Successive dilution problem

  • #1
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Homework Statement:
Due to limitations at the lab I am in, we used atomic absorption spectrometry to measure the amount of Mg, Ca and K in a sample of marine salt.
First 1 g of salt was measured, and was left to dry for 1 day for water loss, and then 1 extra day in a 500ºC oven. The sample was then digested with 3x10mL 3N HCl which were left to evaporate in a hot plate. the remainder dissolved material was filtered and the filtrate was then diluted to 100mL flask with deionized water.
From this 100mL solution, the procedure was this depending on the element being analysed
Mg: 5 mL of solution to 50mL volumetric flask, together with 1mL of SrCl2 and deionized water)
K: 2.5 mL of solution to 50mL volumetric flask, together with 1mL of CsCl2 and deionized water))
Ca: 5 mL of solution to 50mL volumetric flask, together with 1mL of SrCl2 and deionized water))

The diluted results from atomic absorption analysis, were as follow
Mg: 0.2757 mg/L
K: 0.4712 mg/L
Ca: 0.9684 mg/L
How do I account for the 2 dillutions correctly?
Ideally another method should have been tried but this was the one whcih I didn't had to pay by my own pocket or wait months to be able to do it.
Relevant Equations:
CiVi= CfVf
I am afraid I am doing something really wrong, I would thank if somene could see what may I be doing wrong.

I started with the second dilution, the number of moles of Mg, K and Ca present in 5, 2.5 and 20 mL is assumed to be the same after the dilution to 50mL right, only concentration changed. right?
so for Mg: Ci*5*10-3L= 0.2757 mg/L*0.05 L <=> Ci= 2.757 mg/L
for K: Ci*2.5*10-3L= 0.4712 mg/L*0.05 L <=> Ci= 2.421 mg/L
for Ca: Ci*20*10-3L= 0.9684 mg/L*0.05 L <=> Ci= 9.424 mg/L

then for the first dilution to 100mL, for what I understood we must assume all salt was diggested and the transfer was complete to the 100 mL flask.
so for Mg: Ci= 2.757 mg/L = 2.757 ug/mL for 100 mL we have 275,7 mg

Those this even make sense? or there is something I am really missing?

Thank you in advance if yo ucould shed me some lights.
 

Answers and Replies

  • #2
35,518
11,994
so for Mg: Ci*5*10-3L= 0.2757 mg/L*0.05 L <=> Ci= 2.757 mg/L
for K: Ci*2.5*10-3L= 0.4712 mg/L*0.05 L <=> Ci= 2.421 mg/L
for Ca: Ci*20*10-3L= 0.9684 mg/L*0.05 L <=> Ci= 9.424 mg/L
I agree with the Mg value but I don't understand where the other concentrations would come from. If you take less of your sample for K (2.5 ml instead of 5) and still get a larger concentration then certainly the original concentration has to be higher. For Ca you said 5 ml above but now you seem to use 20 ml, and the result doesn't agree with either value (could be a rounding error if it was 5 ml, but then you should use more digits).
then for the first dilution to 100mL, for what I understood we must assume all salt was diggested and the transfer was complete to the 100 mL flask.
so for Mg: Ci= 2.757 mg/L = 2.757 ug/mL for 100 mL we have 275,7 mg
That should be 275.7 micrograms for the result, not milligrams.
 
  • #3
26
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I agree with the Mg value but I don't understand where the other concentrations would come from. If you take less of your sample for K (2.5 ml instead of 5) and still get a larger concentration then certainly the original concentration has to be higher. For Ca you said 5 ml above but now you seem to use 20 ml, and the result doesn't agree with either value (could be a rounding error if it was 5 ml, but then you should use more digits).That should be 275.7 micrograms for the result, not milligrams.

That you pointed is what is annoying me. that is why I think I must be doing something wrong. Ca must be way less than Mg and K. K is much more present in salt than Mg.

first Bold: My mistake in the first mention, it was indeed 20 ml not 5 mL. we indeed started with 5 for Ca, but Ca was too low for the initial

K and Ca were redone, because K was too high at 5mL of sample(out of curve), and Ca was too low, below the curve.

seconds Bold, another mispelling error
 
  • #4
35,518
11,994
That you pointed is what is annoying me. that is why I think I must be doing something wrong. Ca must be way less than Mg and K. K is much more present in salt than Mg.
It looks like you made a simple calculation error for K and Ca.
 
  • #5
26
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It looks like you made a simple calculation error for K and Ca.
Yeah silly me, I need new glasses
the real results are for first dilution: Ca = 2.421 mg/L and K = 9.424 mg/L. I switched them around for some reason
I guess I was my biggest enemy in all this process.
 
  • #6
35,518
11,994
That looks good.
 

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