Unit Conversions

Divide by 1000.

thank u for reply

bt its not like that u r thnkng

bcoz i m extracting frm soil and diluting with water for ICP analysis.

so there will be somethng differnt bt i am unable to get it..

if u know help please

ppb can be by volume or by mass (weight). If the ppb is by weight then

1000 ppb = 1 ppm = 1 mg/kg

therefore, 1 ppb = 1/1000 mg/kg

or divide ppb by 1000 to get in mg/kg

thank u for reply

bt its not like that u r thnkng

bcoz i m extracting frm soil and diluting with water for ICP analysis.

so there will be somethng differnt bt i am unable to get it..

if u know help please

Are you trying to relate the results back to the soil sample, instead of to the diluted mixture? If so, you need to adjust for the 0.5g of soil, vs. the 50 ml [essentially, 50 gm, since water is 1gm per ml, and aqua regia very close - a bit denser depending upon the concentration, but a good first approximation]. So, this would mean that the soil contained 100 times as many PPB as the analysis showed, and now you only need to multiply by 10: the other posts have already told you that the conversion between PPB and mg/kg is 1000. They are correct.

If you want to refine the results further, weigh the 12 ml aqua regia and the roughly 37.75 ml of DI water [assumes that soil is twice as dense as water], add 0.5 gm for the soil, and use that total vs 0.5 gm to get a better approximation than my 100:1. That rough approximation is going to be closer than some analysis systems will come anyway. (and it's closer than your "approximation" of English spelling - are you missing keys, or do you only type with your thumbs? )

hello ron

thank u for ur reply

bt as i got information there is no need of density only i hav to use the sample is diluted to 50 mL after digestion so whatever the metal extracted from soil will be diluted to 50mL, now the determined conc. by ICP is conc. in 50mL water but i hav to convert it into soil basis as mg of metal in kg of soil.

if u got understand my problem please help me..

thank u

thank u very much sabri

may be ur answer is right bcoz i m also thnkng like that

but how i will confirm that this is surely right

but thank u for ur reply..

This is the correct answer (I gave a GA): multiply the ppb value of the liquid by 0.1 to get the concentration in mg/kg of sample. The math above is correct.

As an analytical chemist, this is a very easy calculation. I've done thousands of similar calculations, so I was surprised at the variety of answers. It got me to thinking. Much of the confusion is due to the ambiguity of ppb (and other dimensionless units).

In analytical chemistry, if a liquid concentration is given in ppb, it is assumed to mean ug/l or ng/ml. I'll admit that this is not technically correct, as this is not a dimensionless unit, but this is the convention. For clarity, it really should be written as ppb w/v (ppb weight by volume).

If the concentration is given for a solid, it is always assumed to be on a weight/weight basis. For clarity, it should be written as ppbw.

It gets worse. For gases, ppb is assumed to be on a volume/volume basis. Assuming the ideal gas law, this is the same as on a mole/mole basis. For clarity, this should be written as ppbv (ppb by volume). Some companies designate this unit as ppbm, (for ppb by mole).

However, concentrations of gases can also be given on a weight/weight basis. Again, this should be written as ppbw. ppbv may be considerably different than ppbw, depending on the MW of the gases.

I hope this clears things up.

Thank u Tad I also thnkng that sabri's answer is right but i thnkng of confirmation but as u also said this is right but i m not getting why others are not understanding the problem and giving answers

Every branch of science has it's quirks and assumptions. If you would have said that the concentration of the solution was 5 ug/l instead of 5 ppb, I think that you would have gotten more straight forward answers.

Hi Tad,

I appreciate your reply post, thank you.

The way this test was carried out using liquid up to 50 grams instead of a much easier to calculate 100 grams was not making things easy for himself.

I am not blowing my own trumpet here but, it seems to have taken me to point out what, exactly the ppb/w equates to the the OP. I wish you had explained it sooner!

