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1. <br />2. <br />3. <br />4. <br />5. <br />6 <br />]T OF NEUTRALIZATION POTEt7TIAL OF A <br />)IL OR OVERBURDEN MATERIAL <br />/, <br />Weigh 2.00± .O1 grams of sample, ground to pass a 60 mesh <br />(250 mm) sieve, into a 250 ml Erlenmeyer flask. <br />Carefully pipet 20.00 ml of 0.1 N_ HCL (the normality of which <br />is known exactly) into the flask <br />Heat nearly to boiling until reaction (acid + carbonates) <br />is complete, 5 minutes is usually sufficient. <br />Add H2O to a total volume of 150 ml, boil 1 minute; cool. <br />Titrate using 0.1 N NaOH (concentration exactly known), <br />to pH 7.0 using pH meter. <br />a) If the pH of the suspension is greater than 7.0 <br />prior to beginning the back titration with NaOH, <br />it can be assumed that there is a CaCO3 equivalent <br />of over 50 tons per thousand tons of material. <br />b) If an exact value of this high neutralizing capacity <br />is desired, rerun the sample using a greater amount <br />of acid initially, or using above procedure but <br />substituting 1.0 N NaOH. <br />Calculate N.P. using equation (c)2. below. <br />1, <br />a) ml acid consumed by sample = ml of acid to sample, minus <br />ml base req'd to neutralize sample X ml of acid (or1=~) in a f:Lask <br />ml of base req'd to neutralize <br />b) parts CaCO equivalent/million parts of soil - mi acid <br />consumed by sample X N_ of acid X 100 X <br />grams of sample used <br />10,000 X 50 g of CaCO3 <br />1,000 1 g of H+ <br />c) for a 2.0 gram sample: <br />1. Tons CaCO equivalent/1000 tons = ml acid consumed <br />by sample3X 25,000 X N of acid <br />1,000 <br />2. Tons CaCO equivalent/thousand tons of soil = <br />ml X 25.03X N of acid. <br />d) Maximum CaCO3 requirement for neutralization of acid <br />developed from total sulfur = $S X 31.25 (assuming all " <br />sulfur occurs as pyrite or marcasite). <br />6-59 <br />