Analitical Chem: Percentage of Fluoride Ions in Water

[Idealistic]

I'm having a little trouble with this question on my assignment; I believe my method may be wrong. I`ll breifly explain the question and what I did to get my answer.

A 2.5 mL sample of tap water was placed in a 25 mL volumetric flask. Next, 12 mL of a buffer solution was added and then distilled water to dilute the solution and fill the rest of the flask. This solution gave a fluoride molarity of 2.11x10^(-6). What is the concentration of fluoride ions in water in ppm?

1) First I multiplied my concentration by the total volume I assume to be 25 mL:

(2.11x10^(-6)mol/L)(0.025 L) = 5.275x10^(-8)mol ** My total mols of F- present.

2) Next I divided these mols by the amount of tap water added to the beaker initially:

(5.275x10^(-8)mol)/(0.0025 L) = 0.0000211 mol/L ** the molarity of Fluoride in tap water

Next I multiple by the molar mass of Flouride:

(0.0000211 mol/L)(18.998 g/mol) = 0.0004008578 g/L ** the weight of Fluoride in tap water per litre

Finally, I Multiple my "g/L" by 1000 to get the ppm (mg/L)

= 0.401 ppm??

The steps seem correct, unless Flourdie has to be diatomic and im off by a factor of 2 somewhere? but flouride is an ion...

Please help.

[Russ]

PPM is mg/L and Molarity is moles/L

I would have multiplied the molarity by the molar mass (now we're in g/L) and then multiplied by 1000 (to get to mg/L).

That gets me .0401 and possibly one of us has messed up a power of 10 somewhere.

Hope that helped, ask your teacher/lecturer/tutor for a general conversion table if you need it.

[Idealistic]

PPM is mg/L and Molarity is moles/L

I would have multiplied the molarity by the molar mass (now we're in g/L) and then multiplied by 1000 (to get to mg/L).


You mean the molarity given in the question? but wouldn't that refer to the mols of fluoride present in the entire 25 mL solution as opposed to the original 2.5 mL sample of tap water? It would account for our factor of ten difference though.