I think if the OP had realised that ppb/w was or could be read as a ng, they would not be asking for an example of how to translate from ppb/w to ng? This is very basic stuff, which the OP should have known before they started, the reason being, that if they do not understand the meaning of ppb/w then perhaps it could be said that any explanation of it after the test may still not have been understood?

I had these figures in my head but then researched and found a site which was able to confirm my meaning of measures down to not just a ng but to a pg, to the OP. And of course I also explained why the PPB definition should not have been used as unless the OP knew, which they didn't, the possible value of ppb/w to gram fractions, they would not of been able to complete the test.

Good luck

Hi Ron,

Your basic premise sounds OK. But, the idea of this test is to get an exact measure of the contaminants. Therefore, no weights or amounts can be 'assumed', otherwise the test might just as well be guest in the first place.

This test was made up to 50ml using .5 gm of soil. It would be a whole lot easier to translate if there was 1 gm of soil and it was made up to 100ml ! Just a thought.

I think if you or I did this test for contaminants, we would start with an idea of how to work things out before we started, hence my saying to use a 100 ml of total mix rather than 50 ml. Trying to translate something another person has done and not knowing what the exact weights were, or whether the soil sample was just humus or included stones makes a huge difference also. The OP would know that but we do not know it, which makes you suggestion of 'roughly' this or that sound bad and unprofessional.

I am not getting at you. But can you imagine the uproar there would be if a statistician went and looked at this result in five years time because people were falling ill through cadmium poisoning or something. He looked at the apparently professionally done test and does some math for himself, and gets a completely different answer or outcome. It would then come out as having being guessed not figured out properly. I would not like to be on the end of the argument that ensued.

Good luck

hey babybear

may be u got confused, soil that i used was dry soil and no stones

i want to convrt results in term of soil bcoz in water they r diluted 50 times

u read sabri anwsers that sounds me right..

Hi Ron,

it was me and my terrible attempt at translating the Italian "Aquaregia" who at first suggested it was Turpentine! I did apologise for that in the next post!

I have read the correct way to do this particular Teat, as has Doorman who first mentioned details from the test site describing how to go about this test. The description's are pretty long and some are generic applying to many types of test, ...... the type of container, the type of Microwave etc, other detail must be more specific but I was not in the mood to find the exact detail, (I fell asleep!LOL!) by reading through several pages yesterday, and the part I did read as I recall, did actually say the results would be given in ppb/ppt (parts per billion/Parts per Trillion).

Now, despite what was said by another post which I believe said they were an Analytical Chemist, there is no written rules or regulation or codes to say how the ppb/ppt are converted to a 'gram fraction'.

It seems a complete nonsense to do the test or do multiple tests without first having the knowledge of how you are going to use the final results. I am not saying the test per se were done badly or wrong necessarily, but there is 'so many hurdles' to jump, and so many pitfalls in doing the test correctly, that, any one of which would make the test worthless. and a total waste of time and effort.

As an example a particular type of Microwave has to be used:

--------------------------------------------------------------------------------------

6.1 Microwave apparatus requirements
6.1.1 The temperature performance requirements necessitate the microwave
decomposition system to sense the temperature to within ± 2.5 EC and automatically
adjust the microwave field output power within 2 seconds of sensing. Temperature
sensors should be accurate to ± 2 EC (including the final reaction temperature of 175 ± 5
EC). Temperature feedback control provides the primary performance mechanism for the
method. Due to the variability in sample matrix types and microwave digestion equipment
(i.e., different vessel types and microwave oven designs), temperature feedback control is

preferred for reproducible microwave heating. For further details consult Ref.

3.

Alternatively, for a specific vessel type, specific set of reagent(s), and sample type,

a calibration control mechanism can be developed. Through calibration of the microwave

--------------------------------------------------------------------------------------

It this or any other detail is not scrupulously cared for and against, any part of the sample ready to test, can be lost through not heating correctly.

Other warning are to use the correct vessel.

To use the correct acid and acid mix

In certain circumstance's to allow and take care when and if a test sample which has been made ready with acid and 'DOUBLE 'DE-IRONISED' water used, and this could possible ................................... (And it goes on).

