Soil carbon and fertiliser decisions on a degraded paddock
A 200 ha paddock has been cropped continuously for 40 years. Soil tests show organic carbon of 0.9% in the top 10 cm (bulk density 1.4 g/cm³), pH 4.8 (CaCl₂), and available phosphorus at 12 mg/kg. The recommended range for the crop is pH 5.5–7.0 and phosphorus above 20 mg/kg. The farmer can afford one major intervention this year: 2 t/ha of lime, 60 kg/ha of phosphorus fertiliser, or converting 40 ha to a perennial pasture. A nearby uncleared woodland soil has 2.4% organic carbon and pH 6.1. Rainfall is 480 mm/year and the paddock slopes 3–5%.
1. Calculate the organic carbon stock in the top 10 cm of the paddock and compare it with the woodland, then explain the significance.
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Paddock:
- Soil mass per hectare in top 10 cm = 0.10 m × 10,000 m² × 1,400 kg/m³ = 1,400,000 kg/ha.
- Carbon = 0.9% × 1,400,000 = 12,600 kg C/ha = 12.6 t C/ha.
- Over 200 ha: 2,520 t C.
Woodland:
- Same bulk density assumed: 2.4% × 1,400,000 = 33,600 kg C/ha = 33.6 t C/ha.
Difference: The paddock has lost about 21 t C/ha, or roughly 4,200 t C over the paddock - a substantial depletion.
Significance: Loss of organic carbon reduces cation exchange capacity, water-holding capacity, aggregate stability and nutrient supply. Combined with pH 4.8 (below the crop range), the soil is both acidic and carbon-poor, which limits yields and increases erosion risk on a 3–5% slope. The woodland provides a realistic target for what the soil could support. Note that the comparison assumes equal bulk density, which should be verified.
2. Evaluate the three intervention options against the soil constraints and the budget.
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Lime (2 t/ha): Directly addresses the most limiting factor, pH 4.8. Raising pH toward 5.5–6.0 improves nutrient availability (especially phosphorus and molybdenum) and reduces aluminium toxicity. However, it does nothing for the low carbon or the low phosphorus, and 2 t/ha may be insufficient on a clay-rich soil - the required rate depends on soil texture and buffer capacity.
Phosphorus (60 kg/ha): Addresses the low P (12 mg/kg vs >20 target), but at pH 4.8 much of the added P will be fixed by aluminium and iron oxides and become unavailable. Applying P before correcting pH is inefficient.
Perennial pasture (40 ha): Builds organic carbon over years, improves structure and infiltration, and reduces erosion on the slope. But it takes years to show benefit, removes 40 ha from cropping, and does not correct acidity.
Judgement: Lime is the highest-priority single intervention because pH is the master variable controlling nutrient availability and because P applied at low pH would be largely wasted. The pasture conversion is the best long-term investment for carbon and erosion but is slower. With one year's budget, lime first, then P once pH rises, then pasture. The 40 ha conversion could be staged on the steepest land first.
3. Predict the likely effect of the 3–5% slope on soil carbon and fertility over the next decade if nothing changes, and explain the mechanism.
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Prediction: Organic carbon and fertility will continue to decline, and the decline will be uneven - greatest on the steeper parts.
Mechanism:
- Low organic carbon means weaker aggregates, so the surface is more easily detached by raindrop impact.
- On a 3–5% slope, runoff gains enough energy to transport detached soil downslope. Water erosion preferentially removes the light, carbon-rich surface layer.
- Loss of topsoil exposes subsoil with lower carbon and fertility, reducing plant growth and therefore litter return - a positive feedback of decline.
- Reduced infiltration increases runoff, worsening the erosion.
- Acidification may continue if the cause (e.g. removal of alkalinity in produce, or fertiliser use) persists.
Implication: The paddock is on a trajectory toward the degraded state, and the woodland soil represents the carbon that is being lost. Intervention should target erosion control (contours, perennial cover on steep areas) as well as chemistry, because lime alone will not stop physical loss of topsoil.