Suppose you have sampled the soil from the plow layer of a portion of a field. This volume of soil is presumably the volume of greatest root growth, and you wish to precipitate all exchangeable aluminum throughout the furrow depth, which you consider to be 10 inches. This soil has an average bulk density of 1.18 g/cm³. Suppose you ran the SMP buffer test as described in the lecture notes on a sample of this soil, then filtered the suspension, then used atomic absorption spectroscopy to measure that 602 mg Al/kg soil were extracted into the solution phase (presumably from the cation exchange complex). If we assume this extract is all exchangeable Al³⁺, then how much lime (CaCO₃) in tons per acre would you need to exactly replace and neutralize all the exchangeable aluminum in the top 10 inches of this soil? To do this, you could calculate the amount of limestone needed per kg soil, and then multiply that by the mass of the acre-furrow-slice (AFS) of this soil. The AFS calculation is a bit tedious, so please use the spreadsheet that you made in an earlier homework problem. How much limestone is needed, to the nearest 0.01 ton CaCO₃/AFS?

2,000,000 lbs/AFS