Question

a 2.65g sample of soil was placed in a solution of ammonium acetate (NH4C2H3O2) and shaken for five minutes. After shaking the mixture was filtered, and the filtrate was analyzed. the filtrate was found to contain the following ions:

0.27 mmol Na+, 0.89 mmol K+ and 0.019 mmol Ca2+. what is the cation exchange capacity of soil? Give results in cmol/kg.

Answer 1

Cation exchange capacity of soil = 43.916 cmol/kg

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No of moles of Na+ ion = 0.27 mmol

No. of moles of K+ ion = 0.89 mmol

No. of moles of Ca2+ ion = 0.019 mmol

Since Ca2+ ions have 2+ charges, it can be considered two equivalent of mono positive ions.

No. of moles of cation equal to Ca2+ ion= 0.0019 x 2 = 0.0038 mmol.

Total moles of monovalent cations = 0.27 + 0.89 + 0.0038 = 1.1638 mmol

Mass of soil = 2.65 g = 2.65/1000 = 0.0027 kg

Cation exchange capacity of soil = 1.1638 mmol / 2.65 g = 1.1638 mmol / 0.00265 kg = 439.1698 mmol/kg

Converting to centimol (cmol) from millimol (mmol), 1 cmol = 10 mmol (1 mmol = 0.1 cmol)

So, 439.1698 mmol/kg = 439.1698 x0.1 cmol/kg = 43.916 cmol/kg