Method 62aii Determination of watersoluble phosphorus manual method

This method and the following (6.2b.) are based on the methods given by Craven and Schwer (1960). The concentrations of standard solutions are expressed in terms of mg P2O5, sometimes incorrectly termed phosphoric acid. We have retained this format in order to give exact increments of concentration for the sequence of standards. It is now conventional, however, for the analytical method to express phosphorus concentration in terms of elemental P, and then convert to % P2O5, which is still used for the labelling of fertilizers.

Reagents.

• Nitric acid, approximately 70% m/m HNO3

• Phosphorus stock standard solution, 1000 pg P2O5 ml-1 (436 pg P ml-1) -dissolve 1.9173 g potassium dihydrogen orthophosphate, previously dried for 1 h at 105°C, in water. Transfer with beaker washings to a 1-l volumetric flask, make up to the mark and mix.

• Phosphorus intermediate standard solution, 200 pg P2O5 ml-1 (87.3 pg P ml-1) - dilute 20 ml of the phosphorus stock standard solution to 100 ml with water.

• Potassium dihydrogen orthophosphate

• Sodium hydroxide solution, 1 M

• Vanadium molybdate reagent - Note: wear PPE to prevent injury from concentrated nitric acid. Separately dissolve 20 g of ammonium molybdate and 1 g of ammonium vanadate in water, transfer to a 250-ml beaker and swirl to mix. Slowly add 140 ml nitric acid (approximately 70% m/m HNO3), carefully transfer to a 1-l volumetric flask, make up to the mark with water and mix.

Procedure (standard curve). Fill a 50-ml burette with the phosphorus intermediate standard solution and dispense 25, 26, 27, 28, 29, 30 and 31 ml into a series of 100-ml volumetric flasks. These will contain 50, 52, 54, 56, 58, 60 and 62 pg P2O5 ml-1 (21.8, 22.7, 23.6, 24.4, 25.3, 26.2, and 27.1 pg P ml-1). Add 25 ml of the vanadium molybdate reagent and make up to the mark with water (both liquids at 20°C), mix and stand for 10 min.

Using a matched pair of cells, place the lowest standard in the reference beam, and measure the absorbance of the other standards at 420 nm. Plot the standard curve relating absorbance to known concentration.

Procedure (sample analysis). Dilute an aliquot of the extract from Method 6.2a. to give (^ 25 ml) a solution containing 220-248 pg P2O5 ml-1 (96-108 pg P ml-1) at 20°C. Thus for super triplephosphate by the AOAC extraction, dilute 13 ml extract to 100 ml; and for the SI 1996 No. 1342 method, dilute 5.2 ml extract to 100 ml. Pipette 25 ml of this solution into a 100-ml volumetric flask, add 25 ml vanadium molybdate (20°C), make up to the mark with water (20°C), mix and stand for 10 min. Simultaneously prepare a fresh 50 pg P2O5 ml-1 (21.8 pg P ml-1) reference standard against which the absorbance of the samples are measured.

Calculation. Divide the concentration of the sample solution read from the standard curve by 3 (if diluted as in above sample analysis procedure), to obtain the % P, or divide by 1.304 to obtain the % P2O5 in the sample.

Explanation for the AOAC methodology extract: if the diluted sample concentration as read from the standard curve was 60 pg P2O5 ml-1, this corresponds to a concentration of 60 x 100/13 x 100/25 = 1846 pg P2O5 ml-1 = 1846 x 0.4364 = 805.7 pg P in the original extract (because 13 ml was diluted to 100 ml, and 25 ml of this solution was taken and diluted to 100 ml). There was 250 ml of original extract, therefore this volume would contain: 250 x 1846 pg P2O5 = 461,500/106 g P2O5 = 0.4615 g per 1.0 g sample, or 46.15% P2O5 in the original sample.

This is equivalent to 46.15 x 0.4364 = 20.14% P. Since 60 pg P2O5 ml-1 in the final sample solution corresponds to 46.15% P2O5 in the original sample, sample concentrations from the standard curve should be multiplied by 46.15/60 = 0.769 to give the % P2O5, or 0.336 to give the % P in the original sample.

For the SI 1996 No. 1342 extract, the condensed calculation is: 60 x 100/5.2 x 100/25 = 4615 pg P2O5 ml-1 = 2014 pg P in the original extract. There was 500 ml of original extract, therefore this volume would contain: 4615 x 500 pg P2O5 = 2.3075 g P2O5 per 5 g sample, or 46.15% P2O5, equivalent to 20.14% P in the original sample. Thus the same factor applies as with the AOAC extract, and sample concentrations from the standard curve should be multiplied by 0.769 to give the % P2O5, or 0.336 to give the % P in the original sample.

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