Question

The solubility of CaF2 (Ksp = 3.4 x 10-11 M3) in 0.1M NaF would be
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Answer 1

Given : CaF₂ of solubility product constant, Ksp = 3.4 × 10¯¹¹ M³ in 0.1 M of NaF solution.

To find : The solubility of CaF₂ in 0.1 M of NaF solution.

solution : dissociation reaction of CaF₂ is given as ... CaF₂ ⇔Ca²⁺ + 2F¯

x 2x

similarly, dissociation reaction of NaF is given as.... NaF ⇔Na⁺ + F¯

0.1M 0.1 M

so, total concentration of F¯ ions = 2x + 0.1

but 2x << 0.1

so, total concentration of F¯ ions = 2x + 0.1 ≈ 0.1

now Ksp = [Ca²⁺] [F¯]²

⇒3.4 × 10¯¹¹ = x × (0.1)²

⇒3.4 × 10¯¹¹ = x × 10¯²

⇒x = 3.4 × 10^-9 M

Therefore solubility of CaF₂ in 0.1 M of NaF is 3.4 × 10^-9 M