Question

Fluoridation is the process of adding fluorine compounds to drinking water to help fight tooth decay. A concentration of 1 ppm of fluorine is sufficient for the purpose. (1 ppm means \(1 \mathrm{~g}\) of fluorine per 1 million g of water.) The compound normally chosen for fluoridation is sodium fluoride, which is also added to some toothpastes. Calculate the quantity of sodium fluoride in kilograms needed per year for a city of 50,000 people if the daily consumption of water per person is 150 gallons. What percent of the sodium fluoride is "wasted" if each person uses only \(6.0 \mathrm{~L}\) of water a day for drinking and cooking? (Sodium fluoride is 45.0 percent fluorine by mass. 1 gallon \(=3.79 \mathrm{~L} ; 1\) year \(=365\) days; density of water \(=1.0 \mathrm{~g} / \mathrm{mL} .)\)

Short answer

To compute the quantity of sodium fluoride needed and the percentage that is wasted, calculate the total annual water consumption in liters, calculate the necessary amount of sodium fluoride to be added to reach 1 ppm of fluorine concentration, compute the amount of sodium fluoride used for drinking and cooking, and finally determine the percentage of sodium fluoride that is wasted. Implement the equations from the steps in Step-by-Step Solution to achieve the numerical answer.

Step-by-step solution

Calculate Total Water Consumption per Year

First, convert the daily consumption of water per person from gallons to liters. Multiply this consumption by the number of people and the number of days in a year to get the total annual water consumption for the city. \n\nGiven: \nDaily water consumption per person = 150 gallons\nNumber of people = 50,000\nNumber of days in a year = 365\n1 gallon = \(3.79 \mathrm{~L}\) \n\nTotal Water Consumption in Liters = Daily consumption per person * number of people * number of days in a year\n= (150 gallons/person/day * 3.79 L/gallon) * 50,000 people * 365 days/year

Calculate Total Amount of Sodium Fluoride Needed

Next, calculate the total amount of Sodium Fluoride to be added to the water. This is done by using the concentration of fluorine needed to fight tooth decay (1 ppm). This means \(1 \mathrm{~g}\) of fluorine per 1 million g of water. Remember that Sodium fluoride is 45.0 percent fluorine by mass. \n\nTotal amount of Sodium Fluoride = Total water consumption in liters * density of water * concentration of fluorine needed * proportion of sodium fluoride to fluorine\n= Total water consumption in L * \(1.0 \mathrm{~g/mL} * 1 \mathrm{~mL/L}\) * \(1 \mathrm{~g of fluorine / 10^6 ~g of water}\) * \(\frac{100}{45.0}\)

Calculate Amount of Sodium Fluoride for Drinking and Cooking

It's given that each person uses only \(6.0 \mathrm{~L}\) of water a day for drinking and cooking. Calculate the total amount of Sodium Fluoride that is consumed for drinking and cooking following the same approach as in Step 2. \n\nTotal amount of Sodium Fluoride consumed for drinking and cooking = Daily water consumption for drinking and cooking * number of people * number of days in a year * density of water * concentration of fluorine needed * proportion of sodium fluoride to fluorine\n= \(6.0 \mathrm{~L/person/day}\) * 50,000 people * 365 days/year * \(1.0 \mathrm{~g/mL} * 1 \mathrm{~mL/L}\) * \(1 \mathrm{~g of fluorine / 10^6 ~g of water}\) * \(\frac{100}{45.0}\)

Compute The Percentage of Wasted Sodium Fluoride

The percentage of wasted Sodium Fluoride can be calculated by subtracting the total amount of Sodium Fluoride used for drinking and cooking from the total amount of Sodium Fluoride added to the water, then dividing by the total amount of Sodium Fluoride added to the water, and finally multiplying by 100. \n\nWasted Sodium Fluoride % = \(\frac{Total ~amount ~of ~Sodium ~Fluoride ~- ~Sodium ~Fluoride ~used ~for ~drinking ~and ~cooking}{Total ~amount ~of ~Sodium ~Fluoride}\) * 100

Key concepts

Understanding Sodium Fluoride

Sodium fluoride is a salt compound used widely for dental health. It comprises of sodium (Na) and fluorine (F), and is often found in water fluoridation processes and toothpaste. Its key feature is to release fluoride ions which are instrumental in preventing tooth decay.

Fluoride works by strengthening the tooth enamel, which makes it more resistant to decay. Additionally, it inhibits the processes that lead to the breakdown of dental structures by bacteria.

As sodium fluoride is 45% fluorine by mass, it is this fluorine component that plays a crucial role in dental care. When added to city water supplies, sodium fluoride helps reduce the prevalence of cavities and strengthens teeth across the population, particularly beneficial for children who are developing their permanent teeth.

Impact of Water Consumption

Water consumption is a crucial factor in determining how much sodium fluoride must be added to a city's water supply. For effective fluoridation, it is essential to know the average amount of water each person uses daily.

To calculate annual water usage, multiply the daily consumption by the total population and then by the number of days in a year. Based on the given data, in our case, each person uses 150 gallons of water per day. Translated to liters, this is about 570 liters per person daily.

Understanding this number helps in precisely calculating the amount of sodium fluoride required to ensure every resident receives the optimal fluoride level to benefit dental health. In our case study, the calculation considers a city with 50,000 residents, ensuring that everyone has access to sufficient fluoride in their daily water usage.

Tooth Decay Prevention through Fluoridation

Fluoridation is a highly recommended public health measure to combat tooth decay. When fluoride is present at a concentration of 1 part per million (ppm) in water, it effectively helps to protect teeth from decay.

Tooth decay is caused by bacteria that produce acid, which can degrade tooth structures. Fluoride helps by slowing down this bacterial activity and beams up the tooth enamel. This process fortifies the enamel, making it less susceptible to acid attacks and reducing cavity formation.

The implementation of water fluoridation ensures everyone, regardless of background, benefits from reduced risks of dental caries. Although not all water usage directly benefits dental health, the systemic nature of water fluoridation means that even indirect benefits, such as in food preparation, are realized. Fluoridation is an easy and effective way to sustainably improve dental health at the community level.