Problem 12 - Analytical Chemistry

Question

A city's water supply is fluoridated by adding NaF. The desired concentration of \(\mathrm{F}^{-}\) is 1.6 ppm. How many \(\mathrm{mg}\) of \(\mathrm{NaF}\) should you add per gallon of treated water if the water supply already is 0.2 ppm in \(\mathrm{F}^{-}\) ?

Step-by-Step Solution

Answer

Add 11.71 mg of NaF per gallon of water.

Step 1: Find the Additional Fluoride Needed

Since the water supply already has 0.2 ppm of \(\mathrm{F}^{-}\), we need an additional \(1.6 \text{ ppm} - 0.2 \text{ ppm} = 1.4 \text{ ppm}\) of \(\mathrm{F}^{-}\) ions to reach the desired concentration of 1.6 ppm.

Step 2: Convert PPM to Milligrams per Gallon

The term ppm (parts per million) for water refers to mg/L as there is 1 million mg in a liter of water. Since 1 gallon equals approximately 3.78541 liters, 1 ppm is equivalent to \(1 \text{ mg/L} \times 3.78541 \text{ L/gallon} = 3.78541 \text{ mg/gallon}\). Therefore, for 1.4 ppm, it will be \(1.4 \text{ ppm} \times 3.78541 \text{ mg/gallon} = 5.29958 \text{ mg/gallon of } \mathrm{F}^{-}\).

Step 3: Calculate the Amount of NaF Needed

The molecular weight of \(\mathrm{NaF}\) is approximately 41.99 g/mol and \(\mathrm{F}^{-}\)'s atomic weight is approximately 19.00 g/mol. Therefore, the fraction of \(\mathrm{F}^{-}\) in \(\mathrm{NaF}\) is \(\frac{19.00}{41.99}\). The required \(\mathrm{NaF}\) for 5.29958 mg of \(\mathrm{F}^{-}\) is \(5.29958 \text{ mg} \div \left( \frac{19.00}{41.99} \right) = 11.710 \text{ mg}\).

Step 4: Round to Two Decimal Places

Finally, round the result to two decimal places for simplicity: \(11.71 \text{ mg of } \mathrm{NaF}\) needs to be added per gallon of water.

Key Concepts

Calculating Fluoride Concentrations

When it comes to fluoridating city water supplies, getting the right concentration of fluoride is essential. Too little won't be effective; too much could be harmful. To find the right balance, we must calculate the needed fluoride concentration accurately.

In our example, the city's water already has 0.2 parts per million (ppm) of fluoride. However, the goal is to increase this to a total of 1.6 ppm. To find the necessary increase, we simply subtract the current concentration from the desired concentration: 1.6 ppm - 0.2 ppm = 1.4 ppm. This means we need an additional 1.4 ppm of fluoride ions to get the water supply to the optimal level.

Parts Per Million (PPM)

Parts per million (PPM) is a common measurement in water treatment and describes the amount of a substance per million parts of solution. For water, this is typically expressed as milligrams per liter (mg/L) because there are a million milligrams in a liter of water.

When dealing with different volumes, like gallons, it's necessary to convert ppm to an appropriate unit. Since one gallon equals approximately 3.78541 liters, we can convert ppm to mg per gallon by multiplying by this factor. Thus, 1 ppm is equivalent to 3.78541 mg per gallon. In our example, needing an extra 1.4 ppm of fluoride translates to 5.29958 mg/gallon (1.4 ppm × 3.78541 mg/gallon). This conversion ensures we calculate the appropriate fluoride amounts for the entire water system.

Molecular Weight Calculations

To determine the precise amount of sodium fluoride (NaF) needed, understanding molecular weight is crucial. Sodium fluoride consists of sodium (Na) and fluoride (F), and its molecular weight is about 41.99 grams per mole. Meanwhile, the atomic weight of fluoride is around 19.00 grams per mole.

An important step is to calculate the fraction of fluoride in NaF by dividing the atomic weight of fluoride by the molecular weight of NaF; this gives us \( \frac{19.00}{41.99} \). This step helps identify how much fluoride is present in a given amount of NaF.

Using this fraction, we can determine how much NaF is needed to provide the 5.29958 mg of fluoride per gallon. This is done by dividing the required fluoride mass by the fraction of fluoride in NaF, resulting in 11.71 mg of NaF per gallon. This ensures that the water supply hits the target concentration of fluoride ions.