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Soil test variation:
Can be minimised by soil sampling cropping soils to a standard depth of 0-15 cm each time
Is mostly caused by errors with laboratory analysis
Can be minimised by soil sampling immediately after fertiliser application
Can be minimised by sampling a different area of a block or paddock at each sampling time.
Can be minimised by soil sampling cropping soils to a standard depth of 0-15 cm each time
Soil Test Variation
Spatial Variation
Soils
Topography
Excreta return
Within paddocks and between paddocks
Significant factor in pastoral farming
Uneven return of nutrients in animal excret
Seasonal Variation/Fluctuations
Influence with K, Sulphate, pH
Less important for Olsen P
Sampling Variation/Error
Sampling depth
Handling/Storage conditions
Laboratory Variation/Error
Rare nowadays
System ensures that the results do NOT vary too much
Can replicate the situation to obtain estimate results
Minimising Soil Test Variation
Dividing land into areas of similar soil type, topography, aspect, management + Sample separately using indicator paddocks
Avoiding areas of abnormal fertility + Use the same sampling transects (record location)
Stock camp sites
Fencelines, Gateways, Water troughs
Waiting several months after the last fertiliser/lime application
Sampling at the same time of the year
Sampling to a standard depth using a tool that samples evenly down the soil depth (soil corer)
Pasture soils = 0 - 7.5 cm
Cropping/Horticultural soils = 0 - 15 cm
Collecting an adequate number of cores randomly per sample (15 - 20 cores per sample)
Storing samples in cool conditions and sending to the lab as soon as possible
Which statement is true?
The Olsen P test involves extracting a soil with a 1 M ammonium acetate solution.
The Anion Storage Capacity (ASC) test involves a phosphate solution with a pH adjusted to pH 8.4.
The Cation Exchange Capacity (CEC) test involves extracting a soil with a solution that has a pH of 7.0.
The soil pH test uses a soil to solution ratio of 1:1
The Cation Exchange Capacity (CEC) test involves extracting a soil with a solution that has a pH of 7.0.
Olsen P test
Measures
The phosphate in soil solution
A small portion of P adsorbed on Fe & Al oxides
A small portion of organic P
Indication of plant-available P (H2PO4-)
Conditions
0.5 M sodium bicarbonate (NaHCO3)
pH = 8.4
*The exchangeable K+ test uses 1 M ammonium acetate (NH4OOCCH3) extracting solution at pH 7
Anion Storage Capacity (ASC) test
Measures the storage capacity of phosphate through absorption onto soil surfaces
pH = 4.6 (Hill Labs)
Cation Exchange Capacity (CEC) test
Measures the sum of all exchangeable cations
Includes basic & acidic cations (H+, Al3+)
Determines the ability of the soil to store cations (+) in an exchangeable form and replenish the soil solution concentration
pH = 7.0 (Cornell University)
Soil pH test
Measures the degree of acidity/alkalinity
Conditions
(Lab) 10 g soil : 25 ml distilled water = 1 : 2.5? ← NOT 1:1 ratio
In soil, sulphate (SO42-):
Can be produced from the reduction of hydrogen sulphides in anaerobic conditions.
Is not leached in drainage water.
Is not supplied from the mineralisation of soil organic matter
Can be produced from the oxidation of elemental S in aerobic conditions.
Can be produced from the oxidation of elemental S in aerobic conditions.
A farm’s average annual drainage depth is 250 mm, which has an average nitrogen (N) concentration of 6 g N/m3. The annual average quantity of N leaching from the soil is:
41.7 kg N/hectare/year
15 kg N/hectare/year
1.5 kg N/hectare/year
150 kg N/hectare/year
15 kg N/hectare/year
Drainage depth = 250 mm → 0.25 m
N leaching (g N/m3/yr) = Drainage depth (m) x N concentration (g N/m3)
= 0.25 m x 6 g N/m3
= 1.5 g N/m3/yr
1 hectare (ha) = 10,000 m3
Annual N leaching = 1.5 g N/m3/yr × 10,000 m3
= 15 000 g N/ha/yr → 15 kg N/ha/yr
A reactive phosphate rock (RPR) is recommended for direct application (i.e. without acidulation) to pastures as a phosphorus fertiliser when:
Less than 30% of its total P content is soluble in 2% citric acid.