You can view the detail at one of the sites below.

• [PDF] METHOD 3051a MICROWAVE ASSISTED ACID DISSOLUTION OF SEDIMENTS ...File Format: PDF/Adobe Acrobat - View as HTML
  nitric acid to hydrochloric acid using proposed EPA Method 3051a. ..... using appropriate elemental analysis techniques and/or SW-846 methods. ...
  www.cem.de/documents/pdf/publikation/digestion/RD130.PDF -
  by E CASRN - Related articles
• Microwave-assisted extraction of metal elements from glass fibrous ...samples in USEPA SW846 Method 3051A.5 However,. * E-mail: gzheng@ns.nint.ac.cn these methods which manipulated less than 1 g sample ...
  www.springerlink.com/index/N3508930343524RR.pdf -
  by L Dong-Mei - 2003 - Related articles - All 4 versions
• Nuclear Instruments and Methods in Physics Research Section B ...[4] US EPA Publication SW-846, Test Methods for Evaluating Solid Waste, Method 3051A, Microwave Assisted Acid Digestions of Sediments, Sludges, Soils, ...
  linkinghub.elsevier.com/retrieve/pii/S0168583X08009233 - Similar -
  by JM Lunderberg - 2008 - Related articles
• 
  
• [PDF] SW-846 Method 3051A: Microwave Assisted Acid Digestion of ...File Format: PDF/Adobe Acrobat - View as HTML
  METHOD 3051A. MICROWAVE ASSISTED ACID DIGESTION OF. SEDIMENTS, SLUDGES, SOILS, AND OILS. SW-846 is not intended to be an analytical training manual. ...
  www.epa.gov/waste/hazard/testmethods/sw846/pdfs/3051a.pdf -
  by CAS Element - Related articles - All 5 versions
• [PDF] SW846 Table of ContentsFile Format: PDF/Adobe Acrobat - Quick View
  Method 3051A: Microwave Assisted Acid Digestion of Sediments, Sludges, Soils, and Oils ... A method reference found within the text of SW-846 methods and ...
  www.epa.gov/waste/hazard/testmethods/sw846/pdfs/toc.pdf -
• [PDF] Comparison of SW-846 Method 3051 and SW-846 Method 7471A for the ...File Format: PDF/Adobe Acrobat - Quick View
  The recent update (Update II) to SW-846 includes an approved method 3051 for ... profiles specified in method 3051. A conventional water bath preparation as ...
  www.ornl.gov/rmal/13236.pdf - Similar -
  by JM Giaquinto - Cited by 11 - Related articles - All 4 versions

Hi monugoyal,

PPB has no definite meaning. It depend on the density, and weight.

I have a suggestion though. When you were measuring the Turpentine and DI water, you used mg/ml? Work it out from there.

You had a soil sample weighing ?

Water weighing?

Turpentine weighing?

By Aquaragia, do you mean Turpentine?

This sounds like a pretty common scientific test. Can you not look up the Test and see what previous results were in gm/ml, and extrapolate the answer?

Are you asking with reference to 'Part Per' definition. Or the SI definition

Is your definition of ppb a ng [10^–9 g] Billionth?

Have you checked on the back of your scales for an approx '~' meaning?

SI Measures Gram.............

These notations are:

Gram> gm [1 g] One

Decigram> dg [10^–1 g] Tenth

Centigram> cg [10^–2 g] Hundredth

Milligram> mg [10^–3 g] Thousandth

Microgram> µg also 'mcg' [10^–6 g] Millionth

Nanogram> ng [10^–9 g] Billionth

Picogram> pg [10^–12 g] Trillionth

-------------------------------------------------------------------------------------------SI-compliant expressions

In the English language, named numbers have a consistent meaning only up to "million".

Starting with "billion", there are two numbering conventions: the "long" and "short" scales, and "billion" can mean either 10⁹ or 10¹².

