R Y
Oct., 1916
is not necessary t h a t an exact amount of MgO be added and for most determinations 0.5 of a gram is entirely sufficient. Large quantities, as I O g., which are often used in distillation over MgO, are not necessary and should be avoided. T h e results obtained from a number of determinations are reported In Table 11. For quantities of TABLE11-MILLIGRAMS AMMONIA NITROGEXFOUND,A N D ERRORS
----
T A K E N ,23.92
.
TAKEX,5 045 _*--
Found 23.91 23.88 23.88 21.91 23.90 23.93 23.90 23,89 23,90 23.95 Deviation.. , Probable error
Error -0.01 -0.04 -0.04 -0.01 -0.02 $0.01 -0.02 -0.03 -0,02 -0.03 -0,013 *0.018
Found
Error -n ,044 -0,035 +0.003 -0.010 +0.035 $0.009 -0.000 $0.009
-0,044 -0.030 -0.01 17 *0.0195
1
I n soil, 12.84 Added, 5.040 17’880 Found Error 17.88 0.00 17.89 $0.01 17.85 -0.03 17.89 $0.01 17.85 -0.03 17.85 -0.03 17.91 f0.03 17.92 +0.04 17.84 -0.04 17.92 $0.04 0.000 kO.02
j mg. or less, N / j o HzS04 was used and for quantities above 5 mg., #/IO H 2 S 0 4 was used. T h e results need no discussion and t h e y show t h e accuracy of t h e method. T h e ammonia in 2 5 0 cc. of t h e soil extract was first determined and t h e n a known quantity of ammonia added and t h e total ammonia determined.
SUMMARY
I-Ammonia can be determined b y aeration over MgO. 11-A standard method of measuring t h e rate of aeration should be adopted. 111-The Folin tubes are of no value for aeration work when t h e rate is as rapid as was used in this work. IV-Complete absorption cannot be obtained unless t h e air is scrubbed well as it passes through t h e absorbing liquid. V-The absorbing tower as used in this work gives complete absorption. VI-One-half gram of MgO gives as satisfactory results as larger quantities. VII-Two a n d one-half hours is sufficient t o recover large quantities of ammonia from 2 j o cc. portions of solution when 1080 liters of air per hour are used. LABORATORY OF SOIL BIOLOGY
OHIOAGRICULTURAL EXPGRIMENT STATION WOOSTER,OH:O
NITRATES AND OXYGEN DEMAND By F. W. BRUCKMILLER Received June 9, 1916
T h e determination of oxygen demand in sewages b y t h e so-called saltpeter method depends upon t h e oxygen value ascribed t o t h e nitrogen in t h e nitrate radical. Two values have so far been proposed, one b y Ledereri and one b y Hale.2 T h e former assumes t h a t I nitrate nitrogen is equivalent t o 2 . j atoms of oxygen, while t h e latter assumes t h a t I nitrate nitrogen is equivalent t o 4 atoms of oxygen. T h e process of nitrate reduction is a biologic one, and in order for i t t o go on successfully three conditions must prevail: namely, presence of organic matter, nitrates and bacteria. The products of reduction depend upon t h e kind of bacteria present. T h e reJ. Infect. ET., 14, 488. *THISJOURNAL,7 (1915), 762. 1
899
duction may be such t h a t ( I ) t h e nitrates are reduced t o nitrites only; or ( 2 ) t h e oxygen is taken from t h e nitrates and nitrites for the formation of ammonia; or ( 3 ) t h e nitrates and nitrites are reduced with evolution of N O and N 2 0 ; or (4) the nitrates and nitrites are reduced t o gaseous nitrogen. T h e energy for carrying on t h e reduction is obtained from easily assimilated carbon compounds a n d t h e amount of reduction is directly proportional t o the organic matter present. A number of such compounds so readily utilized b y denitrifiers are produced b y t h e action of putrefactive organisms upon t h e complex insoluble compounds in animal and vegetable tissues, humus and undigested material in feces. Any equations, therefore, devised t o explain t h e reduction of nitrates must necessarily consider carbonaceous matter. Such equations, therefore, as the following’ hardly represent t h e true course of the reaction since they do not t a k e into consideration t h e reducing action of the carbon compounds: “03 HzO = XH3 202 (1) HzC03 2Hz0 = ( 2 ) 2KN03 KzC03 2”3 402 These equations might possibly represent t h e p a r t nitrates play in t h e electrolytic action in t h e corrosion of iron, b u t they hardly apply t o t h e biologic reduction of nitrates in sewages. If it is necessary t o represent t h e biologic