A ground water source contains 2.5 mg of F- ion per liter.
a) what is the concentration of F- ions in moles per liter?
b) will CaF2 (Ksp = 2E-10) precipitate when the concentration of Ca2+ = 5.0E-3M?
mols F^- = 0.0025g/atomic mass F and since that is is 1 L, that is the concn in mols/L.
Evaluate Qsp and compare with Ksp.
To find the concentration of F- ions in moles per liter, we need to convert the mass of F- ions to moles and divide it by the volume of the solution (1 liter).
a) Calculation for concentration of F- ions in moles per liter:
- The molar mass of F- is 18.9984 g/mol.
- Since there are 2.5 mg of F- ions per liter, we need to convert it to grams by dividing by 1000 (1 mg = 0.001 g).
2.5 mg = 2.5 × 0.001 g = 0.0025 g
- To convert grams to moles, divide by the molar mass:
0.0025 g ÷ 18.9984 g/mol = 0.00013132 mol
- The concentration is given in moles per liter, so the result is:
0.00013132 mol/L
Therefore, the concentration of F- ions is approximately 0.00013132 mol/L.
b) To determine whether CaF2 will precipitate when the concentration of Ca2+ is 5.0E-3M, we need to compare the ion concentrations to the solubility product constant (Ksp) value of CaF2.
First, we need to determine the concentration of F- ions in the solution (assuming all CaF2 dissociates to form Ca2+ and 2F-):
- Since CaF2 dissociates to form 1 Ca2+ ion and 2 F- ions, the concentration of F- ions will be twice the concentration of Ca2+ ions.
- Therefore, the concentration of F- ions in the solution is: 2 × 5.0E-3M = 1.0E-2M
Now, we can compare the concentration of F- ions to the Ksp value:
- If the concentration of F- ions is greater than the Ksp value (1.0E-2M > 2E-10), CaF2 will precipitate.
- If the concentration of F- ions is equal to or less than the Ksp value (1.0E-2M ≤ 2E-10), CaF2 will not precipitate.
In this case, the concentration of F- ions (1.0E-2M) is significantly greater than the Ksp value (2E-10). Therefore, CaF2 will precipitate when the concentration of Ca2+ is 5.0E-3M.