• One part per million (ppm): Denotes one part per 1,000,000 parts, one part in 10⁶, and a value of 1 × 10^–6. This is equivalent to one drop of water diluted into 50 liters (roughly the fuel tank capacity of a compact car), or one second of time in approximately 11½ days.
• One part per billion (ppb): Denotes one part per 1,000,000,000 parts, one part in 10⁹, and a value of 1 × 10^–9. This is equivalent to 1 drop of water diluted into 250 chemical drums (50 m³), or one second of time in approximately 31.7 years.

I am sending this to allow you to read and hopefully find a way round your problem. Is this the first time you have done this experiment? Can you ask one of your Teachers?

Good luck

thank u for ur reply

but i thnk u didnt got understand my problem

and aquaragia means HNO3+HCl its acid digetion procedure for extraction of heavy metals from soil.

i hav wieghed 0.5 gm soil in which i hav to determine heavy metals and acid added and digested in microwave and the final solution is diluted to 50mL for analysis by ICP.

the ICP give results in ppb but that refers to metal in water not soil.

but i want units as mg of metal in kg of soil.

so i want the conversion..

Hi monugoyal,

the ICP give results in ppb but that refers to metal in water not soil.

but i want units as mg of metal in kg of soil.

so i want the conversion..

I am sorry I got the acid thing wrong, OK.

You did not say in your first post that you wanted the amount of metal in soil and not water.

Start by weighing the soil.

You know the measured weight of the other water and acid.

Can I assume the acid will have completely digested the contaminants?

If so boil or dry the acid and weigh the result? That should be 100 of the contamination in your sample of soil? It will probably give an answer in micro-grams or less but will be an exact answer.

Despite other answers saying there is a correlation between PPB and a nanogram, there is no official scale to refer to.

(*1)One was proposed in, I think 1999 but it never was ratified.

In saying this I am assuming as I say, that the acid will contain the contaminants. If not this I will leave it there as I am not a professional soil analyst.

The other thing I would have done is use 1 gm of soil, and make it and the acid up to 100ml. It would make the translation a whole lot easier.

My reference was:

(1) Uno

Because of the cumbersome nature of expressing certain dimensionless quantities per SI guidelines, the International Union of Pure and Applied Physics (IUPAP) in 1999 proposed the adoption of the special name "uno" (symbol: U) to represent the number 1 in dimensionless quantities.^[3] This symbol is not to be confused with the always-italicized symbol for the variable 'uncertainty' (symbol: U). This unit name uno and its symbol could be used in combination with the SI prefixes to express the values of dimensionless quantities which are much less—or even greater—than one.^[8]

Good luck.

1 ppb is 1 part in 1000 000 000 by weight. 1 mg/kg is 1 part in 1000 000 by weight. So X ppb = X/1000 mg/kg

read my other replys

bcoz its not like that u r thnkng

ppb is in terms of microgrm per liter but i want in mg/kg of soil

k

It would help us considerably if you are more specific. Do you mean X parts of heavy metal per billion parts of prepared solution to be converted to Y mg heavy metal per kg of soil? Unless you tell us what you mean the problem cannot be answered.

hey read my post to babybear i described details

Hi monugoyal,

ppb is in terms of microgrm per liter but i want in mg/kg of soil

You should get your fact right before you start to criticize.

PPB is in micro-gram per litre? Really.?

Actually, if you refer to a Litre, then that same weight is called a micro-litre, because it is a liquid measure. It is still exactly the same weight, but liquid rather than solid.

As I have said in an earlier post, ppb and micro-gram are not officially ratified or listed as a chart. ppb is not defined as a weight, so it can weigh almost zero or a whole lot more. This is why the micro-gram is not interrelated or interdependent and switchable with ppb.

You have not read anything I have send in yet so why should I think you are going to read this?

I tried to explain how to do your experiment but you were not interested.

