I conducted a calcium analysis using AAS. The weight of the sample was 0,4647 g and the theoretical Ca-content was 200 mg/g. The stock solution was 100 ml at 200,000 ppm. For sample preparation, I diluted 0.5 ml of the stock solution to 50 ml, resulting in a concentration of 2,000 ppm. Then, I took 5 ml of the 2,000 ppm solution and diluted it to 50 ml, giving me 200 ppm. I further diluted 5 ml of the 200 ppm solution to 50 ml, resulting in 20 ppm. Finally, I diluted 0.5 ml of the 20 ppm solution to 50 ml, achieving a concentration of 0.2 ppm. The absorption was 0.3129, with a calibration curve described by the equation y=−0.026821x^2+0.98767x+0.14971. I'm stuck at how to calculate the content tracing back to the stock solution and sample weight because of the serial dilution.
Having just written everything below and re-read your question, I may have misinterpreted your procedure. If you merely created a calibration curve, then what was the unknown sample? You will need to elaborate.
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The serial dilution is not related to the concentration of the sample - it is used to calibrate the instrument against the known concentrations.
- You prepared a curve with 2000, 200, 20 and 0.2 ppm.
- You derived a formula for the curve: Y = -0.0268x2 + 0.9877x + 0.1497.
- You got a value for the unknown was 0.3129 (?) which is the instrument response Yunknown.
If Y = -0.0268x2 + 0.9877x + 0.1497, you now need to solve for x.
Using an online calculator: https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php?a=-0.002682&b=0.9877&c=0.1497&action=solve,
provides values of x = -0.1515 and 368.4 ppm.
You haven't told us the volume the unknown was dissolved in, so 0.4647g/Volunknown could well yield 368 ppm. Rough calculations suggest 0.4647 of 200mg/g in 4 ml = 386ppm.
Your question is not clear. Does A = 0.3129 refer to the unknown or to the known stock? As commented by @Matthew Wheal the volume of the unknown is also needed for the calculation.
A general commentary on standard curves from serial dilutions: You indicated that twice you used only 0.5mL of the previous 'stock', and other times you used 5mL. FIRST, every dilution introduces some measurement error. That could be quite large for a 0.5mL aliquot! Even one or two additional dilution steps to arrive at your desired concentrations will have less accumulated error if you use the same volumes repeatedly.
Once you have an acceptable calibration curve, then any unknown value should be obtained directly from the AA measurement and the response curve.
Hi,
You may use the web tool to calculate
https://www.alfa-chemistry.com/solution-dilution-calculator.htm
https://www.alfa-chemistry.com/mass-molarity-calculator.htm
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