Soil-Plant Nutrient Cycling and Environmental Quality
Department of Plant and Soil Sciences
Raun, G.V. Johnson,
R.W. Mullen, K.W. Freeman,
and R.L. Westerman
044 N. Ag Hall
Tel: (405) 744-6418
FAX: (405) 744-5269
"In recent years the 'human rights' issue has generated much interest and debate around the world. It is a utopian issue and a noble goal to work toward. Nevertheless, in the real world, the attainment of human rights in the fullest sense can not be achieved so long as hundreds of millions of poverty stricken people lack the basic necessities for life. The right to dissent does not mean much to a person with an empty stomach, a shirtless back, a roofless dwelling, the frustrations and fear of unemployment and poverty, the lack of education and opportunity, and the pain, misery and loneliness of sickness without medical care. It is my belief that all who are born into the world have the moral right to the basic ingredients for a decent, human life."
Norman E. Borlaug
1970 Nobel Peace Prize
"Learning science and thinking about science or reading a paper is not about learning what a person did. You have to do that, but to really absorb it, you have to turn it around and cast it in a form as if you invented it yourself. You have to look and be able to see things that other people looked at and didn't see before. How do you do that? There's two ways. Either you make a new instrument, and it gives you better eyes, like Galileo's telescope. And that's a great way to do it, make such a nice instrument that you don't have to be so smart, you just look and there it is. Or you try to internalize it in such a way that it really becomes intuitive. Working on the right problem is only part of what it takes to succeed. Perseverance is another essential ingredient."
1997 Nobel Prize, Physics
immediate apprehension or cognition; without evident rational thought and
inference; quick and ready insight
Soil-Plant Nutrient Cycling and Environmental Quality
Students, 1992- 2002
Spring 1992 Spring 1994 Spring 1996 Spring 1998 Spring 2000
Mohd Akbar Jeri L. Anderson Justin Carpenter Erna Lukina Elbek Arslanov
Altom Jeffrey B. Ball
Edgar N. Ascencio Andrew C. Bennett Mark Everett Curt Woolfolk Danielle Bradford
Senayet Assefa Jing Chen Mike Goedeken Lori Gallimore Kyle Freeman
Randy K. Boman Francisco Gavi-Reyes Eric Hanke Doug Cossey Jon Karl Fuhrman
T. Ramanarayanan David L. Gay Dale Keahey Bryan Howell Prajakta Ghatpande
C. W. Richardson James P. Johnson Butch Koemel Rick Kochenower Jay Ladd
Hasil Sembiring Tracy D. Johnston Heather Lees Renee Albers Jennifer Lepper
Sonia Morales Fred Kanampiu Alan O'Dell Matthew Barnes Rachelle Moussavou
Xin Li John Ringer Clydette Borthick
Steven Phillips Jerry Speir Wade Thomason Susan Mullins
Shiferaw Gary Strickland
Shannon Taylor Jeremy Dennis Eric Palmer
Derrel White Elena Jigoulina Heather Qualls
Mark Wood Aleksandr Felitsiant Chris Stiegler
Jason Yoder Michelle Franetovich Clemn Turner
Todd Heap Jason Warren
Tyson Ochsner Damon Wright
Steven McGowen Kathie Wynn
TABLE of CONTENTS
1. Organic Matter 4
Nutrient Supplying Power of Soil 4
Composition of Organic Matter 4
C:N Ratios as Related to Organic Matter Decomposition 4
Decomposition of Organic Matter (Mineralization) 4
2. Essential Elements 4
Arnon's Criteria of Essentiality 4
3. The Nitrogen Cycle 4
Inorganic Nitrogen Buffering 4
Ammonia Volatilization 4
Chemical Equilibria 4
Urea Hydrolysis 4
H ion buffering capacity of the soil: 4
Factors Affecting Soil Acidity 4
Acidification from N Fertilizers (R.L. Westerman) 4
4. Nitrogen Use Efficiency 4
N Discussion 4
5. Use of Stable and Radioactive Isotopes 4
Sources of Radiation 4
Agronomic Applications 4
6. Exchange 4
Cation Exchange Capacity (CEC): 4
Effective CEC 4
CEC Problems 4
Base Saturation 4
Anion Exchange (Kamprath) 4
7. Phosphorus Fertilizers 4
Rock Phosphate 4
Calcium Orthophosphates 4
8. Theoretical Applications in Soil Fertility 4
Liebig's law of the minimum (Justus von Liebig 1803-1873) 4
Bray Nutrient Mobility Concept 4
Sufficiency: SLAN (Sufficiency Levels of Available Nutrients) 4
Plant Response to Soil Fertility as Described by the Percent Sufficiency and the Mobility Concept 4
Mitscherlich (applicability of this growth function to soil test correlation studies) 4
Bray Modified Mitscherlich 4
Fried and Dean (1951) 4
Base Cation Saturation Ratio 4
9. Soil Testing / Critical Level Determination 4
Economic and Agronomic Impacts of Varied Philosophies of Soil Testing (Olson et al., 1982) 4
Cate and Nelson (1965) 4
Use of Price Ratios 4
Soil Testing for Different Nutrients 4
Dry Combustion (Dumas 1831) 4
Rittenberg Method (N2 gas from sample) 4
Inorganic Nitrogen 4
Phosphorus Soil Index Procedures 4
Total P ? 4
Nutrient Interactions 4
Soil Testing versus Non-destructive Sensor Based VRT 4
Experimental Design/Soil Testing and Field Variability 4
10. Micronutrients 4
11. Special Topics 4
Method of Placement 4
Saline/Sodic Soils 4
Stability Analysis 4
Stability Analysis: discussion 4
Soil Solution Equilibria 4
Some Rules of Thumb for Predicting the Outcome of Simple Inorganic Chemical Reactions Related to Soil Fertility 4
12. NUTRIENT CYCLES 4
13. Example Exams 4
14. STATISTICAL APPLICATIONS 4
Surface Response Model 4
Procedure for Determining Differences in Population Means 4
Randomized Complete Block Randomization 4
Program to output Transposed Data 4
Contrast Program for Unequal Spacing 4
Test of Differences in Slope and Intercept Components from Two Independent Regressions 4
Linear-Plateau Program 4
Linear-Linear Program 4
In the past 150 years, CO2 levels in the atmosphere have increased from 260 to 365 ppm (Follett and McConkey, 2000) and it is expected to rise 1.5 to 2.0 ppm per year (Wittwer, 1985). This increase is believed to have increased the average temperature of the earth by 0.5 °C and thus various reports of global warming as a result of increased evolution of CO2 into earth's atmosphere (Perry, 1983). It is possible to decrease the release of CO2 to the atmosphere by choosing an alternative energy source. However, total control of the release of CO2 is not easy because there are so many different sources, including the production of cement, gasoline-driven automobiles, burning of fuels for home heating, cooking, etc. (Wallace et al., 1990). There are, however, several benefits associated with increased atmospheric CO2 including increased water use efficiency, nitrogen use efficiency and production in many crops.
If the expected fossil fuel CO2 released for many years could be stored as soil organic matter, vastly enhanced productive soil would result. This option requires increased biomass to produce the needed soil organic matter, but this could be achievable due to increased CO2 supplies in the atmosphere (Wallace et al., 1990). Obstacles to increasing the level of soil organic matter are; 1) needed organic matter supplies, 2) needed nitrogen to give around a 10:1 carbon:nitrogen ratio necessary for stable soil organic matter, and 3) efficiency in microbial activity that can result in more stable soil organic matter, instead of burn out resulting in return of CO2 to the atmosphere (Wallace et al., 1990).
It is seldom understood that organic matter contents in soils can be increased via various management practices. Increased use of no-till management practices can increase soil organic matter. After ten years of no-tillage with corn, soil organic carbon in the surface 30 cm was increased by 0.25% (Blevins et al. 1983). Probably the least understood is increased N rates in continuous crop production on resultant soil organic matter levels. Various authors have documented that N rates in excess of that required for maximum yields result in increased biomass production (decreased harvest index values e.g., unit grain produced per unit dry matter). This results in increased amounts of carbon from corn stalks, wheat stems, etc., that are incorporated back into soil organic matter pools. Although this effect is well documented, the deleterious effects of increased fertilizer N rates on potential NO3 leaching and/or NO3 surface runoff should be considered where appropriate. Use of green manures and animal wastes have obvious impacts on soil organic matter when used on a frequent basis.
native fertility of forest and grassland soils in
Figure 1.1. Influence of cultivation time on relative mineralization from soil humus and wheat residue. (From Campbell et al. (1976)).
the reddish prairie soils of
N removal in the Check (no fertilization) plot of the Magruder Plots
20 bu/acre * 60 lb/bu * 100 years = 120000lbs
120000 lbs * 2%N in the grain = 2400 lbs N/acre over 100 years
8000 lbs N in the soil (1892)
2000 lbs N in the soil (1992)
2400 lbs N removed in the grain
=3600 lbs N unaccounted
The effects that management systems will have on soil organic matter and the resultant nutrient supplying power of the organic pools are well known. Various management variables and their effect on soil organic matter are listed;
Organic Matter Management Effect
1) tillage +/- conventional -
2) soil drainage +/-
3) crop residue placement +/-
4) burning -
5) use of green manures +
6) animal wastes and composts +
7) nutrient management +/- excess N +
The living component which includes soil microorganisms and fauna make up a relatively small portion of total soil organic matter (1-8%). It functions however as an important catalyst for transformations of N and other nutrients (Doran and Smith, 1987). The majority of soil organic matter is contained in the nonliving component that includes plant, animal and microbial debris and soil humus.
Common components of soil organic matter and their relative rates of decay are listed in Table 1.1. Cellulose generally accounts for the largest proportion of fresh organic material. It generally decays rather rapidly, however, the presence of N is needed in order for this to take place. Lignin components decompose much more slowly and thus, any nutrients bound in lignin forms will not become available for plant growth. Although lignin is insoluble in hot water and neutral organic solvents, it can be solubilized in alkali solutions. Because of this, we seldom find calcareous soils with extremely high organic matter. All of the polysaccharides decompose rapidly in soils and thus serve as an immediate source of C for microorganisms. Decomposition of these respective components is illustrated in Figure 1.2.
Table 1.1. Components of soil organic matter, rate of decomposition and composition of each fraction.
Form Formula Decomposition Composition
Cellulose (C6H10O5)n rapid * 15-50%
glucose C6H12O6 moderate-slow
xylose C5H10O5 moderate-slow
Lignin(phenyl-propane) slow 15-35%
Crude Protein RCHNH2COOH** rapid 1-10%
Chitin (C6H9O4.NHCOCH3)n rapid
Starch glucose chain rapid
Pectins galacturonic acid rapid
Inulin fructose units
* - decomposition more rapid in the presence of N
** - amino acid glycine (one of many building blocks for proteins)
Figure 1.2. Decomposition of Miscanthus sinensis leaf litter.
Table 1.2. Composition of mature cornstalks (Zea mays L.) initially and after 205 days of incubation with a mixed soil microflora, in the presence and absence of added nutrients (Tenney and Waksman, 1929)
Initial Composition after 205 days (%)
composition No nutrients Nutrients
Constituents or fraction % added added
Ether and alcohol soluble 6 1 <1
Cold water soluble 11 3 4
Hot water soluble 4 4 5
Hemicelluloses 18 15 11
Cellulose 30 13 6
Lignins 11 23 24
Crude protein 2 9 11
Ash 7 19 26
The composition of mature cornstalks before and after 205 days of incubation with a mixed soil in the presence and absence of added nutrients is listed in Table 1.2. As decomposition proceeds, the water soluble fraction (sugars, starch, organic acids, pectins and tannins and array of nitrogen compounds) is readily utilized by the microflora (Parr and Papendick, 1978). Ether and alcohol-soluble fractions (fats, waxes, resins, oils), hemicelluloses and cellulose decrease with time as they are utilized as carbon and energy sources. Lignin, tends to persist and accumulate in the decaying biomass because of its resistance to microbial decomposition. Decomposition rates of crop residues are often proportional to their lignin content and some researchers have suggested that the lignin content may be a more reliable parameter for predicting residue decomposition rates than the C:N ratio (Alexander, 1977). Vigil and Kissel (1991) included the lignin-to-N ratio and total soil N concentration (in g/kg) as independent variables to predict potential N mineralization in soil. They also noted that the break point between net N mineralization and net immobilization was calculated to be at a C/N ratio of 40.
A simple illustration of the carbon cycle is found in Figure 1.3. The carbon cycle revolves around CO2, its fixation and regeneration. Chlorophyll-containing plants utilize the gas as their sole carbon source and the carbonaceous matter synthesized serves to supply the animal world with preformed organic carbon. Upon the death of the plant or animal, microbial metabolism assumes the dominant role in the cyclic sequence (Alexander, 1977). Without the microbial pool, more carbon would be fixed than is released, CO2 concentrations in the atmosphere would decrease and photosynthesis rates would decrease.
Figure 1.3. Simple illustration of the carbon cycle (from Alexander, 1977). "Higher plants use light to convert water (H2O) and carbon dioxide (CO2) to glucose (C6H12O6) and oxygen (O2)."
In general, the following C:N ratios are considered to be a general rule of thumb in terms of what is expected for immobilization and mineralization.
C:N Ratio Effect
20-30:1 immobilization = mineralization
Unfortunately, C:N ratios say nothing about the availability of carbon or nitrogen to microorganisms. The reason for this is because we are not aware of what makes up the carbon (C) component. In tropical soils, significantly higher proportions of lignin will be present in the organic matter. Even though the percent N within the organic matter may be the same, it would be present in highly stable forms that were resistant to decomposition. Therefore, mineralization rates in organic matter that contain high proportions of lignin will be much smaller. The C:N ratios discussed were generally developed from data obtained in temperate climates. Therefore their applicability to tropical soils is at best minimal.