Soil pH is less than 6.
Mean annual rainfall is less than 800 mm.
A quick release soluble fertiliser is required.
Soil pH is less than 6.
Which statement is true?
Triple-superphosphate is manufactured by reacting phosphate rock with phosphoric acid.
Single-superphosphate is manufactured by reacting phosphate rock with hydrochloric acid.
Potassium chloride is manufactured using he Harber-Bosch process.
Ammonium sulphate is a natural fertiliser material that is sourced by mining.
Triple-superphosphate is manufactured by reacting phosphate rock with phosphoric acid.
Which statement is true?
Mixtures of single-superphosphate and potassium chloride are not compatible.
Mixtures of single-superphosphate and urea are compatible.
Mixtures of agricultural lime and urea are compatible.
Mixtures of reactive phosphate rock (RPR) and potassium chloride are compatible.
Mixtures of reactive phosphate rock (RPR) and potassium chloride are compatible.
Using agricultural lime to increase soil pH from 5.4 to 5.8 will:
Increase the plant availability of soil cobalt.
Increase the plant availability of soil aluminium.
Decrease the plant availability of soil calcium.
Increase soil base saturation (%).
Increase soil base saturation (%).
Allophanic soils:
Are common in the Manawatu region.
Are formed from sedimentary parent materials.
Typically have very high reserves of potassium in parent materials.
Typically have anion storage capacity (P retention) values > 80%
Typically have anion storage capacity (P retention) values > 80%
For a high yielding (i.e. > 22 t DM/ha) maize silage crop planted in a soil that has been continually cropped for greater than 10 years:
All of the crop’s nitrogen (N) requirements can be to supplied by the mineralisation of soil organic matter.
The background soil mineral N supply will likely be less than 150 kg N/ha.
The background soil mineral N supply will likely be greater than 150 kg N/ha.
Will require no more than 50 kg N/ha applied in fertiliser.
The background soil mineral N supply will likely be less than 150 kg N/ha.
Which statement is true?
Dolomite is a liming material and a source of potassium (K).
Gypsum is a liming material that neutralises soil acidify.
Gypsum is a source of sulphur (S).
Agricultural limestone mostly contains calcium chloride.
Gypsum is a source of sulphur (S).
Planktonic algae in rivers:
Are heterotrophic organisms.
Grow roots into the river bed.
Only grow on the river bed.
Mostly float on the water surface.
Mostly float on the water surface.
The livestock urine patches, left in pastures by grazing, commonly:
Contain more phosphorus (P) than dung patches.
Are a source of potassium (K) leaching.
Contain more calcium (Ca) than dung patches.
Causes nitrogen (N) to be recycled evenly to the majority of the grazed area annually.
Are a source of potassium (K) leaching.
A plant root taking up more cations than anions:
Will raise the pH of rhizosphere soil.
Will lower the pH of rhizosphere soil.
Won’t occur when ammonium is the main form of nitrogen taken up by a plant.
Is unlikely to occur with legumes.
Will lower the pH of rhizosphere soil.
A potassic superphosphate fertiliser (0-7.2-10-8.8):
Is a mixture of 80% single superphosphate and 20% potassium chloride.
Is a mixture of 80% single superphosphate and 20% potassium sulphate.
Supplies approximately 25 kg of phosphorus (P) for every 250 kg of fertiliser applied.
Is commonly called 10% potassic super.
Is a mixture of 80% single superphosphate and 20% potassium chloride.
If ammonium sulphate (SOA; 20-0-0-23) costs $544 per tonne and urea (46-0-0-0) costs $845 per tonne, then:
The cost of nitrogen (N) in urea is $ 0.85 per kg N.
The “apparent” cost of N in SOA is approximately $1.80 per kg N, if sulphur (S) is valued at $ 0.80/kg S.
