A water sample saturated in CaF2 has Ca2+ of 115 ppm (115g Ca2+ per 10^6g water sample). What is the F- ion content of the water in ppm?
LOL who else dosnt want to pay for chegg haha. Someone plz post the answer. Thank you
To find the F- ion content in the water sample, we can use the molar ratio between Ca2+ and F- ions in CaF2.
The molar mass of Ca2+ is 40.08 g/mol, and the molar mass of F- is 18.99 g/mol. The ratio between Ca2+ and F- in CaF2 is 1:2.
First, let's calculate the number of moles of Ca2+ ions in the water sample:
Number of moles of Ca2+ = Mass of Ca2+ / Molar mass of Ca2+
= 115 g / 40.08 g/mol
≈ 2.871 mol
Since the ratio between Ca2+ and F- ions in CaF2 is 1:2, the number of moles of F- ions is twice the number of moles of Ca2+ ions:
Number of moles of F- = 2 * Number of moles of Ca2+
≈ 2 * 2.871 mol
≈ 5.742 mol
Finally, let's convert the number of moles of F- ions back to mass and express it in ppm (parts per million):
Mass of F- = Number of moles of F- * Molar mass of F-
≈ 5.742 mol * 18.99 g/mol
≈ 109.009 g
Now, we can calculate the F- ion content in ppm:
F- ion content = Mass of F- / Mass of water sample * 10^6
= 109.009 g / 10^6 g * 10^6
≈ 109 ppm
Therefore, the F- ion content in the water sample is approximately 109 ppm.
To determine the F- ion content of the water in ppm, we need to use the concept of stoichiometry and the molar mass of calcium fluoride (CaF2).
Here's how we can calculate it step by step:
1. Find the molar mass of CaF2:
- Ca (calcium) has a molar mass of 40.08 g/mol.
- F (fluorine) has a molar mass of 18.99 g/mol.
- Since CaF2 has one calcium atom and two fluorine atoms, the molar mass of CaF2 is:
Molar mass of CaF2 = (1 * molar mass of Ca) + (2 * molar mass of F)
= (1 * 40.08 g/mol) + (2 * 18.99 g/mol)
= 40.08 g/mol + 37.98 g/mol
= 78.06 g/mol
2. Convert ppm (parts per million) to mg/L (milligrams per liter):
- Since 1 ppm is equivalent to 1 mg/L for dilute solutions, we can use the same conversion.
- Therefore, 115 ppm Ca2+ is equal to 115 mg/L.
3. Convert the concentration of Ca2+ (115 mg/L) to moles per liter:
- To do this, we divide the concentration by the molar mass of calcium (40.08 g/mol):
Moles of Ca2+ per liter = (115 mg/L) / (40.08 g/mol)
= 2.872 mol/L
4. Determine the number of moles of F- ions present:
- Since calcium fluoride (CaF2) has a 1:2 ratio of calcium ions to fluoride ions, the number of moles of F- ions will be twice that of the Ca2+ ions.
- Therefore, the number of moles of F- ions per liter is:
Moles of F- per liter = 2 * (2.872 mol/L)
= 5.744 mol/L
5. Convert moles per liter to ppm:
- To convert moles per liter to ppm, we multiply the concentration by the molar mass of F- (18.99 g/mol) and then divide by 1 mg/L:
F- ion content in ppm = (Moles of F- per liter * molar mass of F-) / (1 mg/L)
= (5.744 mol/L * 18.99 g/mol) / (1 mg/L)
= 109.00356 ppm
Therefore, the F- ion content of the water sample saturated in CaF2 is approximately 109 ppm.