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### Topic: [As] in sample  (Read 3056 times)

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#### Baileys

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##### [As] in sample
« on: May 06, 2010, 04:28:41 PM »
I need to find the original conc of As in a sediment sample, after carrying out these dilutions:

Of the sediment, 0.5g was taken, treated with various chemicals, and then made up to the mark in a 50mL flask.

A 4mL aliquot of this was taken and transfered into another 50mL flask.

A 5mL aliquot of this was transfered into a 100mL flask, which was made up to the mark.

The absorption reading obtained for As was 2.5mg L-1.

This is what I have done so far, where I have worked backwards in the dilutions.
Please let me know if I am on the right lines or not.

Total As in 100mL flask = 2.5 x 100/1000 =  0.25mg
Total As in the 5mL aliquot = 0.25mg
Total As in 50mL aliquot = 0.25 x 50/5 = 2.5mg
Total As in 4mL aliquot = 2.5mg
Total As in 50mL flask = 2.5 x 50/4 = 31.25mg

Conc in sediment = 31.25/0.5 = 62.5mg g-1

I won't bother converting it into mg/kg. I just want to make sure I have the right jist here

#### JGK

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##### Re: [As] in sample
« Reply #1 on: May 07, 2010, 03:28:14 PM »
AS level 2.5 mg/L

total dilution factor 4 mL in 50 (x 12.5) followed by 5 mL in 100 (x 20) = x 250

therefore the original 50 mL sample has a conc of 625 mg/L and 50 mL will contain 31.25 mg

31.25 mg is present in 0.5 g of the sediment = 62.5 mg/g
Experience is something you don't get until just after you need it.

#### Golden_4_Life

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##### Re: [As] in sample
« Reply #2 on: May 10, 2010, 02:06:46 PM »
As (no pun intended) J G  K your calculations are both clear and lucid. Is this done on a relative % basis or ''as is".
Golden4Life

#### JGK

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##### Re: [As] in sample
« Reply #3 on: May 11, 2010, 03:25:39 PM »
[AS] is basis - definitely
Experience is something you don't get until just after you need it.