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Calculate Qsp for calcium fluoride (Ksp = 3.9 ×10–11) when 140.0 mL of a 6.90×10-3 M...

Calculate Qsp for calcium fluoride (Ksp = 3.9 ×10–11) when 140.0 mL of a 6.90×10-3 M solution of Ca(NO3)2 is added to 255.0 mL of a 7.55×10-3 M solution of KF.

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To calculate the solubility product constant (Qsp) for calcium fluoride (CaF2), we need to determine the concentrations of calcium ions (Ca^2+) and fluoride ions (F^-) in the final solution after mixing the two solutions of Ca(NO3)2 and KF.

The balanced equation for the dissolution of CaF2 is:

CaF2(s) ⇌ Ca^2+(aq) + 2F^-(aq)

Given: Ksp (solubility product constant) for CaF2 = 3.9 × 10^(-11) Volume of Ca(NO3)2 solution (V1) = 140.0 mL Molarity of Ca(NO3)2 solution (M1) = 6.90 × 10^(-3) M Volume of KF solution (V2) = 255.0 mL Molarity of KF solution (M2) = 7.55 × 10^(-3) M

Step 1: Calculate the moles of Ca^2+ and F^- ions from each solution.

For Ca(NO3)2 solution: Moles of Ca^2+ = M1 × V1 = 6.90 × 10^(-3) M × 0.1400 L = 9.66 × 10^(-4) moles Moles of F^- = 0 (Ca(NO3)2 does not contain F^- ions)

For KF solution: Moles of Ca^2+ = 0 (KF does not contain Ca^2+ ions) Moles of F^- = M2 × V2 = 7.55 × 10^(-3) M × 0.2550 L = 1.93 × 10^(-3) moles

Step 2: Determine the initial concentrations of Ca^2+ and F^- ions in the final solution.

Total volume of the final solution (Vf) = V1 + V2 = 0.1400 L + 0.2550 L = 0.3950 L

Initial concentration of Ca^2+ ions (initial [Ca^2+]) = Moles of Ca^2+ / Vf Initial concentration of Ca^2+ ions (initial [Ca^2+]) = 9.66 × 10^(-4) moles / 0.3950 L ≈ 2.45 × 10^(-3) M

Initial concentration of F^- ions (initial [F^-]) = Moles of F^- / Vf Initial concentration of F^- ions (initial [F^-]) = 1.93 × 10^(-3) moles / 0.3950 L ≈ 4.89 × 10^(-3) M

Step 3: Calculate Qsp for CaF2 using the initial concentrations of Ca^2+ and F^- ions.

Qsp = [Ca^2+] × [F^-]^2 Qsp = (2.45 × 10^(-3) M) × (4.89 × 10^(-3) M)^2 Qsp ≈ 6.97 × 10^(-11)

Step 4: Compare Qsp with the given Ksp value.

Ksp for CaF2 = 3.9 × 10^(-11)

Since Qsp (6.97 × 10^(-11)) is greater than Ksp (3.9 × 10^(-11)), we can conclude that calcium fluoride (CaF2) will precipitate from the solution until the concentrations of Ca^2+ and F^- ions reach the equilibrium values required by the Ksp expression.


answered by: Hydra Master
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