Solveeit Logo
Login

Question

Chemistry Question on Ionic Equilibrium In Solution

The molar solubility of CaF2_2 (Ksp_{sp} = 5.3 × 10−11^{-11}) in 0.1 M solution of NaF will be

A

5.3 × 10−11^{-11} mol L−1^{-1}

B

5.3 × 10−8^{-8} mol L−1^{-1}

C

5.3 × 10−9^{-9} mol L−1^{-1}

D

5.3 × 10−10^{-10} mol L−1^{-1}

Answer

5.3 × 10−9^{-9} mol L−1^{-1}

Explanation

Solution

CaF2(s)(a−s′)<=>Ca+2(aq)s′+2F−(aq)2s′{\underset{(a-s')}{CaF_2(s)} <=> \underset{s'}{Ca^{+2}(aq)} + \underset{2s'}{2F^- (aq)}}
NaF(aq)0C−>Na+(aq)C0+F−(aq)C0{\underset{\overset{C}{0}}{NaF(aq)} -> \underset{\overset{0}{C}}{Na^+ (aq)} + \underset{\overset{0}{C}}{F^- (aq)}}
In solution- [F−]=(2s′+C)[F^-] = (2s' + C)
[F−]≈C[F^-] \approx C (due to common ion effect)
Ksp(caF2)=[Ca+2].[F−]2K_{sp(caF_2)} = [Ca^{+2}] . [F^-]^2
Ksp(CaF2)=s′.C2K_{sp(CaF_2)} = s' . C^2
s′=5.3×10−11(10−1)2s' = \frac{5.3 \times 10^{-11}}{(10^{-1})^2}
s′=5.3×10−9 mol L−1s' = 5.3 \times 10^{-9} \,mol\,L^{-1}