Question

3. The molar solubility of CaF2(Ksp = 5.3 * 10-11) in
0.1 M solution of NaF will be
[NEET-2019 (Odisha)]
(1) 5.3 * 10–10 mol L-1 (2) 5.3 * 10-11 mol L-1
(3) 5.3 * 10-8 mol L-1 (4) 5.3 x 10-9 mol L-1​

Answer 1

answer : option (4) 5.3 × 10^-9 mol/L

explanation : CaF2 dissociates into Ca²+ and 2F-

CaF2 (aq) ⇔Ca²+(aq) + 2F- (aq)

(a - s) s. 2s.

for NaF (aq)⇔Na+(aq) + F-(aq)

A1 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

given, Ksp = 5.3 × 10^-11, C = 0.1M

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

hence, molar solubility of CaF2 is 5.3 × 10^-9 mol/L

Answer 2

Answer:

Here is your Answer Dear...

Hope U understand..