RelativeResourceManager(12) - Answer a ppm K = 383.18 b...

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12 of 17 To obtain lbs./acre: soil in Ca ppm 2400 = deep 6" acre 1 soil lbs 10 2 soil lbs 10 1 Ca lbs. 2400 6 6 = deep 6" acre 1 Ca lbs. 4800 Try working these examples yourself using the previous examples as a guide. Example 1 . Given: 50 g of soil are extracted with 200 ml of NH 4 OAc. The extract is diluted by a factor of 25. Upon analysis of the final solution, 4 ppm of K+ are found. Calculate: a. ppm of K in soil b. pp2m of K in the soil c. lbs./acre of K d. meq K/100 g of soil Answer: a. 400 ppm K b. 800 pp2m K c. 800 lbs. K/acre 6" d. 1.02 meq K/100 g soil Example 2 . You receive an analysis from a soil testing lab reporting 0.98 meq K/100 g soil. Convert this to: a. ppm K in soil (using 39.1 g/eq. K) b. lbs. K/acre 6"
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Unformatted text preview: Answer: a. ppm K = 383.18 b. 766.36 lbs. K/acre 6" Example 3 . Convert 100 lbs. P/acre 6" to ppm P. Answer: 50 ppm P Example 4 . A 5 g soil sample is leached with NH 4 OAc for the determination of CEC. After distillation of the sample, 12.5 ml of 0.10 N H 2 SO 4 solution is needed to titrate the boric acid indicator to its endpoint. Calculate the CEC of the soil. Answer: 25 meq/100 g Example 5 . A lab reports a Ca/Mg ratio of 2.1 (meq Ca/meq Mg) and a concentration of Ca in your soil of 1600 ppm. Calculate: a. meq Ca/100 g soil b. meq Mg/100 g soil c. ppm Mg Answer: a. 8 meq Ca/100 g b. 3.81 meq Mg/100 g c. 462.9 ppm Mg...
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This note was uploaded on 06/21/2011 for the course NRES 201 taught by Professor Olson,k during the Spring '08 term at University of Illinois, Urbana Champaign.

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