A linear regression line was obtained; $$\log(\ce{[NaF]})~(\mathrm{mol~L^{-1}})~ \mathrm{vs~EMF}: y = 59.8x +600.8$$ $0.7540~\mathrm{g}$ of toothpaste was diluted in a$50~\mathrm{mL}$ and tested to obtain a EMF value of $390$. what is the ppm of $\ce{F-}$ in $198~\mathrm{g}$ of toothpaste?

I plugged in $390$ into the eqn to get $-3.525$, taking the antilog to obtain $0.0002985$, and times it by the volume $0.05~\mathrm{L}$ to get the moles of $\ce{NaF}$.

Due to stoichiometry, this also equals $1~\mathrm{mol}$ of $\ce{F-}$. I multiplied by $18.998~\mathrm{g~mol^{-1}}$ to obtain $0.000283527~\mathrm{mol}$ of $\ce{F-}$. I then multiplied by the factor: $\frac{198}{0.750}$ to obtain $0.074454015~\mathrm{g}$ of $\ce{F-}$ in the $198~\mathrm{g}$ of toothpaste.

I'm stuck here, how can I calculate the ppm?

The expected value of fluoride in toothpaste is $1000~\mathrm{ppm}$

  • 1
    $\begingroup$ I'm having difficulty getting your expected answer. I follow your method through moles of fluoride, but at that point, your units get messed up. Multiplying by the molar mass gives you 0.0002835 grams of fluoride. That value divided by the original mass of the sample, 0.7540 grams of toothpaste, multiplied by one million, should give you ppm, regardless of sample size. I get 376 ppm. Parts per million is a concentration, so if you figure it out once for any quantity of toothpaste, it should apply to all quantities. $\endgroup$ – Jason Patterson Jul 4 '15 at 1:37
  • $\begingroup$ thanks for your support, i guess it has to deal with the regression line, it must be wrong? i get the same answer. $\endgroup$ – Spencer Trinh Jul 4 '15 at 10:12
  • $\begingroup$ @Spencer-Trinh I'd guess you're past caring at this point, but I'm curious, was this a word problem or a lab experiment? $\endgroup$ – MaxW Oct 30 '15 at 23:25
  • $\begingroup$ @JasonPatterson I think you already pretty much gave the answer. Would you mind writing your comment up and posting it as an answer? $\endgroup$ – tschoppi Feb 3 '16 at 11:39

Working with your numbers gives me the same result through moles of $\ce{NaF}$. However, at this point you can multiply by the molar mass of fluoride, 18.998 g/mol, to get grams of fluoride, not moles, specifically, 0.0002835 g $\ce{NaF}$.

Parts per million is a mass-based concentration measure. It's similar in concept to percent mass, but instead of parts per hundred, it uses parts per million. Since we know the original mass of this sample, we can calculate the concentration in ppm based on that value. It will be the same concentration as in the larger sample.

I get the following:
$\frac{0.0002835 \textrm{ g } \ce{F-}}{0.7540 \textrm{ g toothpaste}}\times 1000000 = 376.0\textrm{ ppm }\ce{F-}$


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