1. percent organic matter
2. organic matter composition
3. cultivation (crop, tillage, burning)
4. climate (moisture, temperature)
5. soil pH
6. N management (fertilization)
7. soil aeration
During the initial stages of decomposition of fresh organic material there is a rapid increase in the number of heterotrophic organisms accompanied by the evolution of large amounts of carbon dioxide. If the C:N ratio of the fresh material is wide, there will be a net N immobilization. As decay proceeds, the C:N ratio narrows and the energy supply of carbon diminishes.
The addition of materials that contain more than 1.5 to 1.7% N would ordinarily need no supplemental fertilizer N or soil N to meet the demands of the microorganisms during decomposition (Parr and Papendick, 1978). Note that the 'demands of the microorganisms' is what is discussed first, with no regard as to what plant N needs might be. The addition of large amounts of oxidizable carbon from residues with less than 1.5% N creates a microbiological demand for N which can immobilize residue N and available inorganic soil N for extended periods. The addition of supplemental inorganic fertilizer N to low N residues can accelerate their rate of decomposition (Parr and Papendick, 1978).
the thousands of years prior to the time cultivation was initiated, C and N had
built up in native prairie soils.
However, the C:N ratio was wide, reflecting conditions for
immobilization of N. The combined influence
of tillage and the application of additional organic materials (easily
decomposable wheat straw and/or corn stalks) is illustrated in Figure 1.4. Cultivation alone unleashed a radical
decomposition of the 4% organic matter in
Figure 1.4. Effect of cultivation and addition of straw materials on immobilization and mineralization of N and associated evolution of CO2
Figure 1.6. Changes in soil mineral N as a function of time, and addition of manure and straw.
Table 1.2. Example calculations of total N in organic matter fractions in soils and expected amounts of N mineralized on a yearly basis.
min max min max
Organic Matter, % 1 2 4 12
1 ha (0-15cm), kg 2241653 2241653 2241653 2241653 (Pb = 1.47)
Organic, matter, kg 22416 44833 89666 268998
kg N in
% N mineralized/yr 0.03 0.03 0.03 0.03
TOTAL (kg N/ha/yr) 33.6 67.2 134.4 ? 403.5 ?
DB= Mass of dry soil/volume of solids and voids
ft3*0.02832 = m3
1 ha = 2.471ac
1 ha = 10000m2
1 ac = 4047m2
2000000 lb = 907184.74 kg = 907.184 Mg
43560 ft2 * 0.5 ft = 21780 ft3 = 616.80m3
907.184Mg/616.80m3 = DB 1.4707
10000m2 * 0.15m = 1500 m3
2241653 kg /1000 = 2241.6 Mg
2241.6/1500 = DB 1.49 (g/cm3 = Mg/m3)
What will happen if
a) bulk density is changed?
b) % N in organic matter?
c) % N mineralized per year?
Organic Matter = 0.35 + 1.80 * (organic carbon) Ranney (1969)
The most important function of the microbial flora is usually considered to be the breakdown of organic materials, a process by which the limited supply of CO2 available for photosynthesis is replenished (Alexander, 1977).
Five major groups of microorganisms in the soil are:
Soil Bacteria: 108 to 1010 / g of soil
Heterotroph: (chemoorganotrophic) require preformed organic nutrients to serve as sources of energy and carbon.
3. Most Bacteria
Autotroph: (lithotrophic) obtain their energy from sunlight or by the oxidation of inorganic compounds and their carbon by the assimilation of CO2.
Photoautotroph: energy derived from sunlight
1. Algae (blue-green, cyanobacteria)
2. Higher Plants
3. Some Bacteria
Chemoautotroph: energy for growth obtained by the oxidation of inorganic materials.
1. Few Bacterial species (agronomic importance)
a. nitrobacter, nitrosomonas and thiobacillus
1. Element required to complete life cycle.
2. Deficiency can only be corrected by the ion in question.
3. Element needs to be directly involved in the nutrition of the plant and not indirectly via the need of another organism.
Any mineral element that functions in plant metabolism, whether or not its action is specific (Tisdale et al., 1985).
C, H, O, N, P, and S (constituent of proteins)
Ca, Mg, K, Fe, Mn, Mo, Cu, B, Zn, Cl, Na, Co, V, Si (essential to one or more plants)
'CHOPKNS CaFe MgB Mn Cl CuZn Mo'
1. deficiency symptoms appear in the lower older leaves
1. can be taken up from a large volume of soil
1. deficiency symptoms appear in the upper younger leaves
1. taken up from a small volume of soil
Deficiency Symptom Element Mobility Mobility Form taken up
Soil Plant by Plants
overall chlorosis seen N Nitrogen Yes Yes NO3-,NO2-,NH4+
first on lower leaves
purple leaf margins P Phosphorus No Yes HPO4=,H2PO4-,H3PO4
chlorotic leaf margins K Potassium No Yes K+
uniform chlorosis, stunting
(younger leaves) S Sulfur Yes Yes(no) SO4=,SO2*
stunting - no root
elongation Ca Calcium No No Ca++
veins remain green Fe Iron No (ls) No Fe+++,Fe++
interveinal chlorosis Mg Magnesium No (ls) Yes/No Mg++
growth = chlorotic tips B Boron (NM) Yes No H3BO3°
interveinal chlorosis Mn Manganese No No Mn++, Mn+++
wilting, chlorosis, reduced
root growth Cl Chlorine Yes Yes Cl -
young leaves, yellow &
stunted Cu Copper No (ls) No Cu++
interveinal chlorosis in
young leaves Zn Zinc No (ls) No Zn++
stunting Mo Molybdenum Yes/No(ls) No MoO4=
dark green color Na Sodium No(ls) Yes Na+
C Carbon CO2
H Hydrogen H2O
O Oxygen H2O
* absorbed through plant leaves
(NM) Non Metal
(ls) Low Solubility
Mo availability increases with soil pH, other micronutrients show the opposite of this.
Immobile nutrients in plant; symptoms of deficiency show up in the younger leaves.
Stage of growth when deficiency symptom is apparent = later stage
· Key building block of the protein molecule upon which all life is based
· Indispensable component of the protoplasm of plants animals and microorganisms
· One of the few soil nutrients lost by volatilization and leaching, thus requiring continued conservation and maintenance
· Most frequently deficient nutrient in crop production
Nitrogen Ion/Molecule Oxidation States
Nitrogen ions and molecules that are of interest in soil fertility and plant nutrition cover a range of N apparent oxidation states from -3 to +5. It is most convenient to illustrate these oxidation states using common combinations of N with H and O, because H can be assumed in the +1 oxidation state (H+1) and O in the -2 oxidation state (O=). The apparent N oxidation state, and the electron configurations involved may be depicted as follows.
The electron configuration in the ground state is 1s1 (the first electron shell has only one electron in it), as found in H2 gas. Since the s shell can hold only two electrons, the atom would be most stable by either gaining another electron or losing the existing one. Gaining an electron by sharing occurs in H2, where each H atom gains an electron from the other resulting in a pair of electrons being shared. The electron configuration about the atom, where: represent a pair of electrons, may be shown as
H:H and the bond may be shown as H-H
Hydrogen most commonly exists in ionic form and in combination with other elements where it has lost its single electron. Thus it is present as the H+ ion or brings a + charge to the molecule formed by combining with other elements.
The ground state of O, having a total of eight electrons is 1s2, 2s2, 2p4. Both s orbitals are filled, each with two electrons. The 2p outer or valence orbital capable of holding six electrons, has only four electrons, leaving opportunity to gain two. The common gain of two electrons from some other element results in a valence of -2 for O (O=). The gain of two electrons also occurs in O2 gas, where two pairs of electrons are shared as
O::O and the double bond may be shown as O=O
The ground state of N is 1s2, 2s2, 2p3. It is very similar to that for oxygen, except there is one less electron in the valence 2p orbital. Hence, the 2p orbital contains three electrons but, has room to accept three electrons to fill the shell. Under normal conditions, electron loss to form N+, N2+ or N3+ or electron gain to form N-, N2-, or N3- should not be expected. Instead, N will normally fill its 2p orbital by sharing electrons with other elements to which it is chemically (covalent) bound. Nitrogen can fill the 2p orbital by forming three covalent bonds with itself as in the very stable gas N2.
The Nitrogen cycle is not well understood, largely because of how it is communicated. Similar to the way we communicate the differences between normal, saline, sodic and saline-sodic soils, we should do the same for response variables in the Nitrogen cycle. In addition to temperature and pH included below, we could add reduction/oxidation, tillage (zero vs. conventional), C:N ratios, fertilizer source and a number of other variables. These mechanistic models would ultimately lead to many 'if-then' statements/decisions that could be used within a management strategy.
1. Aerated environment (need for O2)
2. Supply of ammonium
4. Temperature (30-35C or 86-95F) <10C or 50F
5. Soil pH
6. Addition of low C:N ratio materials (low lignin)
Is oxygen required for nitrification?
Does nitrification proceed during the growing cycle? (low C:N ratio)
N Oxidation States:
oxidized: loses electrons, takes on a positive charge
reduced: gains electrons, takes on a negative charge
NH3 ammonia -3
NH4+ ammonium -3
N2 diatomic N 0
N2O nitrous oxide +1
NO nitric oxide +2
H2S hydrogen sulfide -2
SO4= sulfate +6
N: 5 electrons in the outer shell
· loses 5 electrons (+5 oxidation state NO3)
· gains 3 electrons (-3 oxidation state NH3)
O: 6 electrons in the outer shell
· is always being reduced (gains 2 electrons to fill the outer shell)
H: 1 electron in the outer shell
N is losing electrons to O because O is more electronegative
N gains electrons from H because H wants to give up electrons
1. Yield goal (2lb N/bu)
a. Applies fertilization risk on the farmer
b. Removes our inability to predict 'environment' (rainfall)
2. Soil test
a. For every 1 ppm NO3, N recommendation reduced by 2lbN/ac
1. high pH
2. high NH4 levels (NH4 inhibits nitrobacter)
Inorganic nitrogen buffering is defined as the ability of the soil-plant system to control the amount of inorganic N accumulation in the rooting profile when N fertilization rates exceed that required for maximum yield.
If N rates required to detect soil profile NO3 accumulation always exceeded that required for maximum yields, what biological mechanisms are present that cause excess N applied to be lost via other pathways prior to leaching?
Nitrogen Buffering Mechanisms
1. Increased Applied N results in increased plant N loss (NH3)
2. Higher rates of applied N - increased volatilization losses
3. Higher rates of applied N - increased denitrification
4. Higher rates of applied N - increased organic C, -- increased N in organic pools
5. Increased applied N - increased grain protein
6. Increased applied N - increased forage N
7. Increased applied N - increased straw N
· Urease activity · Air Exchange
· Temperature · N Source and Rate
· CEC (less when high) · Application method
· H buffering capacity of the soil · Crop Residues
· Soil Water Content
NH4+ « NH3 + H+
If pH and temperature can be kept low, little potential exists for NH3 volatilization. At pH 7.5, less than 7% of the ammoniacal N is actually in the form of NH3 over the range of temperatures likely for field conditions.
A+B « AB
Kf = AB/A x B
AB « A+B
Kd = A x B/AB
Kf = 1/Kd (relationship between formation and dissociation constants)
Formation constant (Log K°) relating two species is numerically equal to the pH at which the reacting species have equal activities (dilute solutions).
pKa and Log K° are sometimes synonymous
pH = pKa + log [(base)/(acid)]
when (base) = (acid), pH = pKa
· Urea is the most important solid fertilizer in the world today.
· In the early 1960's, ammonium sulfate was the primary N product in world trade (Bock and Kissel, 1988).
The majority of all urea production in the
· Since 1968, direct application of anhydrous ammonia has ranged from 37 to 40% of total N use (Bock and Kissel, 1988).
· Urea: high analysis, safety, economy of production, transport and distribution make it a leader in world N trade.
· In 1978, developed countries accounted for 44% of the world N market (Bock and Kissel, 1988).
· By 1987, developed countries accounted for less than 33%.
Share of world N consumption by product group
Ammonium nitrate 27 15
Urea 9 37
Ammonium phosphates 1 5
Other N products (NH3) 36 29
Other complex N products 16 8
CO(NH2)2 + H+ + 2H2O --------> 2NH4+ + HCO3-
pH 6.5 to 8
HCO3- + H+ ---> CO2 + H2O (added H lost from soil solution)
CO(NH2)2 + 2H+ + 2H2O --------> 2NH4+ + H2CO3 (carbonic acid)
Potential for gaseous loss from applied urea, both broadcast and incorporated.
During hydrolysis, soil pH can increase to >7 because the reaction requires H+ from the soil system.
(How many moles of H+ are consumed for each mole of urea hydrolyzed?) 2
In alkaline soils less H+ is initially needed to drive urea hydrolysis on a soil already having low H+.
In an alkaline soil, removing more H+(from a soil solution already low in H+), can increase pH even higher
NH4+ + OH- ---> NH4OH ---->NH3 + H2O
pH = pKa + log [(base)/(acid)]
At a pH of 9.3 (pKa 9.3) 50% NH4 and 50% NH3
pH Base (NH3) Acid (NH4)
7.3 1 99
8.3 10 90
9.3 50 50
10.3 90 10
11.3 99 1
Equilibrium relationship for ammoniacal N and resultant amount of NH3 and NH4 as affected by pH for a dilute solution.