The “apparent” cost S in SOA is approximately $ 1.63 per kg S, if the value of N in SOA is the same as the value of N in urea.
200 kg of SOA supplies 46 kg N.
The “apparent” cost of N in SOA is approximately $1.80 per kg N, if sulphur (S) is valued at $ 0.80/kg S.
Cost of N in Urea
Urea (46-0-0-0) = $845/t
46% N = $845/t
460 kg N = $845
1 kg N = $1.84/kg N for urea
Cost of N in ammonium sulphate IF S = $0.80/kg S
Ammonium sulphate (20-0-0-23) = $544/t
1 t SOA = 230 kg S
Cost of S = 230 kg S x $0.80/kg S
= $184 for S
Cost of N = $544 for SOA - $184 for S
= $360 for N
1 t SOA = 200 kg N at $360
= $1.80/kg N for SOA
Cost of S in ammonium sulphate IF $/kg N in SOA = $/kg N in urea
Cost of N in urea = $1.84/kg N
1 t SOA = 200 kg N
Cost of N = 200 kg N x $1.84/kg N
= $368
Cost of S = $544 for SOA - $368 for N
= $176 for S
1 t SOA = 230 kg S at $176
= $0.77/kg S for SOA
N Supply at 200 kg SOA
1 t SOA = 200 kg N
100 kg SOA = 20 kg N
200 kg SOA = 40 kg N
If an Allophanic soil, under grazed dairy pasture, requires a capital fertiliser application of 11 kg P/hectare to increase the Olsen P test by 1 mg P/L soil then:
489 kg/hectare/year of single super phosphate (0-9-0-11) will increase Olsen P by 4 mg P/L soil and provide a the maintenance P fertiliser requirement of 32 kg P/ha/yr.
356 kg/hectare/year of single super phosphate (0-9-0-11) will increase Olsen P by 2 mg P/L soil and provide a maintenance P fertiliser requirement of 32 kg P/ha/yr.
257 kg/hectare/year of triple superphosphate (0-21-0-1) will increase Olsen P by 2 mg P/L soil and provide a maintenance P fertiliser requirement of 32 kg P/ha/yr.
210 kg/hectare/year of triple superphosphate (0-21-0-1) will increase Olsen P by 4 mg P/L soil and provide a maintenance P fertiliser requirement of 32 kg P/ha/yr.
257 kg/hectare/year of triple superphosphate (0-21-0-1) will increase Olsen P by 2 mg P/L soil and provide a maintenance P fertiliser requirement of 32 kg P/ha/yr.
Increasing Olsen P = 11 kg P/ha → +1 mg P/L soil
Single superphosphate (SSP) = 0-9-0-11
Triple superphosphate (TSP) = 0-21-0-1
489 kg SSP/ha/yr
489 kg SSP at 9% P = 44.01 kg P/ha/yr
Increasing Olsen P = 44/11 = +4 mg P/L soil
44.01 kg P/ha/yr application > 32 kg P/ha/yr maintenance
356 kg SSP/ha/yr
356 kg SSP at 9% P = 32.04 kg P/ha/yr
Increasing Olsen P = 32/11 = +2.91 → +3 mg P/L soil
32.04 kg P/ha/yr application = 32 kg P/ha/yr maintenance
257 kg TSP/ha/yr
257 kg TSP at 21% P = 53.97 kg P/ha/yr
P application = Maintenance P + Increase in Olsen P
32 kg P/ha/yr maintenance + (+2 mg P/L soil x 11 kg P/ha)
= 32 + 22
= 54 kg P/ha/yr = 53.97 kg P/ha/yr
210 kg TSP/ha/yr
210 kg TSP at 21% P = 44.1 kg P/ha/yr
Increasing Olsen P = 44/11 = +4 mg P/L soil
44.1 kg P/ha/yr application > 32 kg P/ha/yr
Potential negative effects of fertiliser application can include:
Increased risk of gaseous emissions of phosphorus (P) following applications of P fertiliser.
