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.
