The molar solubility of CaF 2(K sp =5.3 × 10-11) in 0.1 M solution of NaF will be

Question

The molar solubility of CaF 2(K sp =5.3 × 10-11) in 0.1 M solution of NaF will be (A) 5.3 × 1011 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

Let the solubility of CaF 2 in 0.1 M NaF is ' S' mol L -1 CaF 2(s) leftharpoons undersetSCa 2+(a q)+ underset2 S2 F -(a q) NaF ( aq ) leftharpoons underset0.1 MNa ++ underset0.1 MF -(a q) [ F -] =2 S+0.1 K sp of CaF 2 =[ Ca 2+][ F -]2 =[S][2 S+0.1]2 =5.3 × 10-11=[S][2 S+0.1]2 ⇒ 5.3 × 10-11=[S][0.1]2 [∵ 2 S ll 0.1] [S]=(5.3 × 10-11/(0.1)2)=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 0.1 {\,}M solution of NaF will be

5.3×1011molL−1

5.3×10−8molL−1

5.3×10−9molL−1

5.3×10−10molL−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

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

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

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

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