Question

The top 10 cm of a forest soil in VT has a bulk density of 1.2 g/cm3 and 20 centimoles of charge from exchangeable Ca2" per 1 kg of soil. There is 500 kg of Ca2* per hectare in the trees. If the trees are cut and removed, will there be enough calcium in the top 10 cm to regrow the forest (assume no input from mineral weathering or lower horizons)? (hint: because Ca2* is divalent, each mole of charge corresponds to 20 g of calcium)

Answer 1

Bulk density is the weight of soil in a given volume.

Bulk density of Forest Soil : 1.2 g/cm3. = 1200 kg/m3.

20 centimoles of Charge / kg of soil.

It is in form of exchangeable Ca²⁺ .

Since Ca is a divalent cation, it have equivlent wt. = 20 g / eq.

we have, 1 centimole = 0.01 moles.

thus charge , is 0.20 moles/kg = 0.20 equivalent , Ca2+.

thus mass of Ca : 0.20equivalent* 20 g/eq. = 4 g /Kg.

( top soil - 10cm = 0.1 m ; 1 ha = 100 m*100 m)

Volume of 1ha. of top soil : 0.1 m*100 m*100 m = 1000 m³.

thus, amount of Ca in 1 Ha of forest Top soil : soil volume*bulk density* Ca/kg of top soil

= (1000 m3 * 1200 kg/m3 * 4 g/Kg) / 1000 g/ kg = 4800 Kg.

If all trees are cut and removed than also soil contains enough Ca to regrow the same forest.