Janhvi Pandey

Yes mam, I Diluted the precipitates which were in a very little quantity with approximately 5 ml of Distiled water, after that i took the AAS readings.

= {(AAS reading-Blank reading) x sample volume x Dilution Factor}/ sample weight

Hi, Asad,

In the sense of Dr.Jinadasa’s answer, but in more detail:

(i) Prepare 2, better 3 blank solutions: „Filter“ the same amount of „diacid mixture“ (Dr. Pandey), i.e. 1.7 mL through filter paper in a little funnel into volume flasks and make up to 5 mL with water. If possible, use the same filter paper from the same box (series) as for the own samples and the same „diacid mixture“. Measure the content of the actual elements (Cd, Pb) in the blank solutions and calculate their average values in mg/L.

(ii) Content calculation in mg/kg: (AAS sample reading – AAS mean blank reading) x 5.0/0.100 = (......) x 50,

where (......) is difference in mg/L, whole volume (5.0) in mL, sample mass (0.100) in g.

Btw: The total volume of the solution to be analyzed (5 mL) seems to me to be on the edge of applicability for the FAAS method, especially if you want to repeat the measurement (reading) of both elements in the same sample (volume). However, I do know the aspiration rate of your AAS used and the integration time to reading the analytical signal.

Good Luck!

Petr Karel Mrak-Fiala

Dear Sir! Could you share the authentic reference for this formula please

Dear Colleague,

Instead of an "authentic" source for the calculation that interests you I would rather help you derive an appreciate formula on the basic of “Common Sense” and to apply it to the most of the problems may encounter in this area.

Assuming you have soil sample with unknown Cd content, but very heavily contaminated, e.g.about 250 mg/kg = (ug/g = ppm) of Cd. Diacid mixture (HNO3+HClO4) for sample digestion is also contaminated with Cd. Digesting 0.25 g of soil with 15 mL of diacid mixture and filtering this into 25 mL volumetric flask, fill up it with distilled water to volume (V). Simultaneously you repeated this procedure with 15 mL of diacid mixture, including filtering in (as possible) the beakers (3) of the same glass material and shape to get blank samples. By AAS you recorded Cd- signal of sample adequate to, say, 2.30 mg/L of Cd. Simultaneously blank sample has Cd content (mean value of 3 blank solutions) of 0.03 mg/L.

But your calibration range is linear within (0.05 – 1.00) mg/L. To get analytical signal to optimum, and to linear part of curve, you decided to dilute original sample solution (and simultaneously the blank solutions) five times, i.e. 5 mL of original sample (v) and blank samples to next 25 volumetric flasks (V1) and filled up with distilled water. So, now is the AAS signal for diluted sample, say, 0.510 mg/L and for blank solutions (mean of the 3 diluted ones) 0.009 mg/L.

So, 25 mL of the original solution (digested solution filled up to 25 mL) is adequate to…0.25 g of sample,

1 mL of this is adequate to… (0.25/25) g = 0.01 g of sample. You took 5 mL (v) from this original solution to 25 ml volumetric flask and filled up to volume(V1) with distilled water, hence this new (V1) volume is adequate to 5 x 0.01 g of sample = 0.05 g of sample. So, 1 mL of this V1 volume is adequate to (0.05/25) g = 0.002 g of sample. Your calibration is set on units of mg/L (= ug/mL). So, 1 mL of “calibration curve” represents 0.002 g (= 2 mg) of sample, 1 000 mL of calibration curve represents 2 g of sample. “Net” AAS signal (sample minus a blank) is adequate to (0.510 – 0.009) = 0.501 mg/L Cd, hence 0.501 mg Cd in 2 g of sample (1 000 mL of the calibration curve).

In 2 g of sample you found 0.501 mg of Cd; in 1 000 g of sample is 0.501 mg x 1 000/2 = 0.501 mg x 500 = 250.5 mg of Cd.

So, in 1 000 g (1 kg) of sample is 250.5 g of Cd = 250.5 ug/g = 250.5 ppm of Cd.

Concentration calculation by Formula (1):

V = 25 mL (whole volume of digested sample); n = sample mass (weigh) = 0.25 g; v = 5 mL of aliquot volume pipetted into new volumetric flask V1=25 mL to dilution:

mg/kg = (mg/L in sample – mg/L in blank) x (V/n) x (V1/v) = (0.510 – 0.009) mg/kg x (25/0.25) x (25/5) = 0.501 mg/kg x 500 = 250.5 mg/kg of Cd.

When exists curvilinear relationship between Cd concentration and its analytical signal and a dilution is required, you must take into consideration the dilution both of sample solution and of blank solution. So, in this case is “net” signal (0.510– 0.003) x appropriate dilution. Hence, you must substitute into calculation formula the difference (0.510 – 0.009) mg/L, not “original” difference (2.30 – 0.03) mg/L.

For a case of linear relationship between Cd concentration and its analytical signal; if a dilution is not required, you calculate Cd content by Formula (2):

Whole volume of digested sample (mass = 0.25 g), i.e. (V) is 25 mL, hence 1 mL of this volume is adequate to 0.25 g/25 = 0.01 g. So 1 mL of calibration curve represents 0.01 g of sample. 1 000 mL represents 10 g of sample. In 10 g of sample is (2.30 – 0.03) mg of Cd. Hence, in 1 000 g of sample is (2.30 – 0.03) mg of Cd * 100 = 227 mg/kg of Cd.

mg/kg = (2.30 – 0.03) mg/kg x (25/0.25) = 2.27 mg/kg x 100 = 227 mg/kg of Cd.

After substituting the appropriate units into the calculation Formula (2):

mg/kg = (2.30 – 0.03) x (V/n) = 2.27 mg/L *(25 *10-3 L/0.25*10-3 kg) = 2.27 mg/kg * 100 = 227 mg/kg. As can you see, most units “cancel” each other out and only the mg / kg unit remains. The same case is it at Formula (1).

In this case when you measure 5times diluted “original” whole volume (V = 25 mL) and simultaneously the 5times diluted blank solution you must substitute into calculation formula the difference (0.510 – 0.009) mg/L, not “original” difference (2.30 – 0.03) mg/L.

Note, please, that result of Cd content after 5times dilution of original solution is significantly higher due to curvilinearity of calibration curve in a region of higher Cd concentrations behind of 1.0 mg/L. This is not, of course, the case after appropriate dilution of original solution.

Next examples see, please, at: https://www.researchgate.net/post/How_to_calculate_concentration_of_sample_in_AAS

Regards,

P.M.

Petr Karel Mrak-Fiala

Well described sir

Thank you very much

It was a pleasure.

Good Luck!

P.M.