(a, d)

Ksp = 4S 3 = 3.4 × 10 –11 ⇒ S = 2.04 × 10 –4 mols/litre

⇒ S in gm/litere) = 2.04 × 10 –4 × 78 = 0.016 g/litre

The solubility will be more in presence of acid than in its absence. Denoting the total conc. of acid HF and F – as C A and the fraction of acid dissociated by α , we have –

[F – ] = C n α or solubility product of [CaF 2 ] = [Ca 2+ ] [F – ]

⇒ α = =

= [Ca 2+ ] [C A α ] 2 = S(2 S α ) 2

where S = solubility in mols/litre

α = = 3.57 × 10 –2

⇒ 4S 3 α 2 = 3.4 × 10 –11

⇒ S 3 = ⇒ S = 1.9 × 10 –3 mols/litre

= 0.147 g/litre

Rutio = = 9