In many communities, water is fluoridated to prevent tooth decay. In the United States, for example, more than half of the population served by public water systems has access to water that is fluoridated at approximately \(1 \mathrm{mg} \mathrm{F}^{-}\) per liter. (a) What is the molarity of \(\mathrm{F}^{-}\) in water if it contains \(1.2 \mathrm{mg} \mathrm{F}^{-}\) per liter? (b) How many grams of solid KF should be added to a \(1.6 \times 10^{8}\) L water reservoir to give a fluoride concentration of \(1.2 \mathrm{mg} \mathrm{F}^{-}\) per liter?

To calculate molarity, the amount of F- should be in moles, not in milligrams. Thus, first we will convert the given mass (1.2 mg) of F- to moles. The molar mass of F is around 18.998 g/mol.\nSo, \(1.2 \, mg = 1.2 \times 10^{-3} \, g\), and number of moles = \(\frac{1.2 \times 10^{-3} \, g}{18.998 \, g/mol} = 6.32 \times 10^{-5} \, mol\)

Now we use the molarity formula which is number of moles divided by volume in liters. Here the amount of water is 1 liter. So, molarity of F- is \( \frac{6.32 \times 10^{-5} \, mol}{1 \, L} = 6.32 \times 10^{-5} \, M\)

To answer the second part of the question, we must first calculate the total amount of F- needed for the reservoir, then calculate the equivalent amount of KF. We have the molarity and the volume of water hence we can find the amount of F- in moles required. The volume is given in liters so we must convert it to liters first: \(1.6 \times 10^{8} \, L\). Now the number of moles of F- required can be found by multiplying the molarity with volume: \(6.32 \times 10^{-5} \, mol/L \times 1.6 \times 10^{8} \, L = 10.1 \, mol\). \nWe know that in KF, for each molecule of KF there is one ion of F-. So the moles of KF required will be the same as the moles of F-. The molar mass of KF is (39.1 + 18.998) = 58.098 g/mol. So to get the mass in grams we multiply the moles with molar mass: \(10.1 \, mol \times 58.098 \, g/mol = 586.59 \, g\) which is the amount of KF required