Did you read my post where I said the 'Parts Per' measure was going to be linked with Macro-gram but it wasn't ratified so officially they are not interdependent and switchable.

hello babybear its not like that i m not reading ur reply's but u r not getting the problem read the sabris answer i thnk that was right like i again explain u the problem look i had took 0.5 gm soil sample and 12 mL aquaragia added and digested in microwave, for ur knowledge microwave works on microwaves that heats the solution and some of the acid get vapourized, and remaing solution after digestion is filterd and later on volume was make up to 50mL to analyse by ICP. As the acid gets diluted to 50 mL i thnk no need of acid density. And ICP analyses heavy metals in ppb means microgram of metal in liter of solution, so at that time solution is water. But as i m analysing in soil so i want heavy metal concentration with respect to kg of soil. so this is the problem please all others also read this to understand problm fully bye...

Dear,

See:

m = ρ.V

For acquaragia:

V acquaragia = 12 [ml] = 12e-3 [l]

ρ acquaragia = 1.1 g/l, so:

m acquaragia = 1.1 [g/l]*12e-3 [l] = 13.2e-3 [g] = 13.2 [mg]

For water:

v h2o = 50 [ml]

ρ h2o = 1e3 [g/l] => m h2o = 1e3 [g/l]*50e-3 [l] = 50 [g]

Total mixture:

m total = m soil + m acquaragia + mwater = 0.5 + 13.2 + 50 = 63.7 [g]

Concentration (C):

C = m substance / m total [% or ppb or ppm]

If you got the results in ppb, so:

C = 1 [kg] / 1e-9 [kg] = 1 [mg] / 1e9 [mg] = 1 [mg] / 1e3 [kg] = 1 [ppb]

So, suppose you got 5 ppb for heavy metals concentration, then, multiply your result by 63.7e-3 to get:

C = 5*63.7e-3 = 0.3185 [mg/kg]

Wbr

Dear, (correction, please):

See:

m = ρ.V

For acquaragia:

V acquaragia = 12 [ml] = 12e-3 [l]

ρ acquaragia = 1.1 kg/l, so:

m acquaragia = 1.1 [kg/l]*12e-3 [l] = 13.2e-3 [kg] = 13.2e3 [mg]

For water:

v h2o = 50 [ml]

ρ h2o = 1e3 [g/l] => m h2o = 1e3 [g/l]*50e-3 [l] = 50 [g]

Total mixture:

m total = m soil + m acquaragia + mwater = 0.5 + 13.2 + 50 = 63.7 [g]

Concentration (C):

C = m substance / m total [% or ppb or ppm]

If you got the results in ppb, so:

C = 1 [kg] / 1e-9 [kg] = 1 [mg] / 1e9 [mg] = 1 [mg] / 1e3 [kg] = 1 [ppb]

So, suppose you got 5 ppb for heavy metals concentration, then, multiply your result by 63.7e-3 to get:

C = 5*63.7e-3 = 0.3185 [mg/kg]

Wbr

The sample is diluted to 50 ml, not with 50 ml water

Dear,

Excuse my misunderstanding, I gave my comments considering the weight of heavy metals in repsect to the mixture (soil + acquaragia + water), to do this in respect to soil, you have to do:

For instance, suppose the concentration of heavy metals is C = 5 ppb = 5.e-3 mg/kg, so, multiply this by the weight fraction of soil, i.e., 0.5.e-3/63.7.e-3 = 0.00785, then you get 5.e-3*0.00785 = 3.9.e-5 mg/kg (note: see my below reply for the mixture weight)

Wbr

hi acflima kindly read my post for baby bear thank u for ur reply

hello to all

this test i carried out is according to USEPA 3051a method, i can not change any quantity and other thngs...

Investigation of information from monugoyal yields (From the US EPA website):

Method 3051A (SW-846): Microwave Assisted Acid Digestion of Sediments, Sludges, Soils, and Oils

This method may be used for the preparation of sludges, sediments, soils and oils for total recoverable metal determinations by FLAA, GFAA, ICP-AES or ICP-MS. Nitric acid and hydrochloric acid are added to the representative sample in a fluorocarbon digestion vessel and heated in a microwave unit prior to metals determination.

