Determination on zinc?

Question

10g of soil was digested using 50ml ammonium EDTA, shaken for 1hr, filtered. The filtrate was further diluted. 10 ml of this filtrate was diluted to 100ml using ammoniun EDTA. This was further diluted (10ml of this solution to 90 ml ammoniun EDTA). Measurement was done using atomic absorption spectrophotometry. The concentration was y=1.2microgram/Litre. what should be done to obtain concentration of Zinc in soil in mg/kg. show calculation and explantion if possible plz. Thanks

Answer 1

uhm... this is the ag department, not the chemistry lab.  My semester of organic chemistry does not even come close to understanding this.  You might take this question to the chemistry aisle.

Answer 2

This is just a dilution and math problem.

You don't have to understand the chemistry aspect.

Note, it isn't digested with AmEDTA,

it's really extraction and chelation

You could determine it in one very long calculation,

but use 3 or so more understandable to you

and less likely to make an error.

ug = microgram (tho usually see it mcg)

mg = milligram

(1.2 ug/L )(10 )(10)= 120 ug/L in original extract.

120 ug/L (50 mL/1000 mL)= 120 ug/L (0.05 L)

= 6 ug in 10 g soil

= 0.6 ug Zn per g soil

1000 g= 1 Kg

0.6 ug (1000)/ 1(1000)

600 ug/Kg

1000 ug=1 mg

600 (1/1000)/Kg

= 0.6 mg/Kg