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#1
09-23-2012, 02:41 AM
 Unregistered Guest Posts: n/a
mg/L - mg/kg Conversion

Hi,

I would like to ask something about unit conversions with dilution effect.

Assume;
1) Dissolve 1gr dry sample in 50 ml solution
2) Treat with different kind of acid/base solutions such as NH4Cl / HCl - Na2SO4 / NaHCO3 / NaOH
3) Filter sample from 45 micro-meter pore diameter membrane
4) All 50 ml is acquired during filtration

If I take 2 different amounts of (5ml and 10ml) this aliquot, digest them, and dilute both of them to 100 ml and measure concentration as 10 mg/L and 30 mg/L, what will be the final mg/kg concentrations.

If I use 50ml aliquot from step 4 as my sample, the result will be X mg/L*0,1L/1gr sample. The real question is do I need to multiply the results of 5/10 ml samples with 50/5 and 50/10 for dilution effect?

And there is one more thing that might be in effect. I have prepared two different calibration curves for 5ml and 10ml samples. I used stock solutions at different concentrations and drew a calibration graph of Concentration vs Absorbance. I have treated the exact method to my stock solutions too.

For example, if I used 5ml NaOH in experiment, i used the same amount in my stock solution. For measuring 10 ml samples, I prepared my calibration curve with 10 ml stock solution+chemicals in experiment and diluting them to 100ml. Can this also effect results in my experiment?

I am really confused thanks for your help.
#2
09-24-2012, 10:01 PM
 Mrs X can't count, can't spell! Join Date: Feb 2006 Location: New Zealand Posts: 2,294 Rep Power: 11
Re: mg/L - mg/kg Conversion

Quote:
 Originally Posted by Unregistered Hi, I would like to ask something about unit conversions with dilution effect. Assume; 1) Dissolve 1gr dry sample in 50 ml solution 2) Treat with different kind of acid/base solutions such as NH4Cl / HCl - Na2SO4 / NaHCO3 / NaOH 3) Filter sample from 45 micro-meter pore diameter membrane 4) All 50 ml is acquired during filtration If I take 2 different amounts of (5ml and 10ml) this aliquot, digest them, and dilute both of them to 100 ml and measure concentration as 10 mg/L and 30 mg/L, what will be the final mg/kg concentrations. If I use 50ml aliquot from step 4 as my sample, the result will be X mg/L*0,1L/1gr sample. The real question is do I need to multiply the results of 5/10 ml samples with 50/5 and 50/10 for dilution effect? And there is one more thing that might be in effect. I have prepared two different calibration curves for 5ml and 10ml samples. I used stock solutions at different concentrations and drew a calibration graph of Concentration vs Absorbance. I have treated the exact method to my stock solutions too. For example, if I used 5ml NaOH in experiment, i used the same amount in my stock solution. For measuring 10 ml samples, I prepared my calibration curve with 10 ml stock solution+chemicals in experiment and diluting them to 100ml. Can this also effect results in my experiment? I am really confused thanks for your help.
At step 2, please can you clarify why the volume hasn't changed?

Why would you have a different calibration curve for your 5mL and 10mL aliquots? - if you have done it properly, they should have the same concentration. Do you have a reference for the experiment you are doing?
#3
10-15-2012, 12:17 AM
 Makiko Junior Member Join Date: Oct 2012 Posts: 1 Rep Power: 0
Re: mg/L - mg/kg Conversion

Why the volume hasn't changed? It is because the 50 ml sample is the acid/bas solution. 1 gr dry sample is added directly into solution of 25 ml twice and centrifuged after reaction time. This makes a total of 50 ml after each step. Loss of dry sample and aliquot collected are assumed to be none.

These are the two calibration curves for my 5 and 10 ml samples:

y=0,0317x+0,0004
y=0,0653x+0,0003

For the question why I have prepared different calibration curves. The answer is to neglect any errors during experiment, including the dilution effect. I want everything to be uniform. For example if I am reading my absorbance in NaOH solution, I also add NaOH to the calibration standard.

I have prepared stock solutions of 50 ml in different concentrations. 8 different concentrations between 0-1 mg/L. Now if I have to read my absorbance in NaOH, I know that I must dilute the sample at a rate of 1/10 since I cannot read absorbance greater than 1-1,5 mg/L.

My sample was 50 ml. So I am taking 5 ml from this sample and diluting to 50 ml. If I use a calibration curve prepared from 50 ml stock solutions, I had to multiply my results by 10. I have also diluted my stock solution by 1/10 so that my results didn't need extra calculation.

