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Introduction to Solubility Equilibria

AP Chemistry· difficulty 4/5

equilibrium reactants products time [X]

KspK_{sp} for PbCl2PbCl_2 is 1.6×10−51.6\times 10^{-5}. What is the molar solubility of PbCl2PbCl_2 in 0.10 M NaCl?

  • A

    1.6×10−51.6\times 10^{-5} M

  • B

    4.0×10−34.0\times 10^{-3} M

  • C

    1.6×10−41.6\times 10^{-4} M

  • D

    1.6×10−31.6\times 10^{-3} M

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Explanation

KspK_{sp} = [Pb2+][Cl−]2[Pb^{2+}][Cl^-]^2. With [Cl−]≈[Cl^-] \approx 0.10 (common ion dominant): [Pb2+][Pb^{2+}] = 1.6e-5/(0.10)2^2 = 1.6e-3 M = molar solubility.

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