Reef Chemistry Question of the Day #56 Fluoride mg/L = ? ppm

[Randy Holmes-Farley]

Reef Chemistry Question of the Day #56

Suppose that fluoride is present in seawater with a salinity of 35 ppt at a concentration of 1.000 mg/L. When the sample is at 77°F, which of the following is closest to the fluoride concentration in ppm?


A. 1.000 ppm
B. 0.977 ppm
C. 1.026 ppm
D. 35 ppm

Good luck!










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[Randy Holmes-Farley]

No fluoride fans today?

 

[Cory]

I'm going to guess A. But only a guess. But just because my alk test kit says ppm is the same as mg/l. But now there is all these other factors like atomic weight, temperature etc. haha.

 

[DFW]

A!!!!!!

 

[spspirate]

I guess A too

 

[Randy Holmes-Farley]

Anyone else want to try before I give the answer?

 

[Andrew.concepcion]

A. I believe mg/L is equal to ppm since ug/L is equal to ppb. I hope I got this right. I work for a lab testing company.

 

[Randy Holmes-Farley]

And the answer is... B. 0.977 ppm

The point of this question is that ppm (parts per million) is not exactly the same as mg/L unless the fluid weighs exactly 1.000 g/L, and since seawater does not, ppm is a bit different than mg/L. Parts per million means exactly what is sounds like: parts of the chemical being measured per million parts of the total material. That could be mg/kg (= ppm by weight) or ul/L (= ppm by volume) or other related sorts of things, but the units must match for it to always be true. You cannot mix a weight unit (mg) and a volume unit (L) unless the weight is exactly 1.

The density of seawater is higher than pure water due to the density added by the dissolved salts. The density (not the specific gravity) of 35 ppt seawater at typical reef aquarium temperatures is about 1.023 grams per cm³ (=1.023 kg/L).

So if you have 1.000 mg/L of fluoride, that equals 1.000 mg of fluoride in `1 liter of seawater. That 1 liter of seawater weighs about 1.023 kg, so

1.000 ppm / 1.023 kg/L = 0.977 mg/kg, which equals 0.977 ppm.

Happy Reefing.