reduction b y a chemical equation t h e following2 can be used t o picture t h e facts in t h e case, although t h e exact nature of t h e reaction in actual conditions might be altogether different. qKn’03 2Hz0 = 4 K H C 0 3 2x2 COz (3) j C 3COz (4) 4C 2KN03 3H20 = 2”s Analyzing this last equation b y splitting it into its supposed successive reactions as they might occur we get t h e following: C = 2KN02 COS (5) 2 K N 0 3 3C 3 H 2 0 = 2KH3 2COz ( 6 ) 2KNOz I n Equation 5, z nitrogens give u p 2 oxygens. I n Equation 6, 2 nitrogens give u p only 3 oxygens (one of t h e m staying with the potassium t o form the K z C O ~ ) , and 3 oxygens are taken from t h e water in order t o carry t h e reaction t o completion. I n t h e complete reduction, therefore, 8 oxygens are required; t h a t is, I nitrogen is equivalent t o 4 oxygens. T h e same conclusion can be obtained without the aid of equations and without recourse t o t h e idea t h a t oxygen must always be associated with oxidation and reduction. Oxidation can be defined as an increase in positive valence or a decrease in negative valence on t h e element oxidized; reduction as the c o n v e ~ s e . ~ I n going from NO3 t o N O 2 or N H 3 , nitrogen changes in valency. T h e nitrogen in nitrate has j positive valences; in ammonia 3 negative; in nitrite 4 positive; and in nitrogen gas no valence. The change from nitrate t o ammonia is 8, t o nitrite is 2 , and t o nitrogen 5, which being interpreted in terms of oxygen with 2
+
+
+
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+
+
+
+
Hale, THIS J O U R N A L , 7 (1915), 763 Marshall, “Microbiology,” 1912, 264. a Cady, “Inorganic Chemistry,” 1912, 255 1 2
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+
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+ +
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T H E J O U R N A L OF I N D U S T R I A L A N D ENGINEERING C H E M I S T R Y
900
valences means 4, I and zl/*; t h a t is, if the reduction all proceeded t o gaseous nitrogen t h e equivalents of oxygen available for each nitrogen in nitrate is ~ 1 , ’ ~ ; if only ammonia was formed, 4 equivalents would be available; a n d if only nitrites I equivalent. Assuming t h a t t h e reduction goes t o nitrites and ammonia, t h e oxygen available will be dependent upon t h e quantity of nitrites and of ammonia produced. If we s t a r t with one equivalent of nitrogen as nitrate, and assume varying percentages of NO2 formed we can get t h e oxygen value for t h e different percentage combinations of KO2 and NH3 present. TABLEI E q v t . N O ? f o r m e d. . . . . . . 0 . 0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1 . 0 E q v t . N H 3 f o r m e d ....... 1 . 0 0 . 9 0 . 8 0 . 7 0 . 6 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 Eqvt.Oavailable. . . . . . . 4 . 0 3 . 7 3.4 3 . 1 2 . 8 2 . 5 2 . 2 1.9 1.6 1 . 3 1 . 0
If 5 0 per cent of t h e nitrogen goes t o S H 3 ,t h e oxygen equivalent is 2 . j . Since bacteria in sewage vary, the percentages of reduction products obtained will not only vary with different sewages b u t will vary from d a y t o d a y with the same sewage. T h a t this is true is shown b y the experiments of various investigators. Letts, Blake, a n d Totton’ found t h a t nitrogen is evolved, sometimes free and sometimes as nitric oxide. Guth and Keim2 report t h e same products. Both of these investigators were unable t o demonstrate a n increase in free ammonia. Hoover3 reports a n increase of free ammonia and nitrites. Lederer* reports a n increase of nitrites and free ammonia, the latter varying from 2 0 t o 50 per cent. TABLE11-FORMATION OF FREEAMMOSIADURINGSALTPETER OXYGEN T e m p of Incubation 1 . . . . . . . . . . 370 c. 2 . . . . . . . . . . 37 3 . . . . . . . . . . 37 4 . . . . . . . . . . 37 5 . . . . . . . . . . 37 . . . . . . . . . . 37 I . ......... 20 8 . . . . . . . . . . 20 9 . . . . . . . . . . 20 10 . . . . . . . . . . 20 1 1 . . . . . . . . . . 20 12 . . . . . . . . . . 20
iY0.
6
COhSLMPTION ( a ) N as free Time of Incubation Before 7.2 3 8.8 3 12.4 3 10.8 3 14.0 3 12.0 3 7.2 5 8.8 5 5 12.4 10.8 5 14.0 5 12.0 5
NH?