The accumulation of metal contaminants in the soils from applications of nitrogen (N) fertilisers.
Increased risk of gaseous emissions of potassium (K) following applications of K fertiliser.
The increased risk of nitrate leaching in drainage water following applications of N fertilisers.
The increased risk of nitrate leaching in drainage water following applications of N fertilisers.
Which following statement is true?
The conversion of ammonium to nitrate in the soil generates acidity.
The conversion of elemental sulphur to sulphate in the soil increases soil pH.
The conversion of nitrate to nitrous oxide in the soil generates acidity.
With carbon dioxide, from respiration in the soil, combines with soil water then there is an increase in soil pH.
The conversion of ammonium to nitrate in the soil generates acidity.
Copper:
Is an essential micronutrient for plants, but not for animals.
Deficiency is more common in sheep than it is in deer.
Deficiency is a common cause of reduced fixation of atmospheric N by rhizobium in clover.
Is recommended to be applied to soils at a rate of 10 kg copper sulphate/ha.
Is recommended to be applied to soils at a rate of 10 kg copper sulphate/ha.
In plants, the role of:
Nitrogen (N) is strongly involved with the transfer of chemical energy.
Phosphorus (P) is strongly involved with protein synthesis.
Sulphur (S) is strongly involved with charge balancing to maintain electrical neutrality.
Potassium (K) is strongly involved with charge balancing to maintain electrical neutrality.
Potassium (K) is strongly involved with charge balancing to maintain electrical neutrality.
The Law of the Minimum:
States that the increase in crop yield per unit of fertiliser applied becomes smaller as fertiliser application rate increases.
States that the amount of yield attainable is controlled by the factor that is present in the most limiting amount in a soil.
States that the increase in crop yield per unit of fertiliser applied is always linear as fertiliser application rate increases
None of the above.
States that the amount of yield attainable is controlled by the factor that is present in the most limiting amount in a soil.
A soil has a total phosphorus (TP) content of 0.24% and a bulk density of 900 kg soil/m3. How much P is contained in a hectare in the top 0-7.5 cm soil depth:
16.2 kg P/hectare
162 kg P/hectare
1,620 kg P/hectare
216 kg P/hectare
1,620 kg P/hectare
Soil depth = 0 - 7.5 cm = 7.5 cm → 0.075 m
1 ha = 10,000 m2
Volume of Soil per ha
Volume = Area x Depth
= 10,000 m2 × 0.075 m
= 750 m3
Mass of Soil
Mass of Soil = Volume x Bulk density
= 750 m3 × 900 kg soil/m3
= 675,000 kg soil
Total P
Total P = Total P content x Mass of Soil
= 0.24% P x 675,000 kg soil
= 0.0024 × 675,000 kg soil
= 1,620 kg P/ha
Which statement is true for a fertiliser product, supplied from an overseas supplier using the following grading system (N-P2O5-K2O-S), with a fertiliser grade of 10-10-16-5 (assume the molar mass of P = 31, K = 39, O = 16).
The fertiliser provides:
About 43.7 kg phosphorus (P) per tonne of product.
About 160 kg of potassium (K) per tonne of product.
The same amount of nitrogen (N) and (P).
More (P) than (S).
About 43.7 kg phosphorus (P) per tonne of product.
N-P2O5-K2O-S = 10-10-16-5
Nutrient Grade Conversion
Rule of thumb = % P = 2.3
P grade = 10/2.3 = 4.35
Rule of thumb = % K = 1.2
K grade = 16/1.2 = 13.3
NZ Nutrient Grade = 10-4.35-13.3-5
P per tonne of product
1 t product = 4.35% P
= 43.5 kg P ← Very close to 43.7 kg P
K per tonne of product
1 t product = 13.3% P
= 133 kg K ← NOT 160 kg K
N & P
1 t product = 10% N & 4.35% P
= 100 kg N & 43.5 kg P
Ans: NOT equal
More P than S?
P = 43.5 kg
S = 5% S in 1 t product
= 50 kg S in 1 t product
Ans: S > P
Which statement is true?
