This is not the correct answer. The ion product expression for CaF_2 arises from

Question

This is not the correct answer. The ion product expression for CaF_2 arises from the equilibrium CaF_2(s) Ca^2+(aq) + 2F^- (aq). K_sp = [Ca^2+][F^-]^2, where the values in brackets are the molar concentrations of the ions. The equilibrium tells us that [F^-] is twice [Ca^2+]. Did you put these values into the K_sp expression or did you just multiply the two numbers? Look at the K_sp expression and the power for F. The molar solubility of calcium fluoride, CaF_2, is 2.1 times 10^-4 mol L^-1. From this we can calculate the K-sp of CaF_2 to be

Explanation / Answer

Dissociation of CaF2 in water:

CaF2 <-> Ca+2 + 2F-

Ksp = [Ca+2][F-]2

CaF2 will dissociate and have a 1 on 1 molar relation on calcium ions and a 2 on 1 on fluorine ions.

Ksp = [2.1 x 10-4][4.2 x 10-4]2

Ksp = 3.7 x 10-11

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