Method 711b Determination of potassium in plant material by flame photometry dry ashing extract

Standards

• Potassium stock standard, 1000 pg K ml-1 - dissolve 0.953 g KCl (dried for 1 h at 102°C) in water, add 1 ml hydrochloric acid, approximately 36% m/m HCl, dilute to 500 ml and add 1 drop of toluene.

• Potassium intermediate standard, 100 pg K ml-1 - pipette 25 ml of the potassium stock standard into a 250-ml volumetric flask, make up to the mark with water and mix.

• Potassium working standards, 0-50 pg K ml-1 - pipette 0, 10, 20, 30, 40 and 50 ml of the potassium intermediate standard into a series of 100-ml volumetric flasks, make up to the mark and mix.

Procedure. Switch on and set up the flame photometer according to the manufacturer's instructions. After a sufficient warm-up time, aspirate the 0 and 50 pg K ml-1 standards and adjust the zero and bring the maximum reading on-scale. Measure the other standards and construct a graph relating emission reading to concentration; check it is a straight line or smooth slight curve. Pipette 2.5 ml sample solution into a 100-ml volumetric flask, dilute to the mark with water and mix. Analyse the blank extract and the sample solutions in batches, and repeat the analysis of standards approximately every 10 min. If the output is on a chart-recorder, use a chart reader to assist the reading of the peaks (see Chapter 1, 'Chart reader'). Re-draw the standard curve if it changes during the analysis.

Calculation. Subtract the blank reading from the sample reading in pg K ml-1 read from the graph. This is equal to the g kg-1 of potassium in the sample. Divide the pg K ml-1 by 10 to obtain the % K in the sample. If the sample was not oven-dried, use the separate moisture determination to correct the result to K in DM.

Explanation: 2 g sample was dissolved in 50 ml solution (x 25 v/m dilution), which was further diluted by a factor of 40 (2.5 ml to 100 ml). Thus the 2-g sample would have been diluted to 50 ml x 40 = 2 l, which is 1 g sample l-1. Therefore, 1 kg sample would be present in 103 l sample solution. If the measured concentration is y pg K ml-1, this is equivalent to 103 y pg K l-1, or 106 y pg K in 103 l, which is y g K in 103 l sample solution, which we have shown to contain 1 kg sample.

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