Calculate the equilibrium concentration of Pb ⁺⁺ ions in a solution of PbS containing 1 10⁻¹¹ ~mol~ dm ⁻³ of sulphide ions. (Given K _ sp for PbS =8.0 10⁻²⁸ )

Step-by-step solution

The dissolution equilibrium for lead(II) sulfide is described by: PbS(s) Pb²⁺(aq) + S²⁻(aq) The solubility product constant expression is: K_ sp = [ Pb²⁺ ][ S²⁻ ] Given K_ sp = 8.0 10⁻²⁸ and [ S²⁻ ] = 1 10⁻¹¹ mol dm⁻³ , the lead ion concentration follows from the equilibrium expression: [ Pb²⁺ ] = K_ sp [ S²⁻ ] = 8.0 10⁻²⁸ 1 10⁻¹¹ = 8.0 10⁻¹⁷ mol dm⁻³ This value corresponds to option C .