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.