Expected concentration of fluoride released from PFAA

  • #1
rwooduk
762
59

Homework Statement


You have 200 mL of perfluorooctanesulfonic acid with a concentration of 10 mg/L. If the PFOS is completely mineralised in solution what is the expected concentration of fluoride in solution?

The experimentally determined fluoride concentration is 5.6 mg/L

Homework Equations


n=m/M
C=n/v
No. of molecules = nNA

The Attempt at a Solution


I seem to have forgotten my basic chemistry, kind of worrying but would appreciate if someone could quickly advise.##m = CV = (200 mL)* 10 \frac{mg}{L} = 0.002g##

##n = \frac{m}{M} = \frac{0.002}{500\frac{g}{mol}} = 4 * 10^{6} mol##

##no. \enspace of \enspace molecules = N_{A}*n = (4*10^{6}mol)(6.022*10^{23})=2.4*10^{30}\enspace PFOS\enspace molecules##

PFOS has 17 fluoride atoms per molecule therefore:

##No.\enspace of\enspace fluoride\enspace atoms\enspace in\enspace solution = 17 * 2.4*10^{30} = 4*10^{31} fluoride\enspace atoms##

Now if I work backwards to find the fluoride concentration in solution...

##
n = \frac{no.\enspace of \enspace molecules}{N_{A}} = \frac{4 * 10^{31}}{6*10^{23}}= 6.67*10^{7} mols####m= nM = (6.67*10^{7} mols)(19\frac{g}{mol})= 1.27*10^{9}g##

Oh dear, that's a lot of fluoride, any pointers where I went wrong appreciated.
 
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  • #2
What is 0.002 divided by 500? Is it more or less than 1?
 
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  • #3
TeethWhitener said:
What is 0.002 divided by 500? Is it more or less than 1?
Ooops good point! Late Friday!
 
  • #4
Yep, get around 6.5 mg/L so within the answer value. Thanks for the spotted mistake!
 

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