Lead(ii) sulfide (pbs) has a ksp of 3.0 × 10–28. what is the concentration of lead(ii) ions in a saturated solution of pbs? 1.7 × 10–14m 1.5 × 10–14m 3.0 × 10–14m
PbS partially dissociates into ions as Pb²⁺ and S²⁻ in water. The balanced reaction is PbS(s) ⇄ Pb²⁺(aq) + S²⁻(aq) In the saturated solution, the reaction is at equilibrium. Let's assume that solubility of PbS in saturated solution is "X". Then according to the stoichiometry, solubility of PbS = equilibrium concentration of Pb²⁺(aq) = equilibrium concentration of S²⁻(aq) = X
Ksp = [Pb²⁺(aq) ][S²⁻(aq)] Ksp = X * X 3.0 × 10⁻²⁸ = X² X = 1.7 x 10⁻¹⁴ M.
Hence the concentration of lead(ii) ions in a saturated solution of PbS is 1.7 x 10⁻¹⁴ M.
