A water treatment plant adds fluoride ion to the water to the extent of 1.5mg F-/L H2O. a)What is the molarity of fluoride ion in this water? b)If the fluoride ion in the water is supplied by calcium fluoride, what mass of calcium fluoride is present in 1.00 x 10^6 L of this water?

1.5 mg/L = 0.0015 g/L

mols F = 0.0015 g/atomic mass F = ? and that is mols/L so = ? M = approx 8E-5 mols/L but that is an estimate.

In 1E6 L there will be approx 8E-5 mols/L x 1E6 L = approx ? mols F.

You will need 1/2 that mols CaF2 since CaF2 contains 2 mols F for every mols CaF2. Then g CaF2 = mols CaF2 x molar mass CaF2.

To find the molarity of fluoride ion in the water, we need to calculate the moles of fluoride ion per liter of water.

a) Molarity (M) is defined as moles of solute per liter of solution.

Given: Fluoride ion concentration = 1.5 mg/L.

We first need to convert the given concentration from mg/L to moles/L. The molar mass of fluoride ion (F-) is 18.998 g/mol.

1.5 mg/L * (1 g / 1000 mg) * (1 mol / 18.998 g) = 0.079 mol/L

Therefore, the molarity of fluoride ion in the water is 0.079 M.

b) To find the mass of calcium fluoride present, we need to calculate the moles of calcium fluoride based on the molar ratio between fluoride ion (F-) and calcium fluoride (CaF2).

From the balanced chemical equation for the dissociation of calcium fluoride:
CaF2 -> Ca2+ + 2F-

We see that for every 1 mole of CaF2, we get 2 moles of F- ions.

Given: Volume of water = 1.00 x 10^6 L.

Number of moles of F- ions = Molarity * Volume = 0.079 mol/L * 1.00 x 10^6 L = 7.90 x 10^4 mol.

From the molar ratio, we know that 1 mole of CaF2 produces 2 moles of F- ions.

Therefore, the moles of CaF2 present in the water = (7.90 x 10^4 mol) / 2 = 3.95 x 10^4 mol.

The molar mass of CaF2 is 78.074 g/mol (40.08 g/mol for Ca + 2 * 18.998 g/mol for F).

Mass of calcium fluoride = moles * molar mass = (3.95 x 10^4 mol) * (78.074 g/mol) = 3.08 x 10^6 g.

Therefore, there are 3.08 x 10^6 grams of calcium fluoride present in 1.00 x 10^6 L of water.

To find the answers to both parts of the question, we will need to use some basic concepts of stoichiometry and molarity.

a) To find the molarity of fluoride ion in the water, we need to know the chemical formula of fluoride ion. Fluoride ion is denoted as F-. It means that it carries a single negative charge.

The molarity (M) of a solution is defined as the moles of solute divided by the liters of solution. In this case, the solute is the fluoride ion (F-) and the solution is the water (H2O).

To convert the given amount of fluoride ion (1.5 mg F-/L H2O) into moles, we need to know the molar mass of fluoride, which is 18.9984 g/mol.

First, convert the given amount from milligrams to grams:
1.5 mg = 0.0015 g

Now, use the molar mass to convert grams of fluoride ion (F-) to moles:
0.0015 g F- × (1 mol F-/18.9984 g F-) = 7.501 x 10^-5 mol F-

Finally, divide the moles of fluoride ion by the liters of water to find the molarity:
Molarity = 7.501 x 10^-5 mol F-/1 L = 7.501 x 10^-5 M

Therefore, the molarity of fluoride ion in this water is 7.501 x 10^-5 M.

b) To find the mass of calcium fluoride present in 1.00 x 10^6 L of water, we will need to use the stoichiometry between calcium fluoride and fluoride ion.

The chemical formula of calcium fluoride is CaF2. It means that one formula unit of calcium fluoride contains two fluoride ions (F-).

First, we need to calculate the moles of fluoride ion in 1.00 x 10^6 L of water using the molarity we calculated earlier (7.501 x 10^-5 M):
Moles of fluoride ions = Molarity × Volume
Moles of fluoride ions = 7.501 x 10^-5 mol F-/L × 1.00 x 10^6 L = 75.01 mol F-

Since each formula unit of calcium fluoride contains two fluoride ions, we can convert the moles of fluoride ions to moles of calcium fluoride by dividing by 2:
Moles of calcium fluoride = 75.01 mol F-/2 = 37.51 mol CaF2

Now, use the molar mass of calcium fluoride to convert moles to grams:
Molar mass of CaF2 = 40.08 g/mol (for Ca) + 18.9984 g/mol (for F) × 2 = 78.08 g/mol
Mass of calcium fluoride = 37.51 mol CaF2 × 78.08 g/mol = 2,926.6 g

Therefore, the mass of calcium fluoride present in 1.00 x 10^6 L of water is 2,926.6 grams.