As the pH increases from urea hydrolysis, negative charges become available for NH4+ adsorption because of the release of H+ (Koelliker and Kissel)
Decrease NH3 loss with increasing CEC (Fenn and Kissel, 1976)
assuming increase pH = increase CEC, what is happening?
In acid soils, the exchange of NH4+ is for H+ on the exchange complex (release of H here, resists change in pH, e.g. going up)
In alkaline soils with high CEC, NH4 exchanges for Ca,
precipitation of CaCO3 (CO3= from HCO3- above) and one H+ released which helps resist the increase in pH
However, pH was already high.
Soil surface pH and cumulative NH3 loss as influenced by pH buffering capacity (from Ferguson et al., 1984).
Ernst and Massey (1960) found increased NH3 volatilization when liming a silt loam soil. The effective CEC would have been increased by liming but the rise in soil pH decreased the soils ability to supply H+
Rapid urea hydrolysis: greater potential for NH3 loss. Why?
management: dry soil surface, incorporate, localized placement- slows urea hydrolysis
Ferguson et al., 1984
(soils total acidity, comprised of exchangeable acidity + nonexchangeable titratable acidity)
A large component of a soils total acidity is that associated with the layer silicate sesquioxide complex (Al and Fe hydrous oxides). These sesquioxides carry a net positive charge and can hydrolyze to form H+ which resist an increase in pH upon an addition of a base.
H+ ion supply comes from:
2. hydrolysis of water
3. Al and Fe hydrous oxides
4. high clay content
A soil with an increased H+ buffering capacity will also show less NH3 loss when urea is applied without incorporation.
1. hydroxy Al-polymers added (carrying a net positive charge) to increase H+ buffering capacity.
2. strong acid cation exchange resins added (buffering capacity changed without affecting CEC, e.g. resin was saturated with H+).
resin: amorphous organic substances (plant secretions), soluble in organic solvents but not in water (used in plastics, inks)
Consider the following
1. H+ is required for urea hydrolysis.
2. Ability of a soil to supply H+ is related to amount of NH3 loss.
3. H+ is produced via nitrification (after urea is applied): acidity generated is not beneficial.
4. What could we apply with the urea to reduce NH3 loss?
an acid; strong electrolyte; dissociates to produce H+;increased H+ buffering; decrease pH
reduce NH3 loss by maintaining a low pH in the vicinity of the fertilizer granule (e.g. H3PO4)
Acid: substance that tends to give up protons (H+) to some other substance
Base: accepts protons
Anion: negatively charged ion
Cation: positively charged ion
Base cation: ? (this has been taught in the past but is not correct)
Electrolyte: nonmetallic electric conductor in which current is carried by the movement of ions
H2SO4 (strong electrolyte)
CH3COOH (weak electrolyte)
HA --------------> H+ + A-
1. Nitrogen Fertilization
A. ammoniacal sources of N
2. Decomposition of organic matter
CO2 + H2O --------> H2CO3 (carbonic acid)
H2CO3 ------> H+ + HCO3- (bicarbonate)
humus contains reactive carboxylic, phenolic groups that behave as weak acids which dissociate and release H+
3. Leaching of exchangeable bases/Removal
Ca, Mg, K and Na (out of the effective root zone)
-problem in sandy soils with low CEC
a. Replaced first by H and subsequently by Al (Al is one of the most abundant elements in soils. 7.1% by weight of earth's crust)
b. Al displaced from clay minerals, hydrolyzed to hydroxy aluminum complexes
c. Hydrolysis of monomeric forms liberate H+
d. Al(H2O)6+3 + H2O -----> Al(OH)(H2O)++ + H2O+
monomeric: a chemical compound that can undergo polymerization
polymerization: a chemical reaction in which two or more small molecules combine to form larger molecules that contain repeating structural units of the original molecules
4. Aluminosilicate clays
Presence of exchangeable Al
Al+3 + H2O -----> AlOH= + H+
5. Acid Rain
1. Assume that the absorbing complex of the soil can be represented by CaX
2. Ca represents various exchangeable bases with which the insoluble anions X are combined in an exchangeable form and that X can only combine with one Ca
3. H2X refers to dibasic acid (e.g., H2SO4)
(NH4)2SO4 -----> NH4+ to the exchange complex, SO4= combines with the base on the exchange complex replaced by NH4+
Thought: Volatilization losses of N as NH3 preclude the development of H+ ions produced via nitrification and would theoretically reduce the total potential development of acidity.
of N via denitrification leave an alkaline residue (
Table X. Reaction of N fertilizers when applied to soil.
1. Ammonium sulfate
a. (NH4)2SO4 + CaX ----> CaSO4 + (NH4)2X
b. (NH4)2X + 4O2 nitrification >2HNO3 + H2X + 2H2O
c. 2HNO3 + CaX ----> Ca(NO3)2 + H2X
Resultant acidity = 4H+ /mole of (NH4)2SO4
2. Ammonium nitrate
a. 2NH4NO3 + CaX ----> Ca(NO3)2 + (NH4)2X
b. (NH4)2X + 4O2 nitrification >2HNO3 + H2X + 2H2O
c. 2HNO3 + CaX ----> Ca(NO3)2 + H2X
Resultant acidity = 2H+ /mole of NH4NO3
a. CO(NH2)2 + 2H2O ----> (NH4)2CO3
b. (NH4)2CO3 + CaX ----> (NH4)2X + CaCO3
c. (NH4)2X + 4O2 nitrification >2HNO3 + H2X +2H2O
d. 2HNO3 +CaX ----> Ca(NO3)2 + H2X
e. H2X + CaCO3 neutralization >CaX + H2O + CO2
Resultant acidity = 2H+ /mole of CO(NH2)2
4. Anhydrous Ammonia
a. 2NH3 +2H2O ----> 2NH4OH
b. 2NH4OH + CaX ----> Ca(OH)2 + (NH4)2X
c. (NH4)2X + 4O2 nitrification >2HNO3 + H2X +2H2O
d. 2HNO3 + CaX ----> Ca(NO3)2 + H2X
e. H2X + Ca(OH)2 neutralization > CaX + 2H2O
Resultant acidity = 1H+/mole of NH3
5. Aqua Ammonia
a. 2NH4ON + CaX ----> Ca(OH)2 + (NH4)2X
b. (NH4)2X + 4O2 nitrification >2HNO3 + H2X +2H2O
c. 2HNO3 +CaX ----> Ca(NO3)2 + H2X
d. H2X + Ca(OH)2 neutralization > CaX +2H2O
Resultant acidity = 1H+/mole of NH4OH
6. Ammonium Phosphate
a. 2NH4H2PO4 + CaX ----> Ca(H2PO4)2 + (NH4)2X
b. (NH4)2X + 4O2 nitrification >2HNO3 + H2X +2H2O
c. 2HNO3 +CaX ----> Ca(NO3)2 + H2X
Resultant acidity = 2H+/mole of NH4H2PO4
In grain production systems, N use efficiency seldom exceeds 50 percent. Variables which influence N use efficiency include
b. N source
c. N application method
d. Time of N application
f. N rate (generally decreases with increasing N applied)
g. Production system
Olson and Swallow, 1984 (27-33% of the applied N fertilizer was removed by the grain following 5 years)
h. Plant N loss
i. Soil type (organic matter)
Calculating N Use Efficiency using The Difference Method
Applied N Grain Yield N content N uptake Fertilizer Recovery
kg/ha kg/ha % kg/ha %
0 1000 2.0 20 -
50 1300 2.1 27.3 (27.3-20)/50=14.6
100 2000 2.2 44 (44-20)/100=24
150 2000 2.3 46 (46-20)/150=17
Estimated N use efficiency for grain production systems ranges between 20 and 50%. The example above does not include straw, therefore, recovery levels are lower. However, further analysis of forage production systems (Altom et al., 1996) demonstrates that N use efficiency can be as high as 60-70%. This is largely because the plant is harvested prior to flowering thus minimizing the potential for plant N loss. Plant N loss is known to be greater when the plant is at flowering and approaching maturity. It is important to observe that estimated N use efficiencies in forage production systems do not decrease with increasing N applied as is normally found in grain production systems. This is suggestive of 'buffering' whereby increased N is lost at higher rates of applied N in grain production systems, but which cannot take place in forage production systems.
Work by Moll et al. (1982) suggested the presence of two primary components of N use efficiency: (1) the efficiency of absorption or uptake (Nt/Ns), and (2) the efficiency with which the N absorbed is utilized to produce grain (Gw/Nt) where Nt is the total N in the plant at maturity (grain + stover), Ns is the nitrogen supply or rate of fertilizer N and Gw is the grain weight, all expressed in the same units. Other parameters defined in their work and modifications (in italics) are reported in Table 4.2.
Recent understanding of plant N loss has required consideration of additional parameters not discussed in Moll et al. (1982). Harper et al. (1987) documented that N was lost as volatile NH3 from wheat plants after fertilizer application and during flowering. Maximum N accumulation has been found to occur at or near flowering in wheat and corn and not at harvest. In order to estimate plant N loss without the use of labeled N forms, the stage of growth where maximum N accumulation is known to occur needs to be identified. The amount of N remaining in the grain + straw or stover, is subtracted from the amount at maximum N accumulation to estimate potential plant N loss (difference method). However, even the use of difference methods for estimating plant N loss are flawed since continued uptake is known to take place beyond flowering or the point of maximum N accumulation.
Figure 4.1 Total N uptake in winter wheat with time and estimated loss following flowering.
Francis et al. (1993) recently documented that plant N losses could account for as much as 73% of the unaccounted-for N in 15N balance calculations. They further noted that gaseous plant N losses could be greater when N supply was increased. Similar to work by Kanampiu et al. (1997) with winter wheat, Francis et al. (1993) found that maximum N accumulation in corn occurred soon after flowering (R3 stage of growth). In addition, Francis et al. (1993) highlighted the importance of plant N loss on the development and interpretations of strategies to improve N fertilizer use efficiencies.
Consistent with work by Kanampiu et al. (1997), and Daigger et al. (1976), Figure 4.1 illustrates winter wheat N accumulation over time. Estimates of plant N loss are reported in Table 4.1. Harper et al. (1987) reported that 21% of the applied N fertilizer was lost as volatile NH3 in wheat, of which 11.4% was from both the soil and plants soon after fertilization and 9.8% from the leaves of wheat between anthesis and physiological maturity. Francis et al. (1993) summarized that failure to include direct plant N losses when calculating an N budget leads to overestimation of N loss from the soil by denitrification, leaching and ammonia volatilization.
NO3- + 2e (nitrate reductase) NO2- + 6e (nitrite reductase) NH4+
Reduction of NO3- to NO2- is the rate-limiting step in the transformation of N into amino forms.
Does the plant wake up in the morning and turn on the TV to check the weather forecast, to see if it should assimilate NO3 and attempt to form amino acids?
Could we look at the forecast and attempt to communicate with the plant, letting it know that weather conditions will be good (or bad), thus proceeding with increased NO3 uptake?
Major pathways for assimilation of NH3
1. Incorporation into glutamic acid to form glutamine, a reaction catalyzed by glutamine synthetase (Olson and Kurtz, 1982)
2. Reaction of NH3 and CO2 to form carbamyl phosphate, which in turn is converted to the amino acid arginine.
3. Biosynthesis of amides by combination of NH3 with an amino acid. In this way aspartic acid is converted to the amide, asparagine.
Table 4.1. Means over N rate and variety for
protein, NUE components and estimated plant N loss,
Protein N-use Uptake N-utilization Fraction of Grain yield/ N loss
% efficiency efficiency efficiency N translocated grain N (kg ha-1)
(Gw/Ns) (Nt/ Ns) (Gw/Nt) to grain(Ng/Nt) (Gw/Ng) (
N rate, kg ha-1 -------------------------------------------------------- means --------------------------------------------------------
0 14.8 0 0 23.2 0.60 38.8 16.4
45 15.9 23.3 1.0 22.9 0.63 36.5 25.0
90 17.4 11.0 0.6 20.2 0.61 33.2 25.8
180 17.6 7.0 0.4 20.5 0.62 33.5 31.4
SED 0.40 1.1 0.05 1.12 0.03 0.89 6.74
Chisholm 16.3 11.8 0.5 22.4 0.6 35.3 21.8a
Karl 17.5 13.1 0.6 23.0 0.7 33.0 26.6a
2180 17.4 18.1 0.8 22.7 0.7 33.4 27.9a
TAM W-101 15.5 11.7 0.6 21.4 0.6 37.4 24.7a
Longhorn 15.0 14.7 0.8 19.5 0.5 38.5 22.3a
SED 0.45 1.5 0.07 1.27 0.04 1.18 7.33
The ability of the soil-plant system to efficiently utilize N for food production (grain or forage) can be considered in four aspects: (1) efficiency of the plant to assimilate applied N, (2a and 2b) once assimilated, the ability of the plant to retain and incorporate N into the grain, (3) efficiency of the soil to supply/retain applied N for plant assimilation over long periods of time and (4) composite system efficiency.
Uptake efficiency should be estimated using Nf/Ns (Eup) instead of Nt/Ns (Eha) as proposed by Moll et al. (1982). More N is assimilated at earlier stages of growth, therefore, uptake efficiency should be estimated at the stage of maximum N accumulation and not at maturity when less N can be accounted for. The component Nt/Ns as proposed by Moll et al. would be better defined as harvest uptake efficiency or physiological maturity uptake efficiency. We define uptake efficiency as the stage where maximum N is taken up by the plant divided by the N supplied.