The two main contaminants in phosphorus fertilisers are cadmium and mercury.
The fertiliser contaminant cadmium accumulates in soils but is not taken up by plants.
The fertiliser contaminant cadmium is held in soils by sorption onto hydrous oxides of aluminium and iron.
Nitrogen, potassium and sulphur fertilisers contain negligible amounts of contaminants.
Nitrogen, potassium and sulphur fertilisers contain negligible amounts of contaminants.
During the initial stages of soil development:
The rate at which plant available nitrogen (N) is released from soil minerals often influences the rate of natural soil fertility development.
Chemical weathering is an important process influencing the release of phosphorus (P) from soil parent materials.
Sulphur fixed by rhizobium, in legume root nodules, is the main source of sulphur available to plants.
Phosphorus deposited in rainfall is the only source of phosphorus made available to plants.
Chemical weathering is an important process influencing the release of phosphorus (P) from soil parent materials.
In the soil nitrogen (N) cycle, the process of:
Denitrification converts nitrate to nitrous oxide gas.
Hydrolysis converts ammonium to urea.
Fixation converts ammonium to nitrate.
Volatilisation involves the loss of nitrate gas.
Denitrification converts nitrate to nitrous oxide gas.
When a forest is cleared by burning:
There is a decreased risk of nutrient loss via surface runoff and erosion processes.
All of the nitrogen (N) from the forest biomass remains in the ash.
A relatively alkaline ash is produced.
Most of the phosphorus (P) from the forest biomass is lost during combustion.
A relatively alkaline ash is produced.
Good quality agricultural lime made from limestone:
Is a good source of plant available phosphorous (P).
Has at least 50% of its weight containing particle sizes < 0.5 mm.
Is required in greater quantities for a soil with a low CEC compared to a soil with a high CEC, to achieve the same change in soil pH.
Has a Calcium Carbonate Equivalent (CCE) value of < 80%.
Has at least 50% of its weight containing particle sizes < 0.5 mm.
Nitrogen fertiliser applications on pastoral dairy farms:
Are recommended when soil temperature is < 7 oC.
Will increase N fixation by clover in mixed grass and white clover pasture swards.
Are recommended to be mostly applied in summer months on non-irrigated farms.
Are typically applied at applications rates of < 50 kg N/ha/application.
Are typically applied at applications rates of < 50 kg N/ha/application.
List four (4) trace elements (micronutrients) and state whether they are essential for either plants, animals or both.
For three (3) of these trace elements, describe why they are essential, what the main deficiency symptoms in plants or animals are, and the common form of fertiliser used for their application. [5 marks]
Provide the meaning for both ‘Maintenance’ and ‘Capital’ fertiliser requirements.
Then describe three (3) factors that would increase the phosphorus fertiliser maintenance requirements for block of land on a dairy farm. [5 marks]
Describe what two (2) meanings can be given to the statement; “a dairy cow uncouples nutrient cycles”.
Also, explain how each type of “uncoupling” influences nitrogen availability for pasture growth and/or the losses of the different forms of nitrogen from the soil. [5 marks]
Compare and contrast the main differences in the nitrogen (N) cycle between pasture soils and cropped soils, including differences in the inputs and losses of N from the soil. [5 marks]
Describe the main differences and similarities between the phosphorus (P) and potassium (K) cycles in farmed soils.
This comparison can include where these nutrients are mostly found in soils (i.e. soil pools), the main soil transformations/processes and the major inputs/losses. [5 marks]
a) From the following soil test results (see table below), complete the table by providing the missing values for basic cations, exchange acidity, cation exchange capacity (CEC) and percentage Base saturation.
For ‘Soil 3’, the magnesium (Mg2+) value first needs to be converted from 0.312 g Mg/kg soil into units of centimoles charge/kg soil (molar mass of Mg = 24 g/mole; valency of 2).
