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16 February, 14:47

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

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  1. 16 February, 17:29
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    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.
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