(1) Uptake efficiency Eup=Nf/Ns
Unlike the description by Moll et al. (1982), uptake efficiency should be partitioned into two separate components since plant N loss (from flowering to maturity) can be significant (Daigger et al., 1976; Harper et al., 1987; Francis et al., 1993). The fraction of N translocated to the grain should be estimated as Ng/Nf and not Ng/Nt as proposed by Moll et al. (1982) since more N was accumulated in the plant at an earlier stage of growth (Kanampiu et al., 1997). Plants losing significant quantities of N as NH3 would have very high fractions of N translocated to the grain when calculated using Nt instead of Nf. In terms of plant breeding efforts, this could be a highly misleading statistic. A second component, the translocation index is proposed that would reflect the ability of a plant genotype or management practice to incorporate N accumulated at flowering into the grain.
(2a) fraction of N translocated to the grain Et=Ng/Nf
(2b) translocation index Eti=Ng/Nf * (1/Nl)
The ability of the soil-plant system to utilize outside sources of N for food production (grain or forage) depends on the efficiency of storage in the soil. The efficiency of the soil to supply N to plants is strongly influenced by immobilization and mineralization with changing climate and environment.
Over a growing season, storage efficiency will be equal to the difference between fertilizer N added (Ns) minus maximum plant uptake (Nf) plus the difference between total soil N at the beginning and end of the season, all divided by fertilizer N added.
Esg = [(Ns-Nf)-(St1-St2)]/Ns
(3) soil (management system) supply efficiency Es=Ns/(Sv+Sd+Sl) where
Lastly, a composite estimate of efficiency for the entire system (soil and plant) can be estimated as follows
(4) composite system efficiency Ec=Eup*Es=Nf/(Sv+Sd+Sl)
It is important to note that these efficiency parameters can be determined without having to determine total N in the soil. Avoiding total soil N analyses is noteworthy since the precision of present analytical procedures (Kjeldahl or dry combustion) approach ± 0.01%. This translates into approximately ± 220 kg N/ha (depending on soil bulk density) which is often greater than the rate of N applied, thus restricting the ability to detect N treatment differences.
Will increased NUE lead to increased NO3 leaching?
Data from Kanampiu et al. (1995)
NUE Sinks: Increased NUE No Change
------------- kg / ha --------------
Total N Applied 180 180
Plant N uptake (at flowering) 68 71
Final Grain N uptake 42 40
Plant N loss 26 31
Denitrification 10 15
Immobilization 80 80
Balance 22 14
Leaching ? ?
Table 4.2. Components of nitrogen use efficiency as reported by Moll et al. (1982) and modifications (in bold italics) for grain crops.
Component Abbreviation Unit
Grain weight Gw kg ha-1
Nitrogen supply (rate of fertilizer N) Ns kg ha-1
Total N in the plant at maturity (grain + stover) Nt kg ha-1
N accumulation after silking Na kg ha-1
N accumulated in grain at harvest Ng kg ha-1
Stage of growth where N accumulated in the plant
is at a maximum, at or near flowering Nf kg ha-1
Total N accumulated in the straw at harvest Nst kg ha-1
Estimate of gaseous loss of N from the
plant Nl =
Flowering uptake efficiency Eup=Nf/Ns
Harvest uptake efficiency (Uptake efficiency) Eha=Nt/Ns
Translocation index (accumulated N at flowering
translocated to the grain) Eti =Ng/Nf * (1/Nl)
Soil supply efficiency Es=Ns/(Sv+Sd+Sl)
Composite system efficiency Ec=Eup*Es=Nf/(Sv+Sd+Sl)
Utilization efficiency Gw/Nt
Efficiency of use Gw/Ns
Grain produced per unit of grain N Gw/Ng
Fraction of total N translocated to grain Et=Ng/Nt
Fraction of total N accumulated after silking Na/Nt
Ratio of N translocated to grain to N accumulated Ng/Na
1892: 4.0 % organic matter = 0.35+ 1.8 OC
OC = 2.03%
TN = 0.16%
Pb = 1.623 (0-12")
lb N/ac = DB * ppm N * 2.7194
= 1.623 * 1600 * 2.7194
OC = 0.62%
TN = 0.0694%
lb N/ac = 1.623*694*2.7194
Difference: 7061 - 3063 = 3998 lbs N
Grain N removal
14.6 bu/ac * 60 lb/bu = 876 lbs
876 lbs * 105 years = 91980 lbs grain
91980 lbs * 0.022086 %N = 2031 lbs N
Plant N loss
10.7 lb/ac/yr (Kanampiu et al., 1995)
105 * 10.7 = 1130 lbs N
2.85 lb/ac/yr (Aulakh et al. 1984)
105 * 2.85 = 300 lbs N
Balance 537 lbs N
Year 1 denitrification, ammonification
Denitrification, ug/g = 50.0 * OC + 6.2 (Burford and Bremner, 1975)
= 50.0 * 2.03 + 6.2
= 107.7 ug/g
= 107.7 * 1.623 * 2.7194
= 475.34 lb/ac (0-12")
New Balance 61.66 lb N/ac
(0.58 lb N/ac/yr unaccounted)
Not included in this balance sheet is the amount of N that would be deposited via rainfall, and the amount lost via ammonification, both of which would be important.
Denitrification losses the first year were likely much higher since increased NO3-N would have been present as a result of mineralized N from a very large total N pool. Burford and Bremner (1975) applied the equivalent of 800 lb NO3-N/ac and found that denitrification losses were extremely high. Although their work has little relevance to annual denitrification losses expected under field conditions, it does provide some insight into what might have happened in the first year when soils were first tilled.
When adequate inorganic N was present, the incorporation of straw in conventional till or the application of straw on the surface of zero till approximately doubled the accumulative gaseous N losses (increased supply of energy to denitrifying organisms) (Aulakh et al., 1984).
From 71 to 77% of the surface applied fertilizer N remaining in the profiles was in the 0 to 0.1 m soil layers (Olson and Swallow, 1984).
Late N application can be efficiently taken up by plants, and does not decrease soil N uptake. To achieve acceptable grain protein levels for bread wheat in this irrigated cropping system, N should be supplied late in the season to improve N uptake during grain fill (Wuest and Cassman, 1992)
Einstein: Relativity theory (1905), quantum theory
Roentgen: discovered x-rays
Becquerel: first recognition of radioactivity
Bremsstrahlung: identified secondary x-rays
Curie - Joliot: first induced artificial radioactivity (1934)
Isotopes are atoms of the same element that differ in mass. They have the same number of protons and electrons but have a different mass which is due to the number of neutrons.
1. All radio isotopes have a particular kind of radiation emission
2. Energy and mass are equivalent (Einstein)
3. All radio nuclides have a characteristic energy of radiation
4. All radio nuclides possess a characteristic rate of decay
1 mole of X has 6.025 x 1023 atoms
one gram of 14N has (14 g/mole)
6.025 x 1023 atoms/mole * 1 mole/14g = 4.3 x 1022 atoms/g
Avogadros # = # of molecules in one gram molecular weight of any substance.
Dealing with reactions in the outer ring that compromise and produce chemical reactions.
atomic mass units charge
proton 1.007594 +
electron 0.000549 -
neutron 1.008986 none
mZE 11H 42He
m - mass
z - atomic number (# of protons in the nucleus)
All hydrogen atoms have one proton
11H 21H 31H
stable stable radioactive
mass = 1 mass=2 mass=3
no neutron 1 neutron 2 neutrons
1 proton 1 proton 1 proton
1 electron 1 electron 1 electron
126C 136C 146C
stable stable radioactive
mass=12 mass=13 mass=14
6 neutrons 7 neutrons 8 neutrons
6 protons 6 protons 6 protons
6 electrons 6 electrons 6 electrons
Chemical versus Nuclear Reactions:
1. 2Na+ + H2O ----> 2NaOH + 2H+
3-5 eV in this reaction
2. 42He + 94Be ----> 126C + 10n
10 million eV in this reaction
In a nuclear reaction, we have to balance both mass and proton number.
Transmutation: changing one element into another
3517Cl + 10n ------> 3215P + 42He
3216S + 10n ------> 3215P + 11p
Chemical reactions involve changes in the outer electronic structure of the atom whereas nuclear reactions involve changes in the nucleus
erg: work done by a force of one dyne acting through a distance of 1 cm.
= 1.0 dyne/cm of 1.0 g - cm2/sec2
dyne: force that would give a free mass of one gram, an acceleration of one centimeter per second per second
Curie: amount of any radioactive material in which 3.7 x 1010 atoms disintegrate (decay or loss of radioactivity) per second.
1 Bq (becquerel) 1 dps
1 uC = 3.7 x 104 dps
1 mC = 3.7 x 107 dps = 2.22 x 109 dpm
1 C = 3.7 x 1010 dps = 2.22 x 1012 dpm
Rad = 100 ergs/g absorbing material (quantity of radiation equivalent to 100 ergs/g of exposed tissue).
1 Rad = 1/100 Roentgen
eV = electron volt (amount of energy required to raise one electron through a potential of one volt)
1 eV = 1.6 x 10-12 erg
1 MeV = 1.6 x 10-6 erg
specific ionization: # of ion pairs produced/unit distance penetrated.
13755Cs (30 year half life) and 9038Sr (28 year half life) were the major radioactive isotopes of concern in that accident
1. Particle accelerators
2. Nuclear reactors
3. Atomic explosions
Mass Energy Equivalents:
E = MC2
1 amu = 1.66 x 10-24 g
= reciprocal of Avogadro's #
E = energy (ergs)
M = mass (grams)
C = velocity of light (cm/sec)
= 186000 miles/sec
= 3 x 1010 cm/sec
How much energy does 1 amu have?
E = (1.66 x 10-24 g) (3 x 1010 cm/sec)2
=1.49 x 10-3 ergs
= (1.49 x 10-3 ergs)/(1.6 x 10-6 erg/Mev) = 931 MeV
Calculate the amount of energy in 1 gram of 235U?
1g/235g/mole x 6.025 x 1023 atoms/mole x 0.215amu/atom x 931MeV/amu
= 5.12 x 1023 MeV
= 2.3 x 1014 kilowatt hours (12 years of electricity for 1 household)
1 kilowatt hour = 2.226 x 109 MeV
only 1/5 or 0.215 of 235U is converted to energy (split)
Fusion: Making hydrogen atoms combine resulting in released energy
-no remnant radioactivity
-no atmospheric contamination
21H + 31H ---> 42He + 10n
deuterium tritium (alpha)
2½ gallons of tritium would provide the
Fission: "Splitting atoms"
-results in the production of radioactive materials
23592U + 10n ---> 9736Kr + 13856Ba +10n + energy
23592U + 10n ---> 9038Sr + 14454Xe + 2 10n + energy
13856Ba is a fission fragment
Strictly chance of actually knowing what we will have as products from the bombardment of 23592U with neutrons.
23592U "controlled reaction that is a chain reaction" using uranium rods
238U accounts for 99.3 percent of the uranium found on earth
23592U is used for fission, because it splits easier.
neutrons emitted in fission can produce a chain reaction
Nuclear fission taps about 1/1000 of the total possible energy of the atom.
1. Alpha (nucleus of the He atom, mass = 4 and charge = +2)
Charge +2, mass 4 (42He) high specific ionization, limited penetration, come only from high z (# of protons) atoms.
22688Ra --> 22286Rn + 42He + energy
23892U --> 23490Th + alpha + 4.19 MeV
Radionuclides which emit alpha are changed into another nuclide with a mass of 4 units less and 2 fewer protons
Three sheets of paper are sufficient to stop alpha radiation.
· when an alpha particle loses energy it attracts electrons and becomes a neutral helium atom.
· not used in plant biology and soil studies.
2. Beta "negatron" (high neutron:proton ratio, originates from the nucleus like alpha)
· neutron in the nucleus changes to a proton, increasing the atomic # by one.
3215P ---> 3216S + B- + e- + v(+1.71 Mev)
3. Beta "positron" (low neutron:proton ratio, comes from the nucleus which has too many protons)
· proton in the nucleus changes to a neutron, decreasing the atomic number by one.
3015P ---> 3014Si + B+ + e+ + v(+3.3 Mev)
B. Photons (a quantum of radiant energy)
1. Gamma, does not have a mass (electromagnetic radiation with the speed of light)
· is not a mode of radioisotope decay but rather associated with particulate emission.
· can penetrate inches of lead
6027Co ---> 6028Ni + B- +gamma + gamma
0.31MeV 1.17 MeV 1.33 MeV
Radio isotope decay schemes result in transmutation of elements that leave the nucleus in a suspended state of animation. Stability is reached by emitting one or more gamma photons.
2. X-ray emitting by electron capture (too many protons and not enough neutrons)
· emitted when cathode rays of high velocity fall directly on a metallic target (anticathode) in a vacuum tube.
· highly penetrating electromagnetic radiation (photons) with a short wave-length.
· identical to gamma rays if their energies are equal
· electron from K ring is pulled into the nucleus
· chain reaction of K ring pulling electron into K from L and so on.
