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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: dnbwise on October 30, 2005, 10:00:56 PM

Title: determination of F- in toothpaste
Post by: dnbwise on October 30, 2005, 10:00:56 PM

I added 10 mL of toothpaste into a beaker and used a F- detecting electrode to take measurments of potential (mV). I used standards to construct a calibration curve and calculated [F-] of toothpaste by the calibration curve. I have the fluoride concentration in ppm ([F-] = 1.919 ppm) now I need to find grams F-. (1ppm = 1ug/mL)

(1.919 ug/mL)(10^-6g/ug)(10ml) = 1.919*10^-5 g

does this look like the correct calculation for determining F- grams from the [F-]?

Title: Re:determination of F- in toothpaste
Post by: sdekivit on October 31, 2005, 10:20:30 AM

looks correct.

Title: Re:determination of F- in toothpaste
Post by: skyaintsnow on November 15, 2005, 01:24:20 PM

it's right.

An easier way to convert the unit is to use the equivalent unit of ppm, mg/liter. You don't need to worry about the 10⁶ part then.