Heavy metal analysis of soil and water sample. Sample preparation using acid digestion method was explained. AAS analysis using was explained in detail along with a standard curve.
If i have started dilution taking water sample 500ml then after digestion it reduces upto 15 then i add distilled water to make volume 30 ml what is formula to find conc of metals n graph formation
hello I have some questions 1) What is the volume of the beaker and does the distilled water add to it before digestion? 2- what is the volume of the volumetric flask? 3- why did you set the dilution factor equal to 2? 4- you wrote (so in 50 ml of sample = 0.78x50 = 39ug / 50ml), I did not understand where it comes from 50 ml? thank you in advance for your response.
1. The distilled water to added until the total volume becomes 50ml or 100ml(Mostly 50ml or 100 ml volumetric flasks are used.) You use only 50ml for the analysis. In case if you did anything wrong then the remaining solution can be used for further analysis. 2. But I believe he used 50ml volumetric flask and used all the sample. 3. He might have diluted the sample with equal amount of water while doing AAS. 4. In 1ml of sample .78 g is present so .78ug/ml. Thus in 50 ml sample 50 times .78 ug which equals 39 ug in 50 ml. Or 39 ug in 1g of soil initially used. Hope this helps...
Thank you.I want to know as I'm confused.If I have 5ml sample and after digestion final volume is 50ml.Reading is 12.34ppm so what will be final concentration of that mineral in sample.
The standard solutions are made using a stock solution of a known concentration. for example, if you have 1 gram per litre of lead sulphate stock solution ( add 1 g lead in 1 litre water), you can make a standard solution with 0.5 g/l by adding 1ml stock and 1ml water. In a similar way can make 0.1, 0.2,..., 0.9 g/ l standards by adding respective ratios of water and stock solution. Hope it helps...
