Aluminum Ion Detection with AAS: Can 0.00014% Al 3+ be Detected?

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The discussion revolves around determining if a solution containing 0.00014% Al 3+ can be detected using Atomic Absorption Spectroscopy (AAS), given the detection limit of 2 mg/L. The calculation reveals that 0.00014% corresponds to 0.0014 g/L, which is significantly higher than the detection limit of 2 mg/L. Participants clarify that the percentage concentration can be converted to weight per volume (w/v) and compared to the detection limit. Ultimately, it is concluded that the aluminum ion in the solution can indeed be detected by AAS. The final answer confirms that detection is possible.
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Hi, I need help with a question on my lab worksheet.

Homework Statement


A solution is 0.00014% in Al 3+ ion. The detection limit for aluminum ion is 2 mg/L using AAS. Would the aluminum ion in this solution be detected by AAS?


Homework Equations


1 mg/L = 1 mg/1000mL which corresponds to 1 mg/1000g


The Attempt at a Solution



I feel like 0.00014% of Al 3+ means 0.0037772g of Al 3+ (0.00014*26.98g/mol). I then have to compare that to going from 2 mg/L to some percentage? We didn't do any calculations in lab other than finding the volume for Pb 2+ & HNO3 needed for our specified ppm.

I honestly have no clue how to go about this.


Thank you for the help.
 
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Usually, when a concentration is described in a percentage it means w/v or w/w, not molarity. So all you have to do is figure out w/v or w/w (which are practically equal here), and see how it compares to 2 mg/l.
 
What do you mean by w/v or w/w?
 
Weight per volume or weight per weight.
 
So 0.00014% would mean 1.4E-7g per .1 L?
And I would then have to covert 2 mg/L to g/L and compare?
 
It equates to 0.00014g/0.1l, so 0.0014g/l or 1.4 mg/l, if I'm not mistaken.
 
I see now.

And thus the answer would be yes.

Thank you for your help.
 
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