Question

A soil with a CEC of 50 cmol(-)/kg has 10% of its exchange sites occupied by hydrogen ions. There is no aluminum in the system.

a) What is the exchangeable acidity of the soil? (2 pt)

b) If one mol of calcium carbonate can neutralize 2 mol of hydrogen ions, what mass of calcium carbonate would be necessary to neutralize all of the exchangeable acidity in one kilogram of this soil? (2 pts)

Answer 1

A) Exchangeable acidity is given by number of H+ or Al3+ ions in exchange capacity of soil.

As Al3+ are not present, exchangeable acidity is due to H+ only

% of exchange sites = 10%

Cmol (-) of H+ on exchange = 10% * 50 Cmol (-) /kg = 5 Cmol (-) /kg

Molar mass of H = 1 g/mol

So, 1 mol of H+ will have 100 Cmol (-)

Amount of H+ ions = 0.05 g/Kg

Exchangeable acidity of soil = 0.05 g/Kg

B) 1 mol of CaCO3 neutralizes 2 mol of H+, so, 1 cmol of CaCO3 = 2 Cmol (-) of Ca2+.

mass of 1 Cmol of CaCO3 = 1 g

For 2Cmol (-) of Ca2+, required Cmol of CaCO3 = 1/2 = 0.5 g

So, Cmol of CaCO3 required = 0.5g/Cmol (-)

Total H+ = 5 Cmol (-) /kg soil

So, CaCO3 required = 0.5 g/Cmol (-) * 5 Cmol (-) /kg soil i.e. 2.5 g/kg

IN both these solutions, Cmol (-) is centimol charge and Cmol is centimol