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

# If the CEC is 5 cmol/kg of soil and the acidity is 1 mol/Kg of soil , then one-fifth of the exchange sites in the soil are occupied by acidic hydrogen and aluminum ions. Here we have to assume that Al³⁺ is absent

# The remaining 4 cmol/kg of soil (or 80% of the CEC) will be occupied by the basic cations.

# 10% of 50 cmol/kg is nothing but 5 cmol/kg .

  Therefore

If CEC is 5 cmol/kg then acidity of the soil is 1 cmol/kg

1 cmol= 0.01mol

i,e., we have 1 cmol of H⁺ per kilogram of soil

Reaction of CaCO₃ with H⁺

1 CaCO₃ + 2H+ H₂CO₃ + Ca(OH)₂

1mol CaCO₃  is required to neutralise 2 mol H⁺

Therefore , for 0.01mol of  H⁺ per kilogram= (1/2 * 0.01) = 0.005mol of CaCO₃

From the mol of calcium carbonate , we can calculate mass.

Mass of CaCO3 = 0.005 mol x 100 g/mol [ since, Mass = mol x M.Wt ]

   = 0.5 g