Some researchers represents the solution concentration in Molars (like mM, uM, nM etc.,) and others represents in ppb or ppm. How can i convert ppb to molar?
If you use 0.001 g/l as your concentration and desire to convert say 250 ppm Magnesium sulfate to molar concentration, you could actually multiply the ppm concentration by 0.001 g/l, then divide the product by the molecular weight of MgSO4 to obtain the molarity.
Dear Nayan;
Good question because it is quite common to use ppm and ppb in the scientific literature. First we need to say that actually these terms are ambiguous because they express a proportion of quantities of a solute in a solution (they are analogous to a concentration %). When we express concentration as %, in general it is necessary to specify the nature of the quantities we are using, as if it is weight or volume. For example, a 70% solution of EtOH in water can be quoted as "70% (v/v) EtOH solution". So you need to specify if the quantities considered in the concentration parameter correspond to a volume or to a weight (mass). If the solute is a solid and the solvent is a liquid or a gas, then we use (w/v); e.g., 200 ppm (w/v). When quoting gas concentrations it is commont to express them as ppm or ppb without indicating the nature of the mixture; in this case it is assumed that the concentration is given by volume (v/v) or by mol (mol/mol).
Considering this, for the conversion from ppm or ppb (m/v) to M, first we need to say that ppm means 1 part of solute in 1 million parts of solution (10^6), ppb means 1 part of solute in 1 billion parts of solution (10^9), and that 1 M = 1 mol/L. Then we can do the following:
1 ppm (w/v) = 10^(-6) g/mL = 10^(-3) g/L
To convert to M, we divide the above by the molecular weight (MW) in g/mol:
1 ppm (w/v) = 10^(-3) g/L / (MW) [=] mol/L [=] M
Therefore, we have the following:
1 ppm (w/v) = 10^(-3)/(MW) mol/L = 10^(-3)/(MW) M
1 ppm (w/v) = 1/(MW) mM
Por ppb, in an analogou manner we have:
1 ppb (w/v) = 1/(MW) 10^-6 M = 1/(MW) uM
In the case that the solute is expressed by volume (v/v or v/w), first we need to obtain the weight (mass) of the solute by using its density, and then proceed as above.
Hope this helps. Regards.
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