The solubility of lead(II)fluoride, PbF₂, in pure water is 2.1 × 10^-³ moles per liter. Calculate the value of K_sp for lead(II)fluoride from this data.

Question

Quizplus · Accepted Answer

The dissolution of PbF2 in water can be represented as PbF2 ⇌ Pb²⁺ + 2F⁻. The solubility product Ksp is given by [Pb²⁺][F⁻]². If the solubility is 2.1 × 10⁻³ M, then [Pb²⁺] = 2.1 × 10⁻³ M and [F⁻] = 2 × 2.1 × 10⁻³ M = 4.2 × 10⁻³ M. Therefore, Ksp = (2.1 × 10⁻³)(4.2 × 10⁻³)² = 3.7 × 10⁻⁸.

The Solubility of Lead(II)fluoride, PbF2, in Pure Water Is 2

The Solubility of Lead(II)fluoride, PbF₂, in Pure Water Is 2