Question

According to the "Guidelines for Canadian Drinking Water Quality", although fluoride (F) is beneficial in preventing tooth decay, its acceptable concentration in drinking water is limited to 1.5mg/L. Find the maximum amount of sodium fluoride (NaF) in the kg that can be added to 3 MG water with no fluoride. ( MG= million gallons; mg=milligrams).

Answer 1

Given,

The volume of water =3 MG (million gallons)

Acceptable concentration of fluoride(F) in drinking water = 1.5 mg /L

Now, calculating the mass of fluoride(F) for 3 MG water,

= 3 MG x ( 10⁶ Gallons / 1 MG) x ( 3.7854 L / 1 gallon) x ( 1.5 mg F / 1 L) x ( 1 g /1000 mg )

= 17034.3 g of fluoride(F) in 3 MG water.

Now, Calculating the number of moles of fluoride(F),

= 17034.3 g of fluoride(F) x ( 1 mol / 19 g)

= 896.5 mol of fluoride

Now, the dissociation of NaF is,

NaF(aq) Na⁺(aq) + F^-(aq)

Now, using the moles of fluoride and mole ratio from the balanced chemical reaction, calculating the number of moels of NaF,

= 896.5 mol of fluoride x ( 1 mol of NaF / 1 mol F^-]

= 896.5 mol of NaF

Converting the number of moles to kilograms,

= 896.5 mol of NaF x ( 41.988 g / 1 mol) x ( 1 kg /1000 g)

= 37.6 kg of NaF

Thus, 37.6 kg NaF is the maximum amount that can be added to 3 MG water with no fluoride.