Question

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?

Answer 1

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.

   $V_{(water, toxic)} = \frac{\text{toxic amount}}{\text{concentration}}$

                  = $\frac{0.2 g}{10^{-3} g/L}$

                  = 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 = $850 \times 10^{7} gal$

We will convert gallons into liters as follows.

          $850 \times 10^{7} gal \times 3785 L/gal$

         = $32183 \times 10^{8} L$

Concentration = 1 mg/L

Therefore, we will calculate the mass as follows.

          m = $\text{volume \times concentration}$$\text{volume} \times \text{concentration}$

              = $32183 \times 10^{8} \times 1 mg/L$

              = $32183 \times 10^{2} kg$       (As 1 mg = $10^{-6}$ kg)

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

Answer 2

Answer:

$\boxed{\text{(a) 14 L; (b) 711 kg}}$

Explanation:

(a) Litres of water

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

(b) Mass of  NaF

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

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

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