2
$\begingroup$
  • 1 ppm is 1 mg / kg
  • Water density: 999.97 kg/m³
  • methanol density: 791.80 kg/m³

solution 1 with water as solvent: at 1 $\ce{\mu}$g / mL is considered 1 ppm

solution 2 with methanol as solvent: at 1 $\ce{\mu}$g / mL, should it be considered 1 ppm?

If this is adjusted by the density of methanol, it wouldn't be 1 ppm. I heard it is a common practice that at the ppm / ppb levels, the density of the solvent is often omitted by only focus on the volume. Is this true?

$\endgroup$
  • $\begingroup$ A little bit of punctuation and formatting really helps. It gets hard to follow what you write, specially in the 2nd paragraph. $\endgroup$ – M.A.R. Jun 4 '15 at 12:31
  • $\begingroup$ thank you. When I typed it up late last night it wasn't looking like this. Thanks for the heads up. Going to correct it now. $\endgroup$ – pzyxian Jun 4 '15 at 21:40
  • $\begingroup$ Could you give a reference on "I heard it is a common practice..." part? The textbooks I usually rely on mention restrict this practice for dilute water solutions for which density can be approximated to 1 kg/l. The IMPCA reference on quantitation in methanol (methanol.org/Technical-Information/Resources/…) actually relies on weighing the methanol sample for trace analysis of chloride. $\endgroup$ – PLD Jun 5 '15 at 8:23
  • $\begingroup$ @PLD This I heard from my friend who is has a master degree in Chemistry and working in the industry as well. Just want to get a second opinion on it here. $\endgroup$ – pzyxian Jun 5 '15 at 23:36
3
$\begingroup$

I think this is why it's generally not advised to use relative units such as $\%$, $\mathrm{ppm}$ and $\mathrm{ppb}$. If they're going to be used, then it's best to indicate what's the ratio being considered, which is frequently done using $\mathrm{(v/v)}$, $\mathrm{(m/m)}$ or $\mathrm{(m/v)}$, $\mathrm{m}$ being short for mass and $\mathrm{v}$ for volume. For example:

$$\frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{1\ mL_{\ methanol}}} = \mathrm{1\ ppm\ (m/v)} = \frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{0.7918\ mg_{\ methanol}}} = \mathrm{1.263\ ppm\ (m/m)}$$

What's common practice may be heavily dependent on whom we're talking about. Since we usually prepare solutions in the lab by completing the volume up to a certain amount, one could say it's best to ignore the solvent density and use $\mathrm{(/v)}$ as a rule of thumb. The solute unity will depend mostly whether it's a solid being weighed or a liquid being measured.

As pointed out by @PLD, however, there might be cases where it's important to weigh the solvent, probably when very accurate measurements are necessary and specially if the solvent's volume changes considerably given changes in temperature.

One problem of using the $\mathrm{(m/v)}$ notation is that it doesn't use SI units, and might also be confusing. A more verbose way of indicating this is saying mass to volume. Doing some research for this I even found the "units" $\mathrm{ppm_v}$ and $\mathrm{ppm_m}$, which indicates the solvent's unity, but not the solute's, but definetly aren't as common and are also unclear.

The best solution here is simply to use the full unity, so instead of using $\mathrm{ppm}$ (which actually isn't accepted in many cases, for example most scientific journals), use $\mathrm{mg/mL}$ (or $\mathrm{mg\ mL^{-1}}$) or whatever was really measured. This doesn't leave room for double interpretations and indicates how the solution was prepared.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.