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.

Sorry i dont understand the question can you rephase it?