P e r cent Increase 30 41 16 40 15 23 40 24 3 50 29 47
After 9.2 12.4 14.4 15.2 16.4 14.8 10.0 11.6 12.8 1.5.2 18.0 17.6
( a ) Lederer, A m . J. Pub. Heallh, 9, 358
Unfortunately, t h e quantity of nitrites and initial nitrates were not given, so t h e value of t h e nitrogen in terms of oxygen could not be calculated from these products. I n our laboratory, we have been unable t o demonstrate t h e presence of any free nitrogen or nitric oxide. Kitrites and ammonia were found t o be t h e only products of reduction. TABLE 111-REDUCTION PRODUCTS OF SALTPETER TREATMENT 1 2 3 4 5 6 Incubation T e m p . . . . . . . . . . . . . . . . . 37 37 37 37 37 20 Incubation Time . . . . . . . . . . . . . . . . 5 5 5 5 5 0 PERCENTINCREASE: Nitrogen . . . . . 0 0 0 0 0 0 Ammonia ..................... 40 51 47 46 52 49 Saarar. No .
7 8 9 10 20 20 20 20’ C. 10 10 10 10 0
0
0
0
45 42 51 47
2 3
4
Ckem. News, 88 (19031, 182. Gesundh. Ing., 36 (1912), 52. Engineeying A-ews, 68, 192, 452. A m e r . J . Pub. Health, 5, P. 357.
IO
enous organic compounds present in sewages, so t h a t any calculations based on t h e increase in free ammonia are not very dependable. If, however, we can get comparable results on t h e reduction of nitrates and t h e reduction of available oxygen in absence of nitrates, we will have dependable d a t a from which the oxygen equivalent of nitrogen in nitrates can be calculated. Such comparisons have been made by Hale a n d Melia,l b y LedererZand b y t h e ~ r i t e r . ~ Hale and Melia could get consistent results with t h e sewage on which they were working only b y assuming t h a t one nitrogen was equivalent t o 4 oxygens, basing their assumptions on Equations I and 2 above. From the considerations herewith presented, t h e nitrate must have been all reduced t o NH,,no nitrites being formed, in order for this value t o be true. Lederer, however, could get checking results with t h e dilution and nitrates method, by assuming t h a t I nitrate nitrogen was equivalent t o 2 . 5 oxygens. This is strictly in accord with his observations t h a t both nitrites and ammonia are formed in the nitrate reduction. He further assumed t h a t one nitrite nitrogen was equivalent t o 1 l I l 2 atoms of oxygens. According t o t h e consideration t h a t t h e nitrogen is the oxidizing agent, one nitrite nitrogen is equivalent t o one oxygen. Ridea14 in his text book allows 2 . 5 atoms of oxygen for one atom of nitrogen, and Hoover a t the Columbus Sewage Testing Station has employed the same value. T h e weight of evidence, therefore, seems t o be in favor of 2 . 5 atoms of oxygen for every nitrate nitrogen. T h e conclusion, however, can not be drawn t h a t this value is applicable t o all sewages. T h e most t h a t can be said is t h a t the percentages of reduction products in each of the sewages so far worked on were about t h e same, namely, about a 50 per cent increase in free NH3. Other investigators might find a different oxygen value for nitrogen. T h a t is, each sewage, depending upon the bacteria and organic matter present, will reduce nitrates differently. The safest thing, therefore, in oxygen demand work is t o determine experimentally t h e oxygen value for t h e nitrogen in nitrates for t h e sewage in question. This can be done best b y conducting very carefully parallel determinations, using nitrates in one and oxygen in t h e other, as t h e oxidizing agent. All t h e precautions mentioned b y Ledererj and others6 should be followed in running the “Dilution” method. From the data t h u s obtained, t h e oxygen equivalent of nitrogen in nitrate can be readily determined, and from the evidence so far presented t h e value obtained should not differ far from 2 . 5 . XTATER A N D S T W A G E LABORATORY UNIVERSITYO F KAKSAS, LAWRENCE
The increase in ammonia, however, may come from t h e hydrolysis of proteins and other complex nitrog1
Vol. 8, No.
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H(
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~763. ~
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THISJOLTRNAL, 8 (1916), 403.
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“Sewage,” 3rd E d . , 1906, 131.
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TIIISJOURNAL, 6 (1914), 882.
6
Bruckmiller, Ibid., 8 (1916), 403; 7 (1915), 762.
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