· emission as an x-ray is external to the nucleus (come from the outer shell of the atom)
3. Cosmic radiation (radiation from outer space)
· mixture of particulate radiation (neutrons) and electromagnetic radiation.
specific ionization penetration charge nucleus
alpha high low +2 inside 226Ra, 238U, 242Pu*
beta (negatron) medium med +1 inside
beta (positron)@ medium med -1 inside 90Sr, 32P
gamma low high none inside 60Co
X-ray high outside 59Ni
* - naturally occurring
@ - characteristic of the majority of radioisotopes used in biological tracer work
A. ionization takes place in an enclosed sensitive medium between two oppositely charged electrodes (ionization chambers, Geiger-Muller)
B. systems that do not depend on ion collection but make use of the property that gamma-ray photons (also alpha and beta) have for exciting fluorescence in certain substances (scintillation)
C. ionizing radiations affect the silver halide in photographic emulsions which show a blackening of the areas exposed to radiation (autoradiography)
Geiger-Muller Counter: (positron) will not measure gamma.
G-M tube filled with Ar or He. Ionizing radiation passing through the gas in the tube causes electrons to be removed from the atoms of gas; form ion-pairs (pairs of electrons and positive ions). Under the influence of an applied field, some of the electrons move towards the anode and some of the positive ions towards the cathode. Charges collect on the electrodes and initiate pulses; a continuous stream of these pulses constitute a weak electric current.
Positive ions are produced from molecules or atoms by subjecting them to an electric discharge or some other source of high energy. The positive ions are accelerated by means of an electric field and then passed through a slit into a magnetic field. The slit serves to select a beam of ions. The charged particles follow a curved path in the magnetic field which is determined by the charge to mass ratio of the ion. When two ions with the same charge travel through the tube, the one with the greater mass will tend to follow the wider circle.
Block diagram of a double collector mass spectrometer (Vose, 1980)
Scintillation: (alpha, positron, negatron, gamma)
When certain materials (zinc sulfide) are exposed to gamma photons or particulate radiation they emit scintillation's or flashes of light. The scintillation's are produced by a complex process involving the production of an excited (higher energy) state of the atoms of the material. When the orbital electrons of these atoms become deexcited, the excess energy is then given off in an infinitely small time as a flash of light (scintillation).
Limits: 1/10 Rad/week
X-ray (dentist) 1-5 rads
0-25 rads no injury
25-50 rads possible blood change, shortened life span
50-100 rads blood changes
100-200 definite injury (possibly disabled)
200-400 definite disability, possible death
400-600 50% chance of dying
>600 assured fatal
1. Nucleic acid injections: enhance blood manufacturing capabilities of the body (blood cells affected most)
2. Bee sting venom (has R-SH radical)
There are four stable or heavy isotopes of potential interest to researchers in soil and plant studies (18O, 2H, 13C and 15N)
(N2 gas bombarded by electrons) N2 gas
(cryogenic distillation of nitric oxide) (microdiffusion techniques)
1. non radioactive
2. no time limits on experiment (versus half-life problems associated with radioactive materials)
3. less sensitive than for measuring radioactive elements where we can accurately determine 1 atom disintegrating
4. mass spec needs 1012 atoms before it can be measured
5. mass spectrometry is more complicated.
6. high enrichment needed in agricultural work
7. high cost associated with purchasing this isotope $250/g
8. need 3/10 enrichment for 1 year experiments.
9. discrimination of plants for 14N versus 15N
10. more sensitive than total N procedures
Nitrogen: radioactive isotopes of N have extremely short half-lives to be of significant use in agriculture (13N t½ =603 seconds)
% present in
14N 14N 99.634
15N 14N 0.366
Ratio needs to be established before starting the experiment: (e.g., background levels)
100g 15NH415NO3 5% enriched $200
100g 15NH415NO3 10% enriched $400
Instead of the specific activity of a sample used in the case of radioisotopes, the term % abundance is used for stable isotopes. The % 15N abundance is the ratio of 15N to 15N + 14N atoms
Because the natural environment has an 15N abundance of 0.3663%, the amount of 15N in a sample is expressed as %15N atom excess over the natural abundance of 0.3663. (subtracting 0.3663 from the determination of 15N abundance to obtain 15N atom excess).
mass spec: detection to 0.002 atom excess:
Essentially measuring the intensity of ion currents (R)
R = 14N 14N/15N 14N
% 15N abundance = 100/2R + 1
By measuring the height of the 14N 14N and 15N 14N peaks (corrected for a background reading), the R values are determined and the % 15N abundance calculated.
N in plant and soil samples must first be converted into N2 gas.
1. Kjeldahl digestion - distillation into acid - total N determined by titration - aliquot taken for transformation into N2 gas (Rittenberg Method)
2NH4Cl + 3NaBrO* + 2NaOH ----> N2 + 5H2O + 3NaBr + 2NaCl
*alkaline sodium hypobromite
(Vose, p 156)
2. Dumas method (sample heated with CuO at high temperatures (> 600°C) in a stream of purified CO2 and the gases liberated are led over hot Cu to reduce nitrogen oxides to N2 and then over CuO to convert CO to CO2. The N2-CO2 mixture thus obtained is collected in a nitrometer containing concentrated alkali which absorbs the CO2 and the volume of N2 gas is measured.
1. N in grain, N in tissue
2. N in organic fractions (immobilized)
3. Inorganic soil N
4. Plant N loss
5. N leaching
For analysis by mass spectrometer, the analytical error including sub-sampling is approximately 0.01% 15N atom excess for a single sample. Improved instrumentation has taken this to 0.002% 15N atom excess.
Therefore samples should contain at least 0.20 % 15N atom excess. (5% error)
1% atom excess 15N is adequate for fertilizer experiments where the crop takes up a substantial portion of the applied fertilizer.
30-50% atom excess is required for soils experiments where turnover processes are high and where various fates of N exist (plant N loss, leaching, plant uptake, grain uptake, etc.). For this reason, 15N studies are usually small due to the price.
If 80 kg N/ha are to be applied in an experiment where the total N uptake is likely to be 100 kg N/ha and the expected utilization of N fertilizer were 30 %, then 0.33 kg/ha of 15N is required (Vose, p. 165, using Figure X from Fried et al.).
Therefore, the enrichment required for a rate of application could be as low as 0.41% 15N atom excess (0.33/80 * 100)
materials with a greater than natural concentration of 15N
% plant N derived from fertilizer = %15N excess in sample
% 15N excess in fertilizer
materials with a lower than natural concentration of 15N (0.003 - 0.01 atom % 15N) or (< 0.01 atom % 15N)
· use of isotopic 14N
· studies involving residual soil nitrogen are not practical with depleted materials due to the high dilution factor.
% plant N derived from the fertilizer =
(Nu - Nt)/(Nu - (Nf/n))
Nu =atom % 15N in unfertilized plants
Nt = atom % 15N in fertilized plants
Nf = atom % 15N in the fertilizer (for example 0.006%)
n = the plant discrimination factor between 14N and 15N.
If it is assumed that there is no discrimination between 14N and 15N, then n = 1.
Fertilizer N Recovery (Varvel and Peterson, 1991)
1. Difference method
PFR = (NF)-(NC)
NF = total N uptake in corn from N fertilized plots
NC = total N uptake in corn from unfertilized plots
R = rate of fertilizer N applied
PFR = percent fertilizer recovery
2. Isotopic method (Depleted material)
PFR = (NF) x (C-B)/D
NF = total N uptake in corn from N fertilized plots
B = atom % 15N of plant tissue from N fertilized plots
C = atom % 15N of plant tissue from unfertilized plots (0.366%)
D = depleted atom % 15N in applied N fertilizer
R = rate of applied 15N-labeled fertilizer
3. Isotopic method (Enriched material, Sanchez et al., 1987)
F = As-Ar/Af-Ar
F= fraction of total N uptake derived from 15N enriched fertilizer
As = atom % 15N measured in the harvested plant sample
Af = atom % 15N in the enriched fertilizer
Ar = atom % 15N of the reference harvested plant material from non 15N enriched fertilizer treatments
Ef = F x total N uptake
Ef = uptake of 15N enriched fertilizer
Shearer and Legg (1975) found that d15N of wheat plants decreased as the N application rate increased.
d15N = atom % 15N (sample) - atom % 15N (standard) x 1000
atom % 15 N (standard)
15N composition of the total N of grain and leaf samples of corn (Zea mays L.) decreased systematically as N fertilizer rates increased (Kohl et al., 1973). This result was considered to be consistent with increasing contributions of fertilizer N to plants as the rate of applied N increased.
Hauck and Bremner, 1976
percent nitrogen recovered (plant or soil) =
= 100P (c-b)
P = total N in the plant part or soil in kg ha-1
f = rate of 15N fertilizer applied
a = atom percent 15N in the labeled fertilizer
b = atom percent 15N in the plant part or soil receiving no 15N
c = atom percent 15N in the plant part or soil that did receive 15N
unlabeled N uptake = (total N uptake in grain and straw) -
[N rate(% recovery of 15N in grain and straw)]
half-life: time required for half of the radioactive atoms to undergo decay (loss of half of its radioactivity)
32P (t½ = 14.3 days)
14C (t½ = 5568 yrs)
l: Decay constant (fraction of the number of atoms of a radioisotope which decay per unit time)
A: Activity (decay intensity which is proportional to the number of radioactive atoms present)
N: number of radioactive atoms present at time t and l is the decay constant
l = 0.693/t½
N = No e -lt
A = lN
N for 1 g of pure 32P = 6.025 x 1023/32 atoms/g
= 1.88 x 1022 atoms/g
All tracer studies assume that the tracer behaves chemically and physically as does the element to be studied (tracee).
Discrimination of the plant /soil microflora
Isotopic Exchange (42K , cytoplasm, exclusion K2SO4, KCl)
1. mobile in the plant
2. found to concentrate in the grain
3. mobility of P in the plant allows for increased concentration in younger tissue and fruiting bodies.
4. strong beta emitter resulting in acceptable characteristics for autoradiograph techniques.
1. P use efficiency
2. Method of placement
3. P fixation
In general, 32P is no longer useful after approximately 7 half lives or 100.1 days.
1. What will the activity of 5 mC 32P in 5 ml be in 36 days?
N = No e -lt
A = Ao e -lt
l = 0.693/t½ = 0.693/14.3 = 0.04846
t = 36 days
-lt = 1.744
e -lt = 0.1748
A = 5 mC/5ml * 0.1748
= 0.1748 mC/ml
2. You intend to set up a field experiment for evaluating the P delivery capacity of a given soil.
a. P rate= 18.12 kg/ha (18120 g/ha)
b. Crop will utilize 10 % of that applied.
c. Need a count of 1000 cpm at the end of the experiment.
d. Instrument has a 20% counting efficiency for 32P.
e. A 10 gram sample will be used from a total plot weight of 3628 kg/ha.
10/3628000 = 0.000002756
What should the specific activity of the fertilizer be in mC/g P if 110 days will lapse between planting and sample assay?
1000 cpm = Ao e -lt
1000 cpm = Ao * e -(0.693/14.3)(110)
1000 cpm = Ao e -5.33
Ao = 1000/0.0048403 = 2.06596 x 105 cpm
2.0659 x 105 cpm ÷ 60 sec/min = 3.443 x 103 dps
3.443 x 103 dps ÷ 0.10 (crop utilization efficiency) = 3.443 x 104 dps
3.443 x 104 dps ÷ 0.20 (counting efficiency) = 1.7216 x 105 dps
1.7216 x 105 dps ÷ 0.000002756 (dilution) = 6.2468 x 1010 dps
6.2468 x 1010 dps ÷ 3.7 x 107 dps/mC (constant) = 1.688 x 103 mC
1.688 x 103 mC ÷ 18120 g = 9.317 x 10-2 mC/g P
3. How much 32P would you put into a system to assure 500 cpm after 2 months using an instrument with a 10% counting efficiency and 10% P utilization efficiency?
A = Ao e -lt
500 cpm = Ao * e -(0.693/14.3)(60)
Ao = 500/0.0546 = 9.157 * 103 cpm
9.157 * 103 cpm ÷ 0.10 (crop utilization efficiency) = 9.157 * 104 cpm
9.157 * 104 cpm ÷ 0.10 (counting efficiency) = 9.157 * 105 cpm
9.157 * 105 cpm ÷ 2.22 x 109 cpm/mC (constant) = 4.13 x 10-4 mC
1 mC 32P weighs 3.5 x 10-9 g
4.13 x 10-4 mC x 3.5 x 10-9 g/mC = 1.44 x 10-12 g 32P
Adsorption: adhesion in an extremely thin layer of molecules to the surfaces of solid bodies or liquids with which they are in contact.
Soils containing large amounts of mineral clay and organic matter are said to be highly buffered and require large amounts of added lime to increase the pH.
Buffering capacity (BC): represents the ability of the soil to re-supply an ion to the soil solution.
You should never use a buffered solution (fixed pH) for CEC. If a 1 N NH4OAc solution were used to displace the cations on the exchange complex of a soil with a pH of 5.0, CEC would be overestimated as pH dependent charge sites would be included (specifically organic matter) that would not have been present at the soils natural pH.
Ions must exist in soils as solid compounds or adsorbed to cation/anion exchange sites.
Can be described by the ratio of the concentrations of absorbed (D Q) and solution (D I) ions; BC = D Q/D I
The BC in soil increases with increasing CEC, organic matter and other solid constituents in the soil.
For most minerals the strength of cation adsorption or lyotropic series is:
ions with a higher valence are held more tightly than monovalent cations (exception, H+)
The degree of replaceability of an ion decreases as its dehydrated radius increases. Cations are attracted toward, and anions are repelled from, negatively charged soil colloids. These interactions follow Coulomb's law where;
F is the force of attraction or repulsion
q and q1 are the electrical charges (esu, equal to 2.09 x 109 individual electronic charges)
r is the distance of charge separation (cm)
D is the dielectric constant (=78 for water at 25°C)
The strength of ion retention or repulsion increases with increasing ion charge, with increasing colloid charge and with decreasing distance between the colloid surface and either the source of charge or the soluble ion.
