The molar solubility of CaF2 (Ksp = 5.3 × 10-11) in 0.1 M solution of NaF will be

Question

The molar solubility of CaF2 (Ksp = 5.3 × 10-11) in 0.1 M solution of NaF will be (A)  5.3 × 10-11 mol L -1 (B) 5.3 × 10-8 mol L -1 (C)  5.3 × 10-9 mol L -1 (D) 5.3 × 10-10 mol L -1

Tardigrade Admin · Accepted Answer

underset(a-s')CaF2(s) leftharpoons undersets'Ca+2(aq) + underset2s'2F- (aq) underset oversetC0NaF(aq) arrow underset overset0CNa+ (aq) + underset overset0CF- (aq) In solution- [F-] = (2s' + C) [F-] ≈ C (due to common ion effect) Ksp(caF2) = [Ca+2] . [F-]2 Ksp(CaF2) = s' . C2 s' = (5.3 × 10-11/(10-1)2) s' = 5.3 × 10-9 mol L-1

Q. The molar solubility of $C a F_{2}$ (K $_{s p} = 5.3 \times 1 0^{- 11}$ ) in $0.1 M$ solution of $N a F$ will be

$5.3 \times 1 0^{- 11} m o l L^{- 1}$

$5.3 \times 1 0^{- 8} m o l L^{- 1}$

$5.3 \times 1 0^{- 9} m o l L^{- 1}$

$5.3 \times 1 0^{- 10} m o l L^{- 1}$

Q. The molar solubility of CaF2​ (Ksp​=5.3×10−11) in 0.1M solution of NaF will be

5.3×10−11molL−1

5.3×10−8molL−1

5.3×10−9molL−1

5.3×10−10molL−1

(a−s′)CaF2​(s)​⇌s′Ca+2(aq)​+2s′2F−(aq)​ 0CNaF(aq)​→C0Na+(aq)​+C0F−(aq)​ In solution- [F−]=(2s′+C) [F−]≈C (due to common ion effect) Ksp(caF2​)​=[Ca+2].[F−]2 Ksp(CaF2​)​=s′.C2 s′=(10−1)25.3×10−11​ s′=5.3×10−9molL−1

Q. The molar solubility of $C a F_{2}$ (K $_{s p} = 5.3 \times 1 0^{- 11}$ ) in $0.1 M$ solution of $N a F$ will be

$5.3 \times 1 0^{- 11} m o l L^{- 1}$

$5.3 \times 1 0^{- 8} m o l L^{- 1}$

$5.3 \times 1 0^{- 9} m o l L^{- 1}$

$5.3 \times 1 0^{- 10} m o l L^{- 1}$