Question

Based on the sample calculation, the CEC of the soil was 50.8 meq/100g and the equivalent weight of Ca2+ (one of the base cations) extracted by ammonium acetate solution was 3.13 meq/100g.

The ammonium acetate solution also contains 150 mg K+/L, 75 mg Na+/L, and 83 mg Mg2+/L. calculate the equivalent weight of K+, Na+, and Mg2+ in a 5 g soil: 35 mL extractant. Given all these values, compute the base saturation of the soil.

Answer 1

Equivalent weight of Ca^+2 = 3.13 meq/100 g (given) K^+ present = 150 mg/L (given) Na^+ present = 75 mg/L (given) Mg^+2 present = 83 mg/L (given) 35 ml of extractant is used then, k^+ present in extractant = 150 times 35/1000 Meq of K^+ = 150/39 times 35/1000 = 0.134 m.eq. Na^+ present in extractant = 75 times 35/1000 M.eq. of Na^+ = 75/23 times 35/1000 = 0.114 m.eq. Mg^+2 present in extractant = 83 times 35/1000 M.eq. of Mg^+2 = 83/24 times 35/1000 = 0.121 m.eq.