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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: JHobbs on June 10, 2005, 04:33:01 PM

Title: m1v1=m2v2
Post by: JHobbs on June 10, 2005, 04:33:01 PM
Hi, my question deals with dilutions and the m1v1=m2v2 equation.  I want to dilute a 1000ppb (1ppm) solution down to 100, 150, 200, 250, and 300 ppb for a calibration curve.  I have five test tubes for each sample all of which I fill with 10 mL of nanopure DI water.  If I want a 1, 2, 3, 4, 5 ppb calibration I would add 10uL, 20uL, 30uL, and so on.  However, my problem with the dilutions to 100ppb-300ppb is if m1v1=m2v2 still holds true due to the large dilution.  Specifically, for example making the 300pbb, would I use (1000ppb)v1=(300ppb)(10mL) or would I use (1000ppb)v1=(300ppb)(13mL) since 13 would be the final volume.  thanks
Title: Re:m1v1=m2v2
Post by: Borek on June 11, 2005, 03:37:36 AM
V2 should be always final volume. Genrally speaking even making 3 ppb solution final volume will be not 10mL but 10.03mL. Such small differences are usually below the sinificant digits level so they can be neglected. When the volumes get higher situation complicates - adding 3mL you will get 231 ppb solution, not 300ppb.

Refer to any book with significant digits discussion and accuracy discussion, even HS level should be enough to get the idea.
Title: Re:m1v1=m2v2
Post by: Dude on June 11, 2005, 03:13:45 PM
Be careful of your units when doing dilutions.  The use of m1v1 = m2v2 is a mole balance (mol / L multiplied by Liters = mol).  When your concentration is given in weight percent (ppb = g / 1,000,000,000 g), a simple ratio can be used.

1,000 ppb (X) = 300 ppb
X = 3/10, therefore, any dilution combination of 3 parts by weight of your 1,000 ppb standard (volume can be used with water since density = 1) diluted up to 10 parts by weight is appropriate.    
Title: Re:m1v1=m2v2
Post by: JHobbs on June 13, 2005, 08:46:30 AM
Well it seems that I've gotten two different solutions with different answers.  Borek, I see how you came to your answer.  
With 10mL H20 and 3 mL of 1000ppb solution:
1000*10-9*(grams analyte/g solution)*3g solution = 3*10-6gram analyte

3*10-6gram analyte/13grams solution

=2.3*10-7 g analyte/g solution = 0.23ppm

This seems to not follow "Dudes" interpretation or does it?  By dudes method I would get a 1mL to 10 mL solution to analyte for 100ppb and 3mL to 10mL  for 300ppb which is the same (I believe) as for m1v1=m2v2.
Borek, for 300ppb are you saying to add 3.9 mL of 1000ppb solution to 10mL H20 instead of 3mL to 10mL H20?  Is what your suggesting the same as dudes response?  Thanks
Title: Re:m1v1=m2v2
Post by: JHobbs on June 13, 2005, 09:38:00 AM
Maybe the easiest solution would be to (for 100ppb) add 1mL 1000ppb solution to 9mL H20, 1.5mL 1000ppb to 8.5mL H20 and so on...
Title: Re:m1v1=m2v2
Post by: Borek on June 13, 2005, 09:49:41 AM
Maybe the easiest solution would be to (for 100ppb) add 1mL 1000ppb solution to 9mL H20, 1.5mL 1000ppb to 8.5mL H20 and so on...

Very good idea.

If you want to start with 10mL of water:

Final volume = 10mL + V

Final concentration = V/(10mL + V)*original concentration.

Solve for V.
Title: Re: m1v1=m2v2
Post by: sportyl21 on October 06, 2007, 10:18:09 PM
understanding the equations, isn't there a rule with density, what if the density is different or not equal to 1, what other equation could i use?  My problem is I need to dilute a mixture of 100% w/ different density down to 10ppm, how would i go about this?
Title: Re: m1v1=m2v2
Post by: Borek on October 07, 2007, 03:43:44 AM
Mostl likely both ppm and % are w/w. You have two unknowns - mass (or volume) of one solution and mass (or volume) of the second solution. Write down all equations and definitions that you may need - like density definition, solvent and solute mass balances. You have two unknowns, you have set of equations - then it is relatively simple algebra.

Assuming you are diluting your solution with pure water you may use faster approach. Calculate amount of substance that will be prsent in the final solution. You know there is only one source of this substance - that's the concetrated solution. Find out what mass of the  concentrated solution you need - and convert it to volume.

Many dilution questions require several steps of dilution, especially when you need quite minute volume of concentrated solution. Without knowing details of your case it is hard to tell something more.