Final answer:
To decide if calcium fluoride will precipitate, compare the product of the ion concentrations (Qsp) with the Ksp. Since the calculated Qsp (3.98 × 10−11) is greater than the Ksp (3.45 × 10−11), CaF2 will precipitate.
Step-by-step explanation:
To determine if calcium fluoride (CaF2) will precipitate in a given solution, we use the solubility product constant (Ksp) and compare it to the product of the ion concentrations in the solution. The dissolution equation for CaF2 is:
CaF2(s) ⇌ Ca2+(aq) + 2F−(aq)
The Ksp of CaF2 is given as 3.45 × 10−11. The stoichiometry of the equation indicates that for each mole of CaF2 that dissolves, there are 1 mole of Ca2+ and 2 moles of F− produced. Therefore, if we have [Ca2+] concentration, we can calculate [F−] concentration by multiplying [Ca2+] by 2.
In the case of adding 2.0 mL of 0.10 M NaF to 128 mL of 2.0 × 10−5 M Ca(NO3)2, we must first determine the final concentrations in the mixed solution.
The concentration of Ca2+ after mixing is approximately 2.15 × 10−4 M. Using the stoichiometry relationship:
[F−] = 2 × [Ca2+] = 2 × (2.15 × 10−4 M) = 4.30 × 10−4 M.
Now, we calculate the ion product (Qsp): [Ca2+][F−]2 = (2.15 × 10−4 M)(4.30 × 10−4 M)2 = 3.98 × 10−11 which is slightly higher than the Ksp. This indicates that CaF2 will precipitate under these conditions as Qsp > Ksp.