Second reason is to show my curves are fitted properly. All 5-10-20 and 50 ml calibration curves fit into each other.

Experiment is sequential extraction of phosphorus from sediment by Psenner et. al. which is in German, and I am using a modified version from Hupfer et. al. Differences between methods are minor.

Phosphorus determination method is US EPA Standard Method for Phosphorus Determination. Ascorbic Acid method.
#4
10-15-2012, 09:46 AM
 Unregistered Guest Posts: n/a
Re: mg/L - mg/kg Conversion

HELP how can i convert mg/l to ppm . got a water test done and don't know if it's alot or a little thanks clarence
#5
10-15-2012, 12:13 PM
 JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 8,707 Rep Power: 17
Re: mg/L - mg/kg Conversion

Quote:
 Originally Posted by Unregistered HELP how can i convert mg/l to ppm . got a water test done and don't know if it's alot or a little thanks clarence
PPm can be calculated on several basis, weight/weight, weight/volume, volume/volume, molar ratios. Weightvolume is common for wet chemistry and particularly for dilute solutions in water. Either water or dilute solutions in water weigh about 1 kg/L, so weight/volume and weight/weight are the same. Milligrams per kilogram or per liter ARE parts per million.
#6
10-15-2012, 04:03 PM
 Mrs X can't count, can't spell! Join Date: Feb 2006 Location: New Zealand Posts: 2,294 Rep Power: 11
Re: mg/L - mg/kg Conversion

Quote:
 Originally Posted by Makiko Sorry for the late reply, Why the volume hasn't changed? It is because the 50 ml sample is the acid/bas solution. 1 gr dry sample is added directly into solution of 25 ml twice and centrifuged after reaction time. This makes a total of 50 ml after each step. Loss of dry sample and aliquot collected are assumed to be none. These are the two calibration curves for my 5 and 10 ml samples: y=0,0317x+0,0004 y=0,0653x+0,0003 For the question why I have prepared different calibration curves. The answer is to neglect any errors during experiment, including the dilution effect. I want everything to be uniform. For example if I am reading my absorbance in NaOH solution, I also add NaOH to the calibration standard. I have prepared stock solutions of 50 ml in different concentrations. 8 different concentrations between 0-1 mg/L. Now if I have to read my absorbance in NaOH, I know that I must dilute the sample at a rate of 1/10 since I cannot read absorbance greater than 1-1,5 mg/L. My sample was 50 ml. So I am taking 5 ml from this sample and diluting to 50 ml. If I use a calibration curve prepared from 50 ml stock solutions, I had to multiply my results by 10. I have also diluted my stock solution by 1/10 so that my results didn't need extra calculation. Second reason is to show my curves are fitted properly. All 5-10-20 and 50 ml calibration curves fit into each other. Experiment is sequential extraction of phosphorus from sediment by Psenner et. al. which is in German, and I am using a modified version from Hupfer et. al. Differences between methods are minor. Phosphorus determination method is US EPA Standard Method for Phosphorus Determination. Ascorbic Acid method.
I'm so sorry, i'm really busy this week, i'll have to have time to get my head around what you are doing, will get back to you maybe in the weekend, although that is pretty busy too.
#7
10-21-2012, 01:43 PM
 Mrs X can't count, can't spell! Join Date: Feb 2006 Location: New Zealand Posts: 2,294 Rep Power: 11
Re: mg/L - mg/kg Conversion

Hi Makiko, I've read through this several times, and looked at Hupfer's papers involving P from sediment, and keep coming to the conclusion that i've missed something about what you are doing. I agree you should treat your standards the same as your sample. However, you need to make your standards up to a known concentration, measure them via the method you are using, and plot the results of reading on the y-axis, and concentration on the x-axis Draw an even line between the points. It is ok if the line is curved. Your x-axis and y-axis scales need to be in even steps, not your concentration steps. You then measure your sample and put that reading mark on the line on the graph, and measure the concentration by drawing a perpendicular line to the x-axis.

Here's where i've missed something: You shouldn't be getting different results for the same calibration, that usually means you are miscalculating something. In this case, i'm not sure if it is the standards or the sample, but you should be able to measure a 5mL and 10mL sample on the same graph, if your standards are made up correctly. Your 10mL sample should have twice as much in as the 5mL sample, and you have to convert them back to the original 50mL to work out your concentration per gram.

Sorry not to be more help.

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