Question

		Molar Mass	% F-	% Pure Compound in the Commercial Product
NaF	Sodium Fluoride	42	45	95
Na2SiF6	Sodium fluorosilicate	188	61	95
H2SiF6	Fluorosilicic acid	144	79	30

ii. How much 30.0% commercial fluorosilicic acid in kg must be added to 50,000 gallons of water in order to fluoridate at the optimum level?

Answer 1

Ans:

Optimum fluoride level in water: 0.5 to 1.5 mg/L. Meaning, 0.5 to 1.5 mg in 1 litre of water.

Let us keep the optimum fluoride level in water to be 0.6mg in 1 L of water

50000 gallons of water = (50000 x 3.7854) L = 189270 L.          {1 gallon = 3.7854 L}

We expect 1 L of water to contain 0.6 mg of fluoride to reach optimum level.

So, 189270 L of water should contain (189270 x 0.6) = 113562 mg of fluoride = 113.562 g of fluoride.

1 mol of H₂SiF₆ contains 6 mol of fluoride.

So, 144 g of H₂SiF₆ contains (6 x 18.96) =113.76 g of fluoride.

Then the mass of H₂SiF₆ which contains 113.562 g of fluoride = (113.562 x 144)/113.76 = 143.749 g.

Required H₂SiF₆ to treat 50000 gallons of water = 143.749 g

Commercially available H₂SiF₆ is 30% pure.

Meaning, 100 g of commercially available fluorosilicic acid contains 30 g of H₂SiF₆.

Mass of commercially available fluorosilicic acid to be taken to get 143.749 g of H₂SiF₆ = (143.749 x 100)/30 = 476.163 g = 0.476 kg (approximated)

So, we have to take 0.476 kg of commercially available fluorosilicic acid to treat 50000 gallons of water to have the fluoride level at 0.6mg/L.