Calculate Qsp for calcium fluoride (Ksp= 3.9x10^-11) when 125.0 mL of a 0.00615 M solution of Ca(NO3)2 is added to 330.0 mL of a 0.00615 M solution of KF

Qsp = 3.36x10⁻⁸

Explanation:

The Ksp of calcium fluoride is defined as:

CaF₂(s) → Ca²⁺(aq) + 2F⁻(aq)

Ksp = 3.9x10⁻¹¹ = [Ca²⁺] [F⁻]²

Where concentrations of ions are in equilibrium.

Molarity of Ca²⁺ and F⁻ is:

[Ca²⁺] = 0.1250L × (0.00615mol / L) = 7.688x10⁻⁴mol / 0.4550L = 1.690x10⁻³M

[F⁻] = 0.3300L × (0.00615mol / L) = 2.030x10⁻³mol / 0.4550L = 4.460x10⁻³M

Thus, Qsp is:

Qsp = 1.690x10⁻³M × (4.460x10⁻³M)²

Qsp = 3.36x10⁻⁸

i think that the answer is c