Water fluoridation is an important preventative measure carried out in much of t

Question

Water fluoridation is an important preventative measure carried out in much of the western world. It results in some of the hydroxyapatite, Ca5(PO4)3OH, of which human tooth enamel is made being replaced by fluoroapatite, Ca5(PO4)3F - a substance significantly more resistant to decay. Thus, to protect the teeth of the population, water is often fluoridated. This is usually done with one of three fluorine-containing chemicals (sodium fluoride, sodium fluorosilicate and hydrofluorosilicic acid). The following equations describe the reactions that occur on human tooth with and without fluoridation.

Tooth decay: Ca5(PO4)3OH(s) + 4H3O+ 5Ca2+ + 3HPO42- + 5H2O

Fluoridation: 6Ca5(PO4)3OH(s) + H2SiF6 6Ca5(PO4)3F(s) + SiO2 + 4H2O

a)Calculate the percent fluoride by weight in H2SiF6.

b)Minnesota Statutes require all municipal water supplies to maintain a fluoride concentration of 0.9 ppm to 1.5 ppm (measured as mass of F per liter), with an optimum level of 1.2 ppm (measured as mass of F per liter). The city of Mankato, MN uses groundwater wells for water supply. The flow of one of the wells is 1,000 gallons per minute (gpm) and runs for 24 hours. The city aims to achieve optimum concentrations of fluoride. The city uses hydrofluorosilicic acid (H2SiF6) with 25 % purity. How many gallons of acid per day must the city add for each well to keep the concentration to 1.2 ppm?

c)According to the fluoridation reaction given above, how much fluoroapatite, Ca5(PO4)3F, is produced when the concentration of hydrofluorosilicic acid,H2SiF6, is supplied at 1.2 ppm?

Explanation / Answer

a)Calculate the percent fluoride by weight in H2SiF6.

% F = mass of F / Total H2SiF6 *100%

% F = 6*(18.998403 )/(144.09180 ) * 100

% F = 79.109 % is F

b)

0.9 ppm to 1.5 ppm

optimal = 1.2

Q = 1000 GPM, t= 24 h

Q = 1000 gal / min * 3.785 L / gal = 3785 L/min -> 3785 L/min * 60 min / h * 24 H = 5450400 L/day

1.2 ppm --> 1.2 mg of F / liter

5450400 L/day

we need

mass of F = 1.2 mgF / L * 5450400 L /day= 6540480 mg F / day = 6540.48 g F/day

if

% F = 79.109 % is F

6540.48 g F/day 100/79.109 H2SiF6

mass of H2SiF6 = 5174.108 g of H2SiF6

but this is at 100% purity, so.. for 25% purity

H2SiF6 = 5174.108 * 100%/25%= 20696.432 g of H2SiF6 mix at 25%

for gallons (volume) of acid, add density/concentration

c)

Ca5(PO4)3F

6Ca5(PO4)3OH(s) + H2SiF6 6Ca5(PO4)3F(s) + SiO2 + 4H2O

ratio is

1 mol of H2SiF6 = 6 mol of fluoroapatite

5174.108 g of H2SiF6 --> mol = mass/MW H2SiF6 = 5174.108   /144.09180 = 35.9084 mol of H2SiF6

then,

mol of fluoroapatite --> 6*35.9084 = 215.4504 mol of fluoroapatite

mass = mol*MW = 215.4504 *504.3025 = 108652.17 g of fluoroapatite = 108.65 kg of fluoroapatite

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