Question

In: Chemistry

A 0.400 g sample of toothpaste was boiled with a 25 ml solution TISAB buffer to...

A 0.400 g sample of toothpaste was boiled with a 25 ml solution TISAB buffer to extract the fluoride. After cooling, the solution was diluted to exactly 100.0ml with more TISAB solution. The potential of fluoride ion-selective electrode/reference electrode in an aliquot of the sample was measured to be 186.6 mV. This value corresponding to a log[F-] of -3.5016 on a working curve of mV versus log [F-]. Calculate the weight percent F- and NaF in the toothpaste sample.

Solutions

Expert Solution

According to the given data:

Log[F-] = -3.5016

i.e. F- = 3.15*10-4 M

i.e. The moles of F- = 3.15*10-4 mol/L * (100/1000) L = 3.15*10-5 mol

Therefore, the mass of F- = 3.15*10-5 mol * 19 g/mol = 6*10-4 g

Therefore, the mass percent of F- = (6*10-4 g/0.4 g)*100 = 0.15 %


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