Can someone please help me answering this question regarding soil chemistry?
A sandy soil has a CEC = 5 meq/100g and a K saturation = 6%.
An alfalfa crop yields 5 ton/ac/yr at 3% K.
Calculate initial exchangeable K and estimate final exchangeable K. Where did the additional K in the crop come from?
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To answer your question, let's go step by step:
1. Calculate initial exchangeable K:
- The K saturation is given as 6% in a sandy soil.
- The CEC (cation exchange capacity) is given as 5 meq/100g.
- Since K saturation represents the percentage of the CEC occupied by K, we can calculate the initial exchangeable K as follows:
Initial exchangeable K = K saturation * CEC
Initial exchangeable K = 6% * 5 meq/100g
Initial exchangeable K = 0.06 * 5 meq/100g
Initial exchangeable K = 0.3 meq/100g
2. Estimate final exchangeable K:
- The alfalfa crop yields 5 tons/ac/yr at 3% K.
- To estimate the final exchangeable K, we need to calculate the amount of K removed by the crop and then reconcile it with the initial exchangeable K.
- First, convert the yield from tons/ac/yr to grams/hectare/yr. Since 1 acre is approximately 0.4047 hectares, we have:
Yield (grams/hectare/yr) = yield (tons/ac/yr) * 1000 kg/ton * 10000 g/hectare / (acre to hectare conversion factor)
Yield (grams/hectare/yr) = 5 tons/ac/yr * 1000 kg/ton * 10000 g/hectare / (0.4047 acre/hectare)
Yield (grams/hectare/yr) ≈ 123,305 kg/hectare/yr
- Now, calculate the amount of K removed by the crop:
K removed (grams/hectare/yr) = Yield (grams/hectare/yr) * K concentration of the crop
K removed (grams/hectare/yr) = 123,305 kg/hectare/yr * 0.03 (3% expressed as a decimal)
K removed (grams/hectare/yr) ≈ 3,699.15 g/hectare/yr
Note: It's important to remember that this estimate assumes that all the K uptake by the crop is from the exchangeable K in the soil.
- Finally, reconcile the initial and final exchangeable K:
Final exchangeable K = Initial exchangeable K - K removed
Final exchangeable K = 0.3 meq/100g - (3,699.15 g/hectare/yr / 100 g/kg) / (10,000 g/kg) * 1000 meq/g
Final exchangeable K ≈ 0.3 meq/100g - 0.369915 meq/100g
Final exchangeable K ≈ -0.069915 meq/100g
The estimated final exchangeable K is approximately -0.069915 meq/100g, which means it is depleted. This suggests that the crop removed more K than was initially available in the soil.
3. Where did the additional K in the crop come from?
Since the final exchangeable K is negative, it means that the crop took up more K than was present in the initial exchangeable K. In this case, the additional K in the crop most likely came from non-exchangeable K sources, such as mineral weathering or organic matter decomposition. These sources release K into the soil, which can then be taken up by the crop.