Calculate PPM of Fluoride: NaF Solution

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Discussion Overview

The discussion revolves around calculating the parts per million (PPM) of fluoride in a sodium fluoride (NaF) solution. Participants explore the implications of concentration calculations, particularly focusing on the distinction between the mass of NaF and the mass of fluoride ions (F-) in the solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the PPM of NaF in a solution and questions whether to divide by two to find the PPM of fluoride, considering the 1:1 ratio of NaF to fluoride.
  • Another participant argues that dividing by two does not make sense, as the molarity of NaF is equal to that of fluoride ions.
  • A later reply clarifies that the PPM calculation must consider the mass of fluoride, suggesting the use of the molar mass of fluoride (19 g/mol) instead of that of NaF (41.99 g/mol).
  • Participants discuss the analogy of cars and steering wheels to illustrate the relationship between the number of molecules and their respective masses.
  • One participant mentions that after dilution, they calculated the PPM of the diluted solution and questions whether it is correct to simply divide the original PPM by 25, given the dilution factor.
  • Another participant challenges this approach, indicating that the method may not be valid.

Areas of Agreement / Disagreement

Participants express differing views on how to calculate the PPM of fluoride from the NaF solution, particularly regarding the treatment of mass in the context of dilution. There is no consensus on the correct method for calculating the PPM of the diluted solution.

Contextual Notes

Participants reference various assumptions about molarity, mass ratios, and the implications of dilution, but these assumptions remain unresolved and are subject to interpretation based on the specific context of the problem.

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Homework Statement



Calculate PPM of *Fluoride* in standard solution. NaF (dissolved in solution) = 1.0534g in 0.250L distilled water.

Homework Equations



PPM = Mol/L * g/mol * 1000
c=n/V
n=m/M

The Attempt at a Solution



C=n/V, n = m/M

n = 1.0534/41.99 = 0.02509
c = n/V = 0.02509/0.250 = 0.1M

PPM = 0.1M * 41.99g/mol * 1000 = 4213.59PPM

BUT! Sure that's the PPM of NaF in the solution, but to get the PPM of FLUORIDE do I need to divide it by 2? (NaF, 1:1 ratio) or do I use the mW of F in the PPM equation? (19 instead of 41.99)?

Thanks for any help.
 
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Dividing by two doesn't make sense - 0.1M solution of NaF is 0.1M in both Na+ and F-.
 
Oh right! So the PPM of Fluoride still will be 4213.59?

So it's like saying out of 50 cars you'll still only have 50 steering wheels?

Do my calculations look ok?

Thanks!
 
Bonesmasher said:
Oh right! So the PPM of Fluoride still will be 4213.59?

No. ppm is mass per mass and mass of F- is different from mass of NaF.

So it's like saying out of 50 cars you'll still only have 50 steering wheels?

Yes, but it all depends on the question asked. If you are asked - how many steering wheels in 50 cars, there is 1 steering wheel in each car, so there are 50 steering wheels. But if you are asked - what part of the mass of 50 cars are their steering wheels - you can't say half of the mass of the cars (even if there are 50 cars and 50 steering wheels), because steering wheel is much lighter than the car. Same with NaF - 50 atoms of F per 50 molecules of NaF, but F is not half of the mass.
 
Ah yeah I get what you mean. The question is asking to calculate the concentration of Fluoride in mg/L (PPM).

I'm guessing this would lean to where the steering wheels mass is much less than the cars mass? So I need to use 19 (MW of F-) in the PPM equation?

Since it's a 0.1M conc. of NaF (essentially 0.1M conc. of F also)?
 
Looks like you got it :smile:
 
Thanks!

Just another thing, the answer came to be 1900ppm in 0.250L. I had to dilute it so that I took 0.01L from that solution and made it up to 0.1L. To find the ppm of the diluted solution is it as simple as dividing the ppm by 25? Because I took 1/25 of the solution?

I'm not sure since PPM is mg/L, so it must be related to the volume somehow.

Dividing 0.1M by 25 brings me to 0.004M and putting this into the PPM equation gives me 76ppm.
 
Last edited:
Bonesmasher said:
Just another thing, the answer came to be 1900ppm in 0.250L. I had to dilute it so that I took 0.01L from that solution and made it up to 0.1L. To find the ppm of the diluted solution is it as simple as dividing the ppm by 25? Because I took 1/25 of the solution?

No. But you already know that.
 
Lol thanks Borek :embarrassed:
 

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