• Sodium hydroxide, 0.5 M, carbonate free
• Phenolphthalein indicator - dissolve 0.25 g phenolphthalein in 150 ml 95% v/v ethanol and dilute to 250 ml with water.
Procedure (moisture determination). From about 2 kg of bulk sample, cone and quarter or by other means obtain a representative sample of about 200 g. If determination of fineness is to be carried out, duplicate approximately 100-g samples should be taken and any soft lumps disintegrated by lightly crushing. The one for fineness should not be ground further; the other should be rapidly ground until it completely passes a 1.0 mm sieve (test sieves conforming to British Standard 410: 1986 are suitable). Mix the ground sample well and form into a flattened cone. Taking random portions with a spatula, weigh 5 g of the prepared sample to the nearest 0.001 g, and transfer to a previously weighed container with airtight lid. Place the uncovered container and the lid in an oven and maintain at 100°C (AOAC: 110°C) for 2-3 h. Replace the lid on the container, remove from the oven and allow to cool in a desiccator and weigh. Reheat for another hour, cool and reweigh. If the difference in weight exceeds 0.01 g continue the heating and cooling procedure until a weight constant within 0.01 g is attained. Calculate the total loss of weight and express it as a percentage of the original weight, which gives the percentage moisture in the fertilizer sample as received.
Procedure (neutralizing value). At the same time as weighing the sample for moisture determination, weigh 0.5 g (AOAC: 1 g, but 0.5 g for CaO or Ca(OH)2), or x g, of the prepared sample to the nearest 0.001 g, and transfer to a 300-ml (AOAC: 250-ml) conical flask. Add 50 ml of 0.5 M hydrochloric acid, cover the flask with a watch glass and boil the contents gently for 5 min. Cool the mixture to room temperature, add 2 or 3 drops of the phenolphthalein indicator and titrate with 0.5 M sodium hydroxide solution to the end point of the indicator.
Calculation. Determine the amount of hydrochloric acid (y ml) consumed by the sample. This is done by subtracting the titre of 0.5 M NaOH from 50 (the volume of 0.5 M HCl added to the liming material). The reactions for HCl on limestone and the subsequent back-titration of the excess HCl are:
CaCO3 + 2HCl = CaCl2 + H2O + CO2t ; HCl + NaOH = NaCl + H2O
From the reaction equations, 1 mol of CaCO3 (molar mass 100.087) is neutralized by 2 mol of HCl. A volume of 1 l of 1 M HCl will neutralize 100.087/2 g CaCO3 = 50.04 g CaCO3. Therefore, 1 l of 0.5 M HCl will neutralize 25.02 g CaCO3, and 1 ml 0.5 M HCl will neutralize 0.02502 g CaCO3. The result has to be expressed in terms of CaO (molar mass 56.077). Therefore, 1 ml 0.5 M HCl will neutralize the equivalent of 0.02502 x 56.077/100.087, or 0.01402 g CaO.
The neutralizing value is expressed as a percentage by weight of calcium oxide (CaO) and refers to the undried sample as received. Thus the formula becomes:
Neutralizing value = y x 0.01402/x x 100% CaO, where x is the sample weight and y is the titre.
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