# Problem: Calculate the molar solubility of CaF2 in a solution containing 0.550 M of Ca(NO3)2. The Ksp value for CaF2 is 1.46 x 10–10.

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###### FREE Expert Solution

CaF2(s) will break up into its ions:              CaF2(s)  Ca2+(aq) + 2 F-(aq)

Ca(NO­3)2(s) also break up to its ions:       Ca(NO­3)2(s)  Ca2+(aq) + 2 NO3-(aq)

There is now an initial concentration of Ca2+ from Ca(NO­3)2(s):  this is called the Common Ion Effect

[Ca(NO3)2] = [Ca2+] = 0.55 M

Set up an ICE Table: CaF2(s)  Ca2+(aq) + 2 F-(aq) $\overline{){{\mathbf{K}}}_{{\mathbf{sp}}}{\mathbf{=}}\frac{\mathbf{products}}{\overline{)\mathbf{reactants}}}{\mathbf{=}}\mathbf{\left[}{\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}{\mathbf{\left[}{\mathbf{F}}^{\mathbf{-}}\mathbf{\right]}}^{{\mathbf{2}}}}$

Substitute equilibrium concentrations and solve for x: ###### Problem Details

Calculate the molar solubility of CaF2 in a solution containing 0.550 M of Ca(NO3)2. The Ksp value for CaF2 is 1.46 x 10–10.