Fluoride ion is poisonous in relatively low amounts: 0.2 g of F
- per 70 kg body weight can cause death. Nevertheless, in order to
prevent tooth decay, F - ions are added to drinking water at a
concentration of 1 mg of F - ion per L of water.
How many liters of fluoridated drinking water would a 70 kg person have to consume in one day to reach this toxic level?
How many kilograms of sodium fluoride would be needed to treat a 6.23 x 10^7-gal reservoir?
Calculate mass in mg
0.2 g = 200 mg for a 70 kg man to die
1 mg per Liter so
200 mg = 200 L
a man needs to drink 200 liter to get 200 mg of F; which will kill him
for V = 6.23*10^7 gal
change gal to l
(6.23*10^7)(3.78) = 235494000 Liter
which must have 1 mg per liter so
235494000 Liter = 235494000 mg of F-
change mg to g
235494000 mg = 235494 g
change F- to mol
mol = mass/MW = 235494/(18.99) = 12400.947 mol of F-
ratio is 1 mol of F- per 1 mol of NaF so
12400.947 mol of NaF are needed
MW of NaF = 41.98871
mas sof NaF = 41.98871*12400.947 = 520699.767 g of NaF are needed
we need this in kg so
520699.767308 g = 520.699 kg of NaF are needed to treat that volume with 1 mg of F- per liter
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