Interaction between ions increases with concentration and with the square of the ion charge. The parameter embracing the concentration and charge effects is the ionic strength (I) of the solution.
I = ½ sum Mi Zi2
where M is the molarity, Z is the charge of each ion i.
Ionic strength measures the effective ion concentration by taking into account the pronounced effect of ion charge on solution properties. A solution has only one ionic strength but each of its constituent ions may have a different activity coefficient.
Exchangeable bases: Ca++ Mg++ K+ and Na+
1. H ions obtained from the hydrolysis of exchangeable, trivalent Al
2. Hydrolysis of partially hydrolyzed and nonexchangeable Al
3. Weakly acidic groups, mostly on organic matter
4. Exchangeable H
In the early days of soil science there was no agreement on the pH of the soil at which exchangeable acidity was to be determined. Bradfield, 1923 noted that the usual substance used to increase the pH of acid soils is CaCO3 and that the maximum pH obtainable with CaCO3 is pH 8.3. Therefore base saturation is defined as the quantity of base adsorbed by a soil in the presence of CaCO3 equilibrated with air having a CO2 content of 0.03% (Thomas, 1982).
1. Sum total of exchangeable cations on the exchange complex expressed in meq/100g (Ca++, Mg++, K+, Na+, H+, Al+++)
2. Quantity of readily exchangeable cations neutralizing negative charge in the soil
3. Exchange of one cation for another in a solution phase
4. Soils capacity to adsorb cations from an aqueous solution of the same pH, ionic strength, dielectric constant and composition as that encountered in the field.
Extract sample with neutral 1 N ammonium acetate. (NH4OAc)
· exchange complex becomes saturated with NH4
· extract same soil with 1N KCl (different salt solution), K+ replaces NH4
· quantity of ammonium ions in the leachate is a measure of CEC
-filtrate has 0.054 g of NH4
(20 g of soil extracted)
1 meq of NH4 = (14+4)/1000
= 0.018g/meq or 18g/eq
0.054/0.018 = 3 meq
3 meq/20g = 15meq/100g
increase clay, increase CEC
increase 2:1 clays, increase CEC
1:1 clays: 1-10 meq/100g
2:1 clays: 80-150 meq/100g
Extraction with an unbuffered salt which would give a measure of the CEC at the soils normal pH.
Use of neutral N ammonium acetate (7.0) will result in a high CEC on acid soils because of the adsorption of NH4 to the pH dependent charge sites.
1.At high pH, H+ are weakly held and may be exchanged; pH dependent charge
2.Deprotonation (dissociation of H from OH groups at the broken edges of clay particles which is the prime source of negative charge in 1:1 clay minerals) occurs only at high pH (7.0 and up)
Kamprath: unbuffered salt solution, 1.0 N KCl will extract only the cations held at active exchange sites at the particular pH of the soil. The exchangeable acidity is due to Al and H.
1. Presence of CaCO3 and/or CaSO4 (dissolution) and the presence of salt in arid type soils. Dissolution of CaCO3 and/or CaSO4 will cause Ca to exchange for Mg, K and Na instead of NH4 replacing all of these. When 1 N KCl is then added to displace the NH4 (from NH4OAc) less NH4 is detected in the filtrate than what should have been present.
2. Variable charge soils (high content of more difficult exchangeable aluminum-hydroxy "cations"). Exchangeable Al and its hydroxy forms are not readily exchanged with monovalent cation saturation solutions. This error results in an underestimation of CEC.
1.Polemio & Rhoades (1977) arid soils containing carbonates, gypsum and zeolites.
a. Saturation of exchange sites with Na (pH 8.2) 0.4N NaOAc + 0.1N NaCl
b. Extraction with 0.5N MgNO3
c. Na determined (soluble Na from saturation step deducted from total Na to obtain exchangeable Na)
d. Method will determine CEC as a result of permanent charge but not for variable charged soils (pH)
2. Gillman (1979) acid soils
a. Saturation of exchange sites with BaCl2 (solution of a concentration approximately equivalent in ionic strength to the soil solution)
b. Extraction with MgSO4 to replace Ba with Mg (MgSO4 concentration is adjusted to achieve an ionic strength comparable with that of the soil solution)
c. Ba determined
The use of unbuffered solutions throughout ensures that natural soil pH is not significantly altered.
The underlying factor which has caused various researchers to develop alternative methods for determining CEC was how to deal with pH dependent charges (pH of the saturating solution and replacement solution). This is important considering the pH is a logarithmic function of H+ where 10 times as much H occurs in solution at pH 5 as pH 6.
Reflects the extent of leaching and weathering of the soil.
It is the percentage of total CEC occupied by cations, Ca++, Mg++, Na+ and K+, where each is determined separately from the NH4OAc extract (Atomic Absorption - interception of radiant energy)
Amount present in soil
Meq of each cation (amount present/g per meq)
Ca = 0.03/0.02 = 1.5
Mg = 0.008/0.012 = 0.66
Na = 0.021/0.023 = 0.91
K = 0.014/0.039 = 0.36
CEC = 20 meq/100g
BS = 17.15/20 = 85.85%
BS = CEC - (H+ + Al+++) / CEC * remember this is exchangeable H+ and Al+++
pH and BS are positively correlated
Why would pH and BS be positively correlated if pH and CEC were not?
Adsorption of anions to + charged sites in hydrous oxide minerals where the hydrous oxides are amphoteric (have - and + charge depending on pH and therefore have AEC and CEC).
Order of adsorption strength H2PO4- > SO4= >NO3- = Cl-
pH < 7.0
More in weathered soils (1:1) containing hydrous oxides of Fe and Al (exposed OH groups on the edges of clay minerals)
Soils which have pH dependent charges.
Anion exchange of 43meq/100g at an acidic equilibrium pH of 4.7.
Can a soil have a net positive charge? (unlikely)
Is H2PO4- adsorption on soils anion exchange? yes
only physically adsorbed initially but soon precipitate as Ca-P in alkaline soils and Fe or Al-P in acid soils.
Can P applications induce S deficiencies in acid soils?
Acid soil: S levels low --> P exchange for S on exchange complex (anion exchange) and SO4= can be leached.
90% of all water soluble bases will be leached as sulfate (Pearson et al, 1962)
Kamprath et al. (1956)
1. Increased P concentration in solution reduced the amounts of SO4= adsorbed by the soil.
2. Amount of sulfate adsorbed decreased as the pH of the soil suspension increased (4 to 6).
Aylmore et al. (1967)
1. Sulfate adsorption on clays possessing positive edge charges + oxides of Fe and Al (highly resistant to leaching and less available for plant growth)
2. Sulfate adsorbed on kaolinite clay is weakly held and easily released
Fox et al. (1964)
Ca(H2PO4)2 best extracting solution for S
AEC negatively correlated with Base Saturation
Ca10(PO4)6F2 or Cl2 or OH2
1. water soluble
2. citrate soluble (dissolves more P than water)
OSP ordinary superphosphate (0-20-0)
· rock phosphate + sulfuric acid
· mixture of monocalcium phosphate and gypsum
· 16-22% P2O5 (90 % water soluble)
· 8-10% S as CaSO4
TSP triple or concentrated superphosphate (0-46-0)
· rock phosphate + phosphoric acid
· essentially all monocalcium phosphate
· 44 to 52% P2O5 (98% water soluble)
· < 3% S
· major phosphate mineral is monocalcium phosphate monohydrate (MCP)
DAP Diammonium phosphate (18-46-0)
· Reacting wet process H3PO4 with NH3
· 46-53% P2O5
MCP monocalcium phosphate monohydrate Ca(H2PO4)2 2H2O (highly water soluble)
DCPD dicalcium phosphate dihydrate CaHPO4* 2H2O - brushite
DCP dicalcium phosphate CaHPO4, 53% P2O5 - monetite
congruent dissolution of Ca(H2PO4)2 2H2O into Ca++ and H2PO4 ions occurs at a pH of 4.68
1. P deficient
2. S deficient
3. pH 5.5
4. anion exchange 20 meq/100g
· Apply triple superphosphate with gypsum
· Supersaturate the band with respect to Ca and precipitate P as DCP and or DCPD which will be slowly available with time.
· including NH4+, K+, Ca++ and Mg++ enables these cations to be included in the initial reaction products.
· MCP contains sufficient Ca to precipitate half of P as DCPD or DCP.
· In acid soils, Fe and Al generally precipitate the additional P.
· Avoid anion exchange interaction (P displacing S from the complex)
Low Soil pH (<5.5) P precipitates as Al and Fe phosphates
a. variscite (ALPO4 * 2H2O)
b. strengite (FePO4 * 2H2O)
Moderate to High pH, P precipitates as Ca phosphates (several)
a. dicalcium phosphate (CaHPO4)
b. dicalcium phosphate dihydrate (CaHPO4 * 2H2O)
c. hydroxyapatite Ca5(PO4)3OH
d. fluorapatite Ca5(PO4)3F (rock phosphate)
Precipitation - Dissolution of phosphate minerals is pH dependent:
Precipitation/Dissolution can be determined by using P solubility diagrams.
1. Soil solution (H2PO4-) and pH above the line (precipitation)
2. Soil solution (H2PO4- and pH below the line (dissolution)
pH 4.5 Event Precipitate Formed
1. add fertilizer soluble P added -
2. 1 - 2 soluble P decreases DCP
3. 2-3 DCP dissolves FA
4. 3-4 FA dissolves Variscite
Example of precipitation/dissolution (1-4)
1. Liebig's law of the minimum
2. Bray's Nutrient Mobility Concept
3. Sufficiency (SLAN)
5. Bray modified Mitscherlich
6. Base Cation Saturation Ratio
He stated that the nutrient present in least relative amount is the limiting nutrient.
soil contained enough N to produce 50 bu/ac
soil contained enough K to produce 70 bu/ac
soil contained enough P to produce 60 bu/ac
N would be the limiting nutrient.
Crop used up all of the deficient nutrient in the soil making the yield directly proportional to the amount of the deficient nutrient present and the crop content of the nutrient.
a. Range of nutrient (insufficient to sufficient)
b. Amount extracted from the soil is inversely proportional to yield increases from added nutrients.
c. Calibrations exist for the changing levels of available nutrients with fertilizer additions and yield response.
d. Concept assumes little if any effect of the level of availability of one ion on that of another.
e. Recognizes that an addition of the most limiting element may cause more efficient utilization of a less limiting element.
Mathematical expression of the law of diminishing returns where increases in yield of a crop per unit of available nutrient decreases as the level of available nutrient approaches sufficiency.
The concept is based on Mitscherlich's equation:
dy/dx = (A-y)c
Yield increases (dy) per unit of available nutrient (dx) decrease as the current yield (y) approaches a maximum yield (A) with c being a proportionality constant.
The derivative was developed for studying tangent lines and rate of change. The first derivative is the slope of the tangent line at xo
d/dx xn = nxn-1
Quadratic: Y = bo + b1x - b2x2
0 = b1-2b2x
2b2x = b1
Simply stated, plants respond to the total amount present of mobile nutrients and to the concentration present of immobile nutrients in soils. Stated this way, yield is directly related (proportional) to the total amount of nutrient present in the soil. However, yield response to immobile nutrients is not related to the total amount of the “available form” present in the soil, but instead is a function of the concentration of available form at, or very near, the root surface. Ideally, the response of crops to mobile nutrients should be linear because mobile nutrients (like water) are not decreased in availability by reaction with the soil. The linear response to mobile nutrients continues with each added increment of the nutrient until the yield potential for that growing environment has been reached, after which it is zero (see figure below)
The ideal situation is not found in soils, only in hydroponics and when the physical phase of the growth media is not reactive, such as with glass beads. However, because the reaction of some nutrients with soil is sometimes minimal (e.g. nitrate-N in cultivated soils with minor potential for immobilization and mineralization of N), they are considered relatively mobile and response tends to follow the ideal. Hence, “rules of thumb” have been developed to guide the use of mobile nutrients like nitrogen, such as “it takes 2 lbs N/bushel of wheat”. The 2 lbs is calculated from the protein or N content (on average) of a bushel of wheat, with the added assumption that measured soil nitrate-N and added fertilizer N will be only 70% utilized. Bray’s mobility concept implies that if available N, for example, is limited to some level below maximum yield potential then a yield plateau will occur at that point. For example, if there is enough total amount of available mobile nutrient to produce the yield potential (20 bu.) and then midway through the season better than average weather conditions result in increasing the yield potential (to 30 bu.), the mobility concept implies the yield will be limited to 20 bu. because the total supply of nutrient will be used up to produce 20 bu. and additional yield can only be obtained if more of the nutrient is added (this is the reason for top dressing wheat midway through the season).
For immobile nutrients, like phosphorus, plants can only extract the nutrient from soil close to the root surface, very little of the nutrient is moved to the root by water in the transpiration stream because soil solution concentrations are minute (< 0.05 ppm for phosphate compared to as high as 100 ppm for nitrate-N). As a plant grows and roots extend out into the soil, roots come in contact with “new” soil from which they can extract phosphate. The amount extracted is limited by the concentration at (or very near) the root-soil interface. If the concentration of phosphate available to the plant at the root -soil interface is inadequate to meet the needs of the plant, then the plant will be deficient in P throughout its development. The deficiency will always be present, and plant growth and crop yield will be limited by the degree to which the immobile nutrient is deficient. Another, perhaps more common way of expressing this nutrient limitation is to state that yield will be obtained according to the sufficiency of the nutrient supply. When this is expressed as a percentage of the yield possibility then the term percent sufficiency may be applied. Whenever the percent sufficiency is less than 100, plant performance is less than the yield possibility provided by the growing environment. Consequently, it does not matter whether the yield possibility is 20 bu. or 30 bu., if the percent sufficiency is 80, then actual yield obtained (theoretically) will only be 80% of that yield possibility.
