The molar solubility of CaF2 is 2.6 × 10-5 M in pure water at a certain temperature. Calculate the Ksp for CaF2 at this temperature.

Answer 1

The molar solubility of CaF2 in pure water is given as 2.6 × 10-5 M. This means that at equilibrium, the concentration of Ca2+ ions and F- ions from the dissociation of CaF2 is 2.6 × 10-5 M.

The balanced equation for the dissociation of CaF2 is:
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)

From the balanced equation, we can see that the molar solubility represents the concentration of Ca2+ ions, so [Ca2+] = 2.6 × 10-5 M.

Using the solubility product expression, Ksp = [Ca2+][F-]^2, we can plug in the values to calculate Ksp:

Ksp = (2.6 × 10-5)(2.6 × 10-5)^2
= 2.6 × 10-5 × 6.76 × 10-10
= 1.756 × 10-14

Therefore, the Ksp for CaF2 at this temperature is 1.756 × 10-14.

Answer 2

Bot, you forgot to multiply by 2 to obtain F^-.
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
solid..........2.6E-5.......2*2.6E-5
Ksp = (Ca^2+)(F^-)^2 = (2.6E-5)(5.2E-5)^2
Ksp = 7.03E-14

Answer 3

You are correct, I apologize for the error. The correct calculation is as follows:

The molar solubility of CaF2 in pure water is 2.6 × 10-5 M, which represents the concentration of Ca2+ ions. Since the balanced equation shows that 2 moles of F- ions are produced for every mole of CaF2 that dissolves, the concentration of F- ions is 2 * (2.6 × 10-5) = 5.2 × 10-5 M.

Using the solubility product expression:
Ksp = (Ca2+)(F-)2 = (2.6 × 10-5)(5.2 × 10-5)2
= 7.03 × 10-14

Therefore, the correct Ksp for CaF2 at this temperature is 7.03 × 10-14.