Provide all values to 1 decimal place: [3.5 marks]
centimoles charge/kg | Soil 1 | Soil 2 | Soil 3 |
K+ | 0.8 | 0.4 | 0.6 |
Na+ | 0.5 | 0.2 | 0.3 |
Ca2+ | 12.8 | A | 8.5 |
Mg2+ | 3.9 | 1.6 | B |
Exchange acidity | C | 12 | 3 |
Cation exchange capacity (CEC) | D | 20 | E |
Base saturation (%) | 60 | F | G |
b) Assuming that the three soils in (a) are the same type of soil with similar soil textures, which soil is likely to have the highest soil organic matter content? [0.5 marks]
Soil 1
Soil 2
Soil 3
c) Assuming that the three soils, in Answer a), are the same type of soil with similar soil textures, which soil is likely to have the lowest soil pH level? [0.5 marks]
Soil 1
Soil 2
Soil 3
d) Assuming that the three soils, in Answer a), are the same type of soil with similar soil textures, which soil is likely to have the greatest ability to buffer a change in soil pH? [0.5 marks]
Soil 1
Soil 2
Soil 3
a) From the following soil test results (see table below), complete the table by providing the missing values for basic cations, exchange acidity, cation exchange capacity (CEC) and percentage Base saturation.
For ‘Soil 3’, the magnesium (Mg2+) value first needs to be converted from 0.312 g Mg/kg soil into units of centimoles charge/kg soil (molar mass of Mg = 24 g/mole; valency of 2).
Provide all values to 1 decimal place: [3.5 marks]
centimoles charge/kg | Soil 1 | Soil 2 | Soil 3 |
K+ | 0.8 | 0.4 | 0.6 |
Na+ | 0.5 | 0.2 | 0.3 |
Ca2+ | 12.8 | (A) 5.8 | 8.5 |
Mg2+ | 3.9 | 1.6 | (B) 2.6 |
Exchange acidity | (C) 12 | 12 | 3 |
Cation exchange capacity (CEC) | (D) 30 | 20 | (E) 15 |
Base saturation (%) | 60 | (F) 40 | (G) 80 |
Soil 1
D
Base saturation = (Sum of base cations/CEC) x 100
60 = ((0.8 + 0.5 + 12.8 + 3.9)/CEC) x 100
0.6 = 18/CEC
0.6 CEC = 18
CEC = 30 cmol charge/kg
C
CEC = Sum of base cations + Exchange acidity
30 = 18 + Exchange acidity
Exchange acidity = 12 cmol charge/kg
Soil 2
A
CEC = Sum of base cations + Exchange acidity
CEC = Sum(K+ + Na+ + Ca2+ + Mg2+) + Exchange acidity
20 cmol charge/kg = Sum(0.4 + 0.2 + Ca2+ + 1.6) + 12 cmol charge/kg
Ca2+ = 20 - 14.2
= 5.8 cmol charge/kg
F
Base saturation = (Sum of base cations/CEC) x 100
= 8/20 × 100
= 40%
Soil 3
B
Mg molar mass = 24 g/mol
Valency = 2 ← Mg2+
0.312 g Mg/kg soil / 24 g/mol
= 0.013 × 2
= 0.026 × 100
= 2.6 cmol charge/kg soil
E
CEC = Sum of base cations + Exchange acidity
= Sum(0.6 + 0.3 + 8.5 + 2.6) + 3
= 15
G
Base saturation = (Sum of base cations/CEC) x 100
= (12/15) x 100
= 80%
b) Assuming that the three soils in (a) are the same type of soil with similar soil textures, which soil is likely to have the highest soil organic matter content? [0.5 marks]
Soil 1
High CEC → High organic matter
c) Assuming that the three soils, in Answer a), are the same type of soil with similar soil textures, which soil is likely to have the lowest soil pH level? [0.5 marks]
Soil 2
Low base saturation → Low pH
d) Assuming that the three soils, in Answer a), are the same type of soil with similar soil textures, which soil is likely to have the greatest ability to buffer a change in soil pH? [0.5 marks]
Soil 1
High CEC → Greater ability to buffer a change in soil pH