The soil test for mobile nutrients is an indicator of the total amount available. If this amount is enough to produce 20 bu/ac, more N would have to be added to the total pool to produce 40 bu/ac. With P, an index is developed that is independent of the environment. If the crop year was good, roots would expand into more volume of soil that had the same level of nutrient supply. Sufficiency is independent of the environment since increased root growth will expand into areas where contact exchange uptake is the same (total amount present in the soil is not greatly affected).
Concept yield goal sufficiency
Environment dependent independent
Sorption Zone root system root surface
on total available large small
Soil test is an
indicator of the
total available yes no
concentrations 0-100 ug/g <0.05 ug/g
Function of conc. in the root syst. conc. at the root surf.
Topdress appl. Yes No
(4081 kg/ha = 60 bu/ac)
2.5%N in the grain
=102.03 kg N
(4081 kg/ha = 60 bu/ac)
0.36%P in the grain
=14.69 kg P
0.1% N*10000=1000 ug/g * 1.47 * 1.524 = 2240 kg N/ha 0-15 cm
NO3-N: 10 ug/g * 1.47 * 1.524 = 22.40 kg NO3-N/ha 0-15 cm
NO3-N soil test is the actual N available at time X
NO3-N soil test is valid for one point in time (1 crop or year)
Some states predict N mineralization
0.1% P*10000=1000 ug/g * 1.47 * 1.524 = 2240 kg P/ha 0-15 cm
P soil test is an index (sufficiency) of availability
P soil test is valid for up to 5 years or more**
10 ug/g P, Mehlich III is not equal to 22.40 kg P/ha
We cannot predict P mineralization
(102.03/2240)*100 = 4.5%
(14.69/2240)*100 = 0.65%
Steps for Using the Sufficiency Concept:
1. Selection or determination of the sufficiency level
a. estimated from results of studies with a crop on similar soils.
2. Computation of fertilizer required for sufficiency
a. amount of soluble P required to raise the available P from the initial level to the sufficiency level.
3. Method of supplying the fertilizers (and/or lime)
a. soil build-up plus crop needs (BUILD-UP) long-term.
b. crop needs (MAINTENANCE) short-term.
Mitscherlich-Baule percent sufficiency concept:
When more than one nutrient was deficient, the final percent sufficiency is the product of the individual sufficiencies.
Maximum yield when N,P and K are
present in sufficient quantities 5000 kg/ha
Yield when N and P are present in
sufficient quantities 4000 kg/ha
4000/5000 = 80% of MAXIMUM
Yield when N and K are present in sufficient
quantities 3000 kg/ha
3000/5000 = 60% of MAXIMUM
What will be the predicted yield when only N is present in
sufficient quantities 2400 kg/ha
5000(0.6 * 0.8)
"present" function of both soil levels and amount applied.
If this percent sufficiency concept is correct, then Liebig's concept of the limiting nutrient is wrong.
Present Field X Field Y * Field Z *
in adequate Yield kg/ha
NP 6400 9600 8000
NPK 8000 12000 10000
NK 7200 10800 9000
PK 7000 7000 7000
N 5760 8640 7200
8000*.8*.9 12000*.8*.9 10000*.8*.9
% sufficiency K NP/NPK = 6400/8000 = 0.8
% sufficiency P NK/NPK = 7200/8000 = 0.9
* - assume that the % sufficiency levels for P and K are the same in field Y and field Z
Mitscherlich was incorrect in his use of c values for N 0.122.
The original work by Mitscherlich showed that the response of plants to nutrients in the soil can be expressed by a curvilinear function and a logarithmic equation, and further concluded that the regression coefficient c in the equation was constant for each nutrient regardless of any change in environment, plant type, soil and other factors (Balba and Bray, 1956).
log (A-y) = log A - cx
A = yield possibility when all nutrients are present in adequate amounts but not in excess
y = yield obtained at a given level of x (dy = dx) and when y is always less than A(99%)
c = proportionality constant
NOTE: some texts use c and others c1, however, it does not matter which one is used, so long as they are defined. Similarly, b and x are used interchangeably
---- = c(A-y) and ----- = dxc
log(A-y) = log A - cx
* A and y can be expressed as actual yield or % of the maximum yield
Experimental locations with different soil test P (b) levels
NPK NK Sufficiency x calc. c
Loc 1 30 20 0.66 12 1.53=2-12c 0.039
Loc 2 40 15 0.375 4 1.79=2-4c 0.051
Loc 3 30 16 0.53 9 1.67=2-9c 0.036
y = 66
x = 12
log(100-66) = log 100-12c
1.53 = 2 - 12c
c = 0.039
Apply value of c where applicable. If the soil pH or soil test K changes over an area, then c has to be altered accordingly.
Now that an average c factor has been determined, we can relate the soil test level of b with yield sufficiency for this element. (CAN determine % SUFFICIENCY)
STEP 3. (Bray Modified Mitscherlich)
Expand Mitscherlich to calculate the amount of fertilizer needed to raise the percent yield from any given starting level to any other desired upper level for which fertilization is desired
Log(A-y) = log A - cb - c1x
c1 = efficiency factor for the method of applying the fertilizer (determined from fertilizer studies). This factor will change accordingly for immobile nutrients (band versus broadcast)
x = quantity of fertilizer that needs to be applied.
Fertilizer studies c1 (broadcast P) = 0.0070
c1 (banded P) = 0.0025
c and c1 vary with
2. planting density/pattern
3. nutrient applied (source)
4. method of application
2. climate, moisture
3. yield potential (hybrid)
4. planting density and pattern
Soil Nutrient Requirement (level determined)
1. when sampled
2. stage of growth
4. form of nutrient applied
5. analytical method
Fertilizer Requirement (x)
2. fertilizer used
Log (A-y) = Log A - cb - c1x
A = maximum yield
y = yield obtained at some level of b
b = soil test index
c = efficiency factor (constant) for b
x = amount of fertilizer added to the soil
c1 = efficiency factor for x (method of placement)
Example: Soil test value for P = 20
N, K and all other nutrients adequate
kg P/ha Yield, kg/ha % Sufficiency
0 2000 40
25 3000 60
50 4500 90
75 5000 100
log (100-40) = log 100 - c(20)
1.778 = 2.00 - c(20)
-0.2218 = -c(20)
c = 0.01109
solve for c
log (5000 - 3000) = log 5000 - 0.01109(20) - c1(25)
3.301 = 3.477 - c1(25)
c1 = 0.00704
log (5000 - 4500) = log 5000 - 0.01109(20) - c1(50)
2.6989 = 3.477 - c1(50)
c1 = 0.0155
average of c1 = (0.00704 + 0.0155)/2
Apply concept (solve for x, determine the amount of fertilizer to be applied)
Log (A-y) = log A - cb - c1x
* The dangers of using % yield: It is difficult to determine amounts of fertilizer to add (e.g., 2.0 Mg/ha yield and 4.0 Mg/ha yield).
Assumes that reliable soil test data is available for good soil test correlation.
Assuming that plants take up nutrients from two different sources in direct proportion to the amount available, the A-value was developed as the expression
A = B(1-y)/y
where; A = amount available nutrient in the soil
B = amount of fertilizer nutrient (standard) applied
y = proportion of nutrient in the plant derived from the standard
"In a true sense, the plant is the only agent that can determine the amount available."
For a specific soil, crop and growing conditions, the A-value is constant, and has been found to be independent of rate of fertilizer application, size of test pot and growth rate.
The A value was primarily developed to determine the availability of P in soil (P supplying power of a given soil).
With the band placement, the A values increased with increasing P rates. This suggests that the availability to plants when P was banded does not remain constant with increasing rates.
Fried and Dean (1951) noted that because it can be assumed that the method of placement does not change the soil phosphorus, the lower A values obtained with the band placement can be attributed to a higher availability of the standard (nutrient applied).
For optimum growth of crops, both a best ratio of basic cations and a best total base saturation exist in a soil.
et al. (1945)
Percent saturation of cations selected as being "ideal". Work originally conducted on alfalfa. Historically, it is interesting to note that this work was being done at the same time Bray developed the mobility concept.
Mg 10% (minimum required for alfalfa)
Ca:Mg > 6.5:1
Ca:K > 13:1
Mg:K > 2:1
Bear et al. (1945) suggested that
1. 10% Mg saturation was minimal for alfalfa
2. Soluble Mg sources were essential for correcting Mg deficiencies in sandy soils
3. Liming above 80% base saturation (20% H) brought about deficiencies of Mn and other micronutrients.
Graham (1959) established ranges or % saturation of the CEC for the 'ideal' soil
· When this proportion exists, you can obtain maximum yield.
· Works only in sandy soils.
1. Bonding of cations to exchange sites differs greatly from one type of cation to another and it differs greatly for the same type of cation at different saturations.
2. Exchangeable cations are not proportional to soluble amounts (plant available)
3. Excess of one type of cation may depress the activity and plant uptake of another
4. Adsorbed ion (x) can have marked effects on the ion in question
5. Capacity (total exchangeable) and intensity (activity) of an adsorbed cation influence the total availability of a cation to the plant
6. Saturation of pH-dependent charges increases the activity and plant availability of divalent basic cations
Steps in USING BCSR:
1. Soil analyzed for exchangeable bases
2. Lime required to raise the soil pH to X
3. CEC is determined by totaling basic cations + acidity (exchangeable H and Al), each expressed as meq/100g or cmol/kg
4. Each basic cation expressed as a % of the total CEC
5. Cations must be added to the extent that the existing saturations of basic cations = ranges chosen (e.g., some must decrease and others must increase)
· Works well on low to moderate CEC soils and coarse textured soils, highly weathered soils of low pH that require major adjustments in fertility.
· Useful where it is important to maintain a fairly high level of Mg in the soil to alleviate grass tetany in ruminants.
Grass tetany (low concentrations of Mg and Ca in cool-season grasses in late fall and early spring).
Grass tetany will occur when forage contains K/(Ca+Mg) > 2.2
(physiological nutrient imbalance which leads to muscle spasms and deficient parathyroid secretion)
1. Assess the relative adequacy of available nutrients (or lime requirements)
To provide guidance on amounts of fertilizers
(or lime) required to obtain optimum growth conditions for plants (
3. Diagnosis of nutrient limitations before a crop is planted so that corrective measures can be taken.
*Must be fast, reliable and reproducible
Philosophical differences exist on interpreting the tests which result in radically different fertilizer recommendations
1. Base Cation Saturation Ratio
2. Nutrient Maintenance
Disregarding the soil test level, a quantity of nutrient should be added to replace the amount expected to be removed by the crop. All required nutrients- not feasible.
3. Nutrient Sufficiency
No yield response to nutrients above a certain soil test level.
a. response assured very low
b. response likely low
c. response possible medium
d. response unlikely high
Depth of Sampling
1. 0-6, 0-8, 0-12, inclusion of subsoil (micronutrients)
1. Cate Nelson
4. Square Root
Field experiments (1973-1980)
Irrigated Corn (Zea mays L.)
5 soil testing laboratories
No differences in yield
No agronomic basis for 'balance' or 'maintenance' concepts
K, S, Zn, Mn, Cu, B, Mg, Fe
1. probability of response to added fertilizer is small
2. probability of response to added fertilizer is large
A. Percent yield values obtained for a wide range in locations (fertilizer rate studies)
· Percent yield = yield at 0 level of a nutrient / yield where all factors are adequate
B. Soil test values obtained (Check Plot)
· Will generate a single % yield and one soil test value for each location
C. Scatter diagram, % yield (Y axis) versus soil test level (x axis)
· Range in Y = 0 to 100%
· overlay moved to the point where data in the +/+ quadrants are at a maximum
· point where vertical line crosses the x = critical soil test level
depends on the extraction method used and crop being grown.
Maximizes the computed chi-square value representing the test of the null hypothesis that the # of observations in each of the four cells (quadrants is equal).
4. Square Root
5. Linear Plateau: obtaining the smallest pooled residuals over two linear regressions.
Equation MR MER (dy/dx = PR)
2. Mitscherlich Log(A-Y) = Log A - C1(x+b) x=log((2.3*A*c)/PR)/c-b
3. Quadratic y = b0 + b1(x) - b2(x2) x=0.5 b1/b2 x=(PR-b1)/(2*b2)
4. Square Root y = bo + b1(x) + b2(sqrt(x)) x=0.25(b2/b1)2 x=(b2/ 2*(PR-b1))2
5. Linear Plateau y = bo + b1(x) when x < joint
y = bo + b1(joint) when x > joint
PR = (price per unit fertilizer) / (price per unit yield)
Optimum rate of fertilizer capable of generating the maximum economic yield is dependent upon the price of fertilizer, the value of the crop and magnitude of fixed production costs. The value of a crop defined as a function of yield and rate of fertilizer can be expressed as:
V = Y * Py = F(x) * Py
where yield (Y) for each fertilizer rate is multiplied by the crop price (Py) per unit of yield. A line describing fertilizer costs per unit area cultivated can be expressed as a function of fixed costs (F) and fertilizer price (Px) times the amount of fertilizer (X)
T = F + Px * X
where total cost (T) is a linear function of fertilizer amount, the slope of the line is given by the price of fertilizer and the intercept by the amount of fixed costs involved (F).
