Skills developed: determining the cation exchange capacity (CEC) and base satura
ID: 493940 • Letter: S
Question
Skills developed: determining the cation exchange capacity (CEC) and base saturation of the soil. The cation exchange capacity (CEC) of a Soil is strongly affected by the nature and amount of mineral and organic colloid presenting the soil. Soils with large amounts of clay and organic matter have a higher CEC than sandy soils low in organic matter. Examples of CEC values for different textures in a Mollisols are as follows: Suppose that a soil contains 900 ppm Ca^2+ (mg/kg soil) 540 ppm AI^3+ (mg/kg soil) 585 ppm K^+ (mg/kg soil) 300 ppm Mg^2+ (mg/kg soil) Calculate the meq/100 g soil for each cation.Explanation / Answer
Answer :
Note some important conversions to be used in required conversions.
1) 1 g = 1000 mg
2) 1 ppm = 1 mg/Kg.
3) Equivalent weight of Metal cation = Atomic weight of metal / # of electrons transferred
so, 1 equivalent cation = Atomic weight of respective metal / Charge on cation.
4) 1 equivalent = 1000 miliequivalent.
5) 1 g/eq = 1000 mg/1000 meq = 1 mg/meq.
6) 1 ppm = 1 mg/Kg = 1mg/1000 g = 0.1 mg / 100 g
7) [Mn+] in (meq/100 g) = Concentration in (mg/100g) / Equivalent weight in mg/meq.
a) Let us first calculate equivalent weights of each cation involved in mg/meq.
i) Ca2+ : Atomic mass of Ca metal = 40.08 g/mole. Charge of Ca2+ = 2 units.
Equivalent weigt of Ca2+ = 40.08 / 2 = 20.04 g/eq = 20.04 mg/meq. (Conversion 5)
ii) Al3+ : Atmoic weight of Al = 27 g/mole. Charge on Al3+ = 3 units.
Equivalent weight of Al3+ = 27/3 = 9 g/eq = 9 mg/meq.
iii) K+ : Atomic mass of K = 39.1 g/mole, Charge on = 1
Equivalent weight of K+ = 39.1 / 1 = 39.1 g/eq = 39.1 mg/meq.
iv) Mg2+ : Atomic weight of Mg = 24.3 g/mole. Charge = 2
Equivalent weight of Mg2+ = 24.3 / 2 = 12.15 g/eq = 12.15 mg/meq.
b)
i) For i) Ca2+ :
i) [Ca2+]= 900 ppm = 900 mg/Kg = 900 mg/ 1000 g = 90 mg/ 100 g.
By conversion formula (7)
[Ca2+] in (meq/100 g) = 90 / 20.04 = 4.49 meq/100g.
ii) [Al3+] = 540 ppm = 54 mg/100 g
So, [Al3+] in (meq/100g) = 54 / 9 = 6 meq/100g.
iii) [K+] = 585 ppm = 58.5 mg/100g.
So, [K+] in (meq/100g) = 58.5 / 39.1 = 1.5 meq/100g.
iv) [Mg2+] = 300 ppm = 30 mg/100g.
So,[Mg2+] in (meq/100g) = 30 / 12.15 = 2.47 meq/100g.
========================XXXXXXXXXXXXXXXXXXXX======================
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.