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Common-Ion Effect

AP Chemistry· difficulty 4/5

AgCl has Ksp=1.8×10−10K_{sp} = 1.8 \times 10^{-10}. Its molar solubility in a 0.10 M NaCl solution is approximately

  • A

    0.100.10 M

  • B

    1.3×10−51.3 \times 10^{-5} M

  • C

    1.8×10−101.8 \times 10^{-10} M

  • D

    1.8×10−91.8 \times 10^{-9} M

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Explanation

[Cl⁻] ≈ 0.10 M from NaCl. [Ag⁺] = Ksp/[Cl⁻] = 1.8e-10/0.10 = 1.8e-9 M (much lower than in pure water).

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