Question

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?

Answer #1

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

b)

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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