Question

The molar solubility of caf2(ksp = 5.3 1011) in 0.1 m solution of naf will be

Answer 1

Answer:

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Answer 2

Thus the molar solubility of CaF2 is 5.3 × 10^-9 mol/L

Explanation:

Given data:

• K sp = 5.3 × 10^-11,
• C = 0.1 M

Solution:

CaF2 dissociates into Ca²+ and 2 F- ions.

Ca F2  ⇔  Ca²+  +  2 F-

For Na F ⇔Na +  + F-

At equilibrium:

0. C. C

In solution, [F-] = (2s + C)

Due to common ion effect, [F-] ≈ C

Now, Ksp = [Ca²+] [F-]²

Ksp = s × C²

s = Ksp/C²

s = 5.3 × 10^-11 / (0.1)^2 = 5.3 × 10^-9 mol/L

Thus the molar solubility of CaF2 is 5.3 × 10^-9 mol/L