What would be the volume of water required to dissolve 0.2 ~g of PbCl ₂ of molar mass 278 ~g / mol to prepare a saturated solution of the salt? ( K _ SP of PbCl ₂=3.2 10⁻⁸)

Step-by-step solution

The solubility equilibrium for PbCl ₂ is given by PbCl ₂(s) Pb ²⁺(aq) + 2 Cl ⁻(aq) . Let s be the solubility of PbCl ₂ in mol/L. Then [ Pb ²⁺] = s and [ Cl ⁻] = 2s . The solubility product constant is K_ sp = [ Pb ²⁺][ Cl ⁻]² = s(2s)² = 4s³ . Given K_ sp = 3.2 10⁻⁸ , we have 4s³ = 3.2 10⁻⁸ s³ = 0.8 10⁻⁸ = 8 10⁻⁹ . Taking the cube root, s = 2 10⁻³ mol/L . The number of moles of PbCl ₂ in 0.2 g is n = 0.2 g 278 g/mol 7.194 10⁻⁴ mol . The volume V required to dissolve this amount is V = n s = 7.194 10⁻⁴ mol 2 10⁻³ mol