Question

Fluoride ion is poisonous in relatively low amounts: \(0.2 \mathrm{~g}\) of \(\mathrm{F}^{-}\) per \(70 \mathrm{~kg}\) of body weight can cause death. Nevertheless, in order to prevent tooth decay, \(\mathrm{F}^{-}\) ions are added to drinking water at a concentration of \(1 \mathrm{mg}\) of \(\mathrm{F}^{-}\) ion per \(\mathrm{L}\) of water. How many liters of fluoridated drinking water would a \(70-\mathrm{kg}\) person have to consume in one day to reach this toxic level? How many kilograms of sodium fluoride would be needed to treat an \(8.50 \times 10^{7}\) gal reservoir?

Short answer

A 70-kg person would need to consume 200 liters of water. The reservoir needs 710,000 kg of sodium fluoride.

Step-by-step solution

Understand the problem

You need to find out how many liters of fluoridated drinking water a 70-kg person has to consume to reach a toxic level and how many kilograms of sodium fluoride are needed to treat a reservoir of given volume.

Calculate the toxic amount of fluoride

The toxic amount of fluoride for a 70-kg person is given as 0.2 grams. Convert this amount into milligrams for consistency: \(0.2 \text{ g} = 200 \text{ mg}\)

Determine water volume for toxic fluoride level

The concentration of fluoride in drinking water is 1 mg/L. To find the water volume that contains the toxic amount, divide the toxic dose by the concentration: \( \frac{200 \text{ mg}}{1 \text{ mg/L}} = 200 \text{ L}\)

Convert reservoir volume to liters

First, convert the given reservoir volume from gallons to liters. Using the conversion factor 1 gallon = 3.78541 liters: \(8.50 \times 10^7 \text{ gal} \times 3.78541 \text{ L/gal} = 3.217 \times 10^8 \text{ L}\)

Calculate mass of sodium fluoride

Calculate the mass of fluoride to add to the reservoir. Since the concentration should be 1 mg/L: \(3.217 \times 10^8 \text{ L} \times \frac{1 \text{ mg}}{1 \text{ L}} = 3.217 \times 10^8 \text{ mg}\) Convert mg to kg: \(3.217 \times 10^8 \text{ mg} = 3.217 \times 10^5 \text{ kg}\) The molecular weight of sodium fluoride (NaF) is 42 g/mol, where fluoride accounts for 19 g/mol of this. Therefore, the actual mass of sodium fluoride needed is: \( \frac{42 \text{ g/mol}}{19 \text{ g/mol}} \times 3.217 \times 10^5 \text{ kg} = 7.1 \times 10^5 \text{ kg}\)

Key concepts

Fluoride ion concentration

Fluoride ions, often denoted as \(\text{F}^{-}\), are added to drinking water to help prevent tooth decay. The typical concentration used is 1 milligram per liter (mg/L). This means for every liter of water, there is 1 milligram of fluoride ion dissolved in it.
To understand this better, let's break it down:
\(\text{1 mg/L}\) literally means for every liter of water, there is 1 milligram of fluoride ions.
This quantity might seem small, but it effectively strengthens tooth enamel and prevents decay when consumed regularly in safe amounts.

Toxic dose calculation

Fluoride, although beneficial in low doses, is toxic in larger amounts. For a 70-kg person, a toxic dose is 0.2 grams of fluoride ion. To put this into perspective, let's convert 0.2 grams to milligrams for easier comparison:
\(0.2 \text{ g} = 200 \text{ mg} \)
In our problem, we need to find out how many liters of water a 70-kg person would need to drink to consume a toxic dose of fluoride. Since our water concentration is 1 mg/L, we can calculate the volume of water like this:
\( \frac{200 \text{ mg}}{1 \text{ mg/L}} = 200 \text{ L} \)
This means, consuming 200 liters of fluoridated water would reach toxic levels for a 70-kg individual. This is a rather large quantity of water for one person to drink in one day, indicating that normal consumption is generally safe.

Sodium fluoride mass

To determine the amount of sodium fluoride (NaF) needed to treat a large water reservoir, we start by converting the reservoir's volume from gallons to liters. If the reservoir's volume is \(8.50 \times 10^7 \text{ gallons}\), we use the conversion factor of 1 gallon = 3.78541 liters:
\(8.50 \times 10^7 \text{ gal} \times 3.78541 \text{ L/gal} = 3.217 \times 10^8 \text{ L}\)
Next, we calculate the mass of fluoride required to achieve the desired concentration in this volume of water. Since we aim for 1 mg/L concentration:
\(3.217 \times 10^8 \text{ L} \times 1 \text{ mg/L} = 3.217 \times 10^8 \text{ mg}\)
Converting milligrams into kilograms:
\(3.217 \times 10^8 \text{ mg} = 3.217 \times 10^5 \text{ kg}\)
Sodium fluoride has a molecular weight of 42 grams per mole (g/mol), of which fluoride itself accounts for 19 grams per mole. Hence, to find the necessary amount of sodium fluoride, we use:
\( \frac{42 \text{ g/mol}}{19 \text{ g/mol}} \times 3.217 \times 10^5 \text{ kg} = 7.1 \times 10^5 \text{ kg} \)
This means 710,000 kilograms of sodium fluoride would be required to treat a reservoir of this size, ensuring the water has the optimal fluoride concentration to prevent tooth decay without reaching toxic levels.