In: Civil Engineering

A soil has the following properties: pH 8.6; ESP = 18%; CEC = 25 meq/100g; CaCO3 = 2%. (Question 32, Chapter 3 in text), acre furrow slice (afs) = 2 x 106 lbs. soil.

a. Calculate the SAR

b. A test lab recommended 4,300 lbs gypsum/afs. Calculate the final ESP if the grower followed the recommendation.

c. If the grower added elemental S instead of gypsum, calculate the final CaCO3 content

Ans a) Sodium adsorption ratio (SAR) = Na^{+} /
[0.5(Ca^{+2} + Mg^{+2})^{0.5}]

where sodium , calcium and magnesium concentrations are in milliequivalents /Litre

Also,

ESP = Na^{+} / CEC

=> Na^{+} = ESP x CEC

= 0.18 x 25

= 4.5 meq/L

Ca^{+2} = 0.02 g/L or 20 mg/L

= (20/50) meq/L

= 0.4 meq/L

SAR = 4.5 / [0.5(0.4 + 0)^{0.5}]

SAR = 10.17

Ans b) Now, 4300 lb gypsum added to 2 x 10^{6} lb
soil

or 0.00215 lb / lb soil

or 0.00215 g/ g soil

or 0.215 g /100g soil

or 215 mg/100 g soil

=> (215/50) meq/100 g soil

=> 4.3 meq/100gm

New CEC = 25 + 4.3 = 29.3 meq/100 gm

ESP = 4.5 / 29.3

ESP = 15.36 %

Ans c) Elemental S added will oxidize to produce 2H^{+},
which will dissolve exactly 2.4 meq CaCO3 /100 g soil to produce
2.4 meq Ca^{+2} /100 g of soil.

=> 4300 lb S / 2 x 10^{6} lb soil

=> 4300 g S/ 2 x 10^{6} g soil

or 0.215 g /100 g soil

or 4.3 meq/100gm

Ca^{+}^{2} produced = 4.3 x 2.4 meq/100gm

= 10.32 meq/100 gm

= 10.32 / 50 mg / 100gm

= 0.2064 mg/100 gm

= 0.0002064 g/100 gm

= 2.064 %

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