ICP-AES allows simultaneous or rapid sequential determination of many elements in a short time. Aerosol samples are introduced into an extremely hot plasma source which vaporizes, atomizes, ionizes and electronically excites the sample components. Upon exiting the plasma, the electronically excited analytes emit characteristic photons that are detected via emission spectrometry. A primary disadvantage of ICP-AES is the occurrence of background radiation from other elements and the plasma gases. Although all ICP-AES instruments utilize high-resolution optics and background correction to minimize these interferences, analysis of trace levels of inorganic analytes in the presence of a large excess of a single analyte is difficult. Examples would be trace levels of inorganic analytes in an alloy or trace metals in a limed (high calcium) waste. ICP-AES and FLAA have comparable detection limits (within a factor of 4) except that ICP-AES exhibits greater sensitivity for refractories (Al, Ba, etc.). FLAA, in general, will exhibit lower detection limits than either ICP-AES or FLAA.

In the various tables, text, and discussions within SW-846, these test criteria will all result in measurement of PPM, PPB, or PPT of the aqueous sample. Are you working to another end?

Also, your test sample quantities and solution volume do not appear to be within recommendations of table 3-2 (RECOMMENDED SAMPLE HOLDING TIMES, PRESERVATION, COLLECTION QUANTITIES, AND DIGESTION VOLUMES FOR SELECTED INORGANIC ANALYTE DETERMINATIONS IN AQUEOUS AND SOLID SAMPLES) .

If my opinions about result quantification or sample size are incorrect (they probably are), please advise... I really don't want to read the entire (+ / -) 3,500 pages! This will also help others to help you.

Another interesting discussion, this.

Hi Doorman,

Appreciate all your hard work!

I have already said the sample as described seems the wrong and inappropriate size. And it would have been a whole lot easier to figure out quantities if the same was made up to 100.

As I have also said, someone doing this kind of test but who has not got the knowledge or foresight to see he would have to do this conversion, or how to do this conversion, IMHO should not be doing this kind of critical test.

But I really appreciate that you actually went to find out exactly what the test wast and how it should have been done correctly. As the OP gave very little detail even when asked several times.

Thanks once again.

hey look i want to tell u that waht procedure i followed is correct, bcoz it is not only followed by me, its already well developed procedure and many researchers worked on it, and i also had read many papers regarding this... but i m not getting for unit conversion, bcoz no one mention unit conversion like thngs in papers the doorman didn't read full information, and for ur kind of knowledge, the vessles in which soil is digested with acid have specific size, u can not take as much quantity u want and it is clearly written in epa guidelines that 0.5 gm sample is more than enough so how u can say this is wrong and for ur info. i already gave full information regarding the test but i thnk u don't know anythng just writing any thng look at sabriahmd's answer which i gave rating was a good answer and tad also supported that one ok

Hi Guest,

I am assuming you are the OP (Original Poster)? Writing as 'Guest' it is not obvious, sorry.

I also see you refer to my post #31.

If you read it carefully, I have not said you did not complete the tests correctly.

I said simply that, if possible, using twice as much to test, would have made the conversion easier.

The real problem I have, which cannot be remedied in the short term, and perhaps you may have learned something of conversion after your posts, ........... But it does seem you start and perhaps completed the test, but could not convert from one measurement to another. I know that the test you did was always going to give a 'part per' result and it does not leave people enthused with confidence, because, if you have no idea how to check any particular ppb/ppt, how can you confidently double check ANY result that involves ppb/ppt figures?

Why did you need the conversion from one to another nomenclature, was it to figure out how much w/w contamination?

Good luck and I hope you perhaps have already sorted your problem out.

u r wright babybear, i want to figure out how much is contamination at site from where i collected samples..

somewht i solved out my problem but i m not getting confirm, the answer given by sabriahmd, i also found in one of paper, it is not written directly but by back calculation i got the same to multiply only by 0.1 in ppb and result will be converted to mg/kg.

thank u for ur posts

bye