Fluoride ion is poisonous in relatively low amounts: 0.2 g of F- per 70 kg of 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 8.5 x107-gal reservoir?

[tex]\text{Mass of F}^{-} = 8.5 \times 10^{7}\text{ gal} \times \dfrac{\text{0.2 mg F}^{-}}{\text{1 kg}} = \textbf{14 mg F}^{-}\\\\\text{Volume of water} = \text{14 mg F}^{-} \times \dfrac{\text{1 L water}}{\text{1 mg F}^{-}} = \textbf{14 L water}\\\\\text{The person would have to consume $\boxed{\textbf{14 L}}$ of water per day}[/tex]

(b) Mass of NaF

[tex]\text{Volume} =8.5 \times 10^{7} \text{gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} = 3.22 \times 10^{8}\text{ L}\\\\\text{Milligrams of F}^{-} = 3.22 \times 10^{8}\text{ L} \times \dfrac{\text{ 1 mg F}^{-}}{\text{1 L}} = 3.22 \times 10^{8}\text{ mg F}^{-}\\\\\text{kilograms of F}^{-} = 3.22 \times 10^{8}\text{ mg F}^{-} \times \dfrac{\text{1 mg F}^{-}}{10^{6}\text{kg F⁻}} = \text{322 kg F}^{-}[/tex]

1 kmol of NaF (41.99 kg) contains 19.00 kg of F⁻.

[tex]\text{Kilograms of NaF}= \text{322 kg F}^{-} \times \dfrac{\text{41.99 kg NaF}}{\text{19.00 kg F}^{-}} = \text{711 kg NaF}\\\\\text{It will take } \boxed{\textbf{711 kg}} \text{ of NaF to treat the reservoir}[/tex]

OlaMacgregor OlaMacgregor · Answer 2 · 2020-03-23T08:48:31+01:00

Explanation:

(a) The given data is as follows.

Concentration = 1 mg/L, Toxic amount = 0.2 g

Now, we will calculate the volume of fluoridated drinking water as follows.

[tex]V_{(water, toxic)} = \frac{\text{toxic amount}}{\text{concentration}}[/tex]

= [tex]\frac{0.2 g}{10^{-3} g/L}[/tex]

= 200 L

Hence, there will be 200 liters of ﬂuoridated drinking water would a 70-kg person have to consume in one day to reach this toxic level.

(b) Now, it is also given that

V = [tex]850 \times 10^{7} gal[/tex]

We will convert gallons into liters as follows.

[tex]850 \times 10^{7} gal \times 3785 L/gal[/tex]

= [tex]32183 \times 10^{8} L[/tex]

Concentration = 1 mg/L

Therefore, we will calculate the mass as follows.

m = [tex]\text{volume \times concentration}[/tex][tex]\text{volume} \times \text{concentration}[/tex]

= [tex]32183 \times 10^{8} \times 1 mg/L[/tex]

= [tex]32183 \times 10^{2} kg[/tex] (As 1 mg = [tex]10^{-6}[/tex] kg)

Thus, we can conclude that there are [tex]32183 \times 10^{2} kg[/tex] of sodium fluoride would be needed to treat a [tex]8.5 \times 10^{7} gal[/tex] reservoir.