A plot of the value and cost functions illustrates the areas where use of fertilizer is profitable. Net profit can only be generated by use of a fertilizer amount equal or greater than 0-x1. Fertilizer should not be used if the value curve is lower throughout than the total cost curve for fertilizer plus fixed costs (F). With fixed costs involved, the amount of fertilizer that can be used profitably is greater than zero or an amount equal to or greater than 0-x1. For fertilizer input greater than 0-x1, crop value exceeds costs and net profit is generated. Profit from fertilizer application can be increased until input reaches the value of 0-x2. This is the level which maximizes profit. At 0-x2 the difference between value and cost is at a maximum.
For each production function the amount of fertilizer which maximizes profit can be found by obtaining the first derivative and setting it equal to the price ratio (PR).
PR = Price per unit of fertilizer / Price per unit of yield
(from Barreto and Westerman, 1985)
Total Nitrogen in Soils:
Surface soils: 0.05 to 0.10%
precision 0.01% = +/- 200 lb/ac
Why would we run total N on soils if the precision is so low?
· long term experiments (differences greater than 200 lb N/ac)
· C:N relationships at the same level of precision
A. Kjeldahl 1883 (organic + inorganic N)
1. digestion to convert organic N to NH4
2. determination of NH4 in the digest
(N pool consists of NO3-, NH4+, NO2-, organic N)
devardas: reducing agent, that is a finely powdered mixture of metals that act as a source of donor electrons to reduce NO3- and NO2- to ammonium
N pool + K2SO4, CuSO4, Se, H2SO4 -----> (NH4)2SO4
(NH4)2SO4 + NaOH ----> NH3 + NaSO4 (catch in boric acid)
K2SO4 is used to raise the temperature of the digest (increases speed and completeness of the conversion of organic N to NH4)
Se, Cu are used as catalysts to promote the oxidation of the organic matter
NO3 and NO2 are not included in the total N analysis from dry combustion, but it does not matter since there will be less than 20 lb N /ac as NO3 and the total N procedure detects to only +/- 200 lbs N/ac
0.01 +/- 200 lbs/ac 20 lbs N/ac as NO3 is lost between 0.01 and 0.02 %total N
0.02 +/- 400 lbs/ac because its small value exceeded the detection limits.
On a KCl extract: (have both NH4 and NO3 in the extract)
1. distill over once (to collect NH4)
2. add devardas alloy (distill over again to collect NO3 and NO2)
devardas alloy: acts as a source of donor electrons to reduce NO2 and NO3 to NH4
problems: N-N and N-O compounds
Sample heated with CuO at high temperature (above 600 °C) in a stream of purified CO2 and the gasses lost are passed over hot Cu to reduce nitrogen to N2 and then over CuO to convert CO to CO2. The N2-CO2 mixture is collected in a nitrometer containing concentrated alkali which absorbs the CO2 and the volume of N2 gas is measured.
2NH4Cl + 4CuO -----> N2 + 4H2O + 2CuCl + 2 Cu (CO2)
problems: heterocyclic compounds (pyridine) are difficult to burn
Sample weighed in a tin (Sn) container
Combustion reactor enriched with pure oxygen (sample oxidation) 1020 °C in combustion tube
Reaches 1700 °C during flash combustion (complete oxidation)
Flash combustion converts all organic and inorganic substances into elemental gases (stable compounds combusted
Combustion products carried by He pass through an oxidation catalyst of Chromium oxide
Combustion Reactor Reduction Reactor
CO + 1/2O2 = CO2 (Cr2O3 is accepting electrons)
Cr2O3 ensures complete combustion (oxidation) of all organic materials
NOx N2 (Cu is donating electrons)
Combustion products (CO, N, NO) and water pass through a reduction reactor (metallic Cu).
Excess O2 is removed in the reduction reactor (Cu at 650 C).
N oxides from the combustion tube are reduced to elemental N2 .
Taking CO, N, NOx and converting them to CO2, N2.
Gases are separated in a chromatographic column and detected using a thermal conductivity detector (TCD) which gives an output signal proportional to the concentration of the CO2 and N2 present.
2NH4Cl + 3NaBrO + 2NaOH ----> N2 + 5H2O + 3NaBr + 2NaCl
Inorganic N may represent only a small fraction < 2% of the total N in soils (Bremner, 1965)
Nitrate testing does not work in
high mineralization potential
consideration of NH4
R-NH2 groups from N cycle
· rapid changes (biological transformations) affect inorganic N analysis
NO3-N and NO2-N
1. Phenoldisulfonic acid or chromotrophic acid
· interference of organic matter, Cl and Fe have affected these colorimetric procedures
2. Selective ion electrodes
· interference of Cl
· (NH4)2SO4, AgSO4 extracting solution: Ag used to precipitate Cl
3. Cadmium reduction
· 2 M KCl extract (colorimetric procedure) - samples are stable for several months if stored at low temperatures
· not subject to interference, extremely sensitive making dilution possible.
· NO3 reduced to NO2 by passing through a column of copperized Cd
· NO2 reacts well with the diazotizing reagent (sulfanilamide) and NO3 does not, thus explaining the need for reducing NO3 to NO2 for analysis using the Griess-Ilosvay method
4. Steam distillation with Devardas alloy (reductant) reduce NO2 and NO3 to NH4
Bremner (1959) stated soils contain a large amount of fixed (non-exchangeable) NH4. Defined as the NH4 that cannot be replaced by a neutral K salt solution present as NH4 ions in interlayer positions of 2:1 type clay minerals.
Air-drying can lead to small but significant changes in NH4-N
1. Steam distillation with MgO (alkaline reagent) color: indophenol blue
2. 2 M KCl (indophenol blue) phenol and NH3 react to form an intense blue color
3. Ammonia gas sensing electrodes
Problems in N analysis:
· -accuracy is measured by the least precise measurement.
· -weight of the soil is the largest error (propagates through to +/- 0.01%N)
0.01% N = +/- 100 ppm (0.01* 10000)
total N in soils 0.10 = 1000 ppm +/- 100 ppm
inorganic N in soils 0.002 = 20 ppm +/- 1 ppm
Total N Inorganic N Organic N?
1000 ppm 20 ppm 980 ppm
1. Inorganic N is not determined on a percent basis because it is done on an aliquot basis.
2. Cannot subtract 20 from 1000 to get organic N (determined on a different basis).
3. Unrealistic because of the incompatibility of error terms.
4. Organic-N is difficult to determine (by subtraction, we have an extremely poor estimate).
Procedures exist, but are unreliable and are not reproducible.
1.Leach with CaCl2 - dissolves all the soluble N (NO3 and NO2)
2.Incubate the soil - over time - to determine the amount of NO3 that has been mineralized (set period of time under set conditions)
3.Leach with CaCl2 again (sample now has NO3)
Bray and Kurtz P-1
Designed to remove easily acid soluble forms of P, largely calcium phosphates and a portion of the aluminum and iron phosphates. The NH4F dissolves aluminum and iron phosphates by its complex ion formation with these metal ions in acid solution. This method has proved to be very successful in acid soils.
In view of the high efficiency of the fluoride ion in dissolving phosphate, Bray (1945) recommended the use of this reagent together with HCl as an extractant (effectively removed sorbed phosphate).
Al reacts with F and inactivates Al leaving P in solution. Use of NH4F will increase extractable P, or stabilize P (restricting Al from precipitating with P because of the solubility constants).
0.20 NH4Cl, 0.2N CH3COOH, 0.015N NH4F and 0.012N HCl
(pH = 2.5)
The concentrations of HCl and NH4F used in Mehlich are half that used in Bray and Kurtz P-1. However this extracting solution also contains NH4Cl and acetic acid which probably buffer the solution (i.e., keeps its acidic strength for a longer period of time). Therefore, it can dissolve more of the P in apatite.
0.2N CH3COOH, 0.015N NH4F, 0.25N NH4NO3, 0.13N HNO3, 0.001M EDTA (pH = 2.4)
Designed to be applicable across a wide range of soil properties ranging in reaction from acid to basic. Can also be used for exchangeable cations (Ca and Mg). Because this extractant is so acid, there is some concern that the soil can be dissolved, increasing exchangeable amounts.
0.5N NaHCO3 (pH = 8.5)
This extracting solution is used to extract phosphorus in calcareous soils. It will theoretically extract the phosphorus available to plants in high pH soils. This extractant decreases the concentration of Ca in solution by causing precipitation of Ca as CaCO3; as a result, the concentration of P in solution increases.
Essentially, increase the activity of CO3 in solution which reacts with Ca, and CaCO3 precipitates.
Nelson et al. (1953) (Mehlich I and or "Double Acid")
0.05N HCl and 0.025N H2SO4 (pH<2.0)
Found to be effective in high P-fixing
Extractable P discussion:
The pH of the extracting solution is an indicator of what forms of P will be extracted. However, this should be used with caution as the shaking time is important in terms of reaching an equilibrium.
Susuki et al. (1963) noted that 0.1N HCl extractable P was positively correlated with Ca-P.
was negatively correlated with Ca-P on 17
What would happen if Bray P-1 was used on a calcareous soil?
The lime in the calcareous soil would neutralize the acidity in the extracting solution thus decreasing its ability to extract the Fe and Al-P forms which would be available at that soils pH.
Calibrations for the Bray-Kurtz P-1, Mehlich III and Olsen soil tests (Tisdale, Nelson, Beaton and Havlin, 1993)
P sufficiency level Bray-Kurtz P-1 Mehlich III Olsen Fertilizer P
lb P2O5/ac kg P/ha
Very low <5 <7 <3 50 25
Low 6-12 8-14 4-7 30 15
Medium 13-25 15-28 8-11 15 8
High >25 >28 >12 0 0
Analysis for total P in soils was abandoned in the early 1900's as scientists recognized that this analysis was not correlated with plant availability. For this reason various strengths of extracting solutions were evaluated for specific soils at selected soil pH that mirrored what the plant would find in soil solution. All of these are indices that determine orthophosphate concentrations (from the dissolution of precipitated forms). Attempts to correlate extractable P (x - procedure) with total P will result in meaningless information. Total P (strong acid digest) will in essence dissolve P forms that will not be available at that soils specific pH.
Bray and Nye:
K applications on soils with high K by mass action displace Al+++ which complexes with P inducing a net P deficiency (pH < 6.0)
P and Zinc
Zinc deficiencies attributed to the immobilization of zinc owing to the increase in the concentrations of P in the roots above the threshold values.
Depression of zinc concentrations in plant tissue by P (interaction occurred in the plant and not in the soil).
Source of N by P
NO3- uptake (increase pH)
NH4+ uptake (decrease pH)
Interaction of light with matter
1nm = 1mu (millimicron) = 10A (angstrom) = 10-7 cm
The interaction of radiation with matter may result in the absorption of incident radiation, emission of fluorescence or phosphorescence, scattering into new directions, rotation of the plane of polarization, or other changes. Each of these interactions can provide useful information about the nature of the same in which they occur (Tinoco et al. 1978).
Color is characteristic of the spectrum (in the visible region) of light transmitted by the substance when white light (or sunlight) shines through it, or when light is reflected from it.
<0.01 Gamma (non particulate photons)
0.01-10 X-Ray (photons)
Wavelength absorbed, nm Absorbed Color Transmitted Color (Complement)
380-450 Violet Yellow-green
450-495 Blue Yellow
495-570 Green Violet
570-590 Yellow Blue
620-750 Red Blue-green
1x106-1x1011 Micro and short
>1x1011 Radio, FM TV
Wavelength: distance of one complete cycle
Frequency: the number of cycles passing a fixed point per unit time
l = c/v
l = wavelength in cm
v = frequency in sec-1 or hertz (Hz)
c = velocity of light in a vacuum (3x1010 cm/sec)
Electromagnetic radiation possesses a certain amount of energy. The energy of a unit of radiation, called the photon is related to the frequency by E = hv = hc/l
where E is the energy of the photon in ergs
h is Planck’s constant 6.62 x 10-27 erg-sec
The shorter the wavelength or the greater the frequency, the greater the energy. Energy of a single photon (E) is proportional to its frequency (v) or inversely proportional to its wavelength.
If a molecule absorbs radiation, it is raised to a higher energy level, with the increase in energy being equal to the energy of the absorbed radiation (hl).
The relative energy levels of the three transition processes are in the order electronic > vibrational > rotational.
If the electromagnetic force results in a change in the arrangement of the electrons in a molecule, we say that a transition to a new electronic state has occurred. The absorbed photon results in the excitation of the molecule from its normal or ground state, G, to a higher energy or excited electronic state, E. The excited electronic state has a rearranged electron distribution.
When considering absorbing substances that are either liquids, solids or gases, each will have a characteristic transmission of light. Suppose that light of intensity Io is incident from the left, propagates along the x direction and exits from the right with decreased intensity It. At any point x within the sample, it has intensity I, which will decrease smoothly from left to right.
If the sample is homogeneous, the fractional decrease in the light intensity is the same across a small interval dx, regardless of the value of x. The decrease for a solution depends linearly on the concentration of the absorbing substance.
· Not all molecules can absorb in the infrared region
· The wavelength of absorption is a measure of the energy required for the transition
· Each molecule will have a complete absorption spectrum unique to that molecule, so a 'fingerprint' of the molecule is obtained
Soil Testing Sensor Based VRT
low resolution high resolution
Chemistry-Site specific Site specific
Reliable and tested untested
Years of correlation/calibration new technology
Economical high potential of being economical
Crop specific untested
Variety specific untested
Management specific untested
Nutrient interactions untested
NA weed recognition
NA time of day
NA direction of travel