school
Acemy

Introduction to Solubility Equilibria

AP Chemistry· difficulty 4/5

For PbCl2_2, Ksp=1.6×10−5K_{sp} = 1.6 \times 10^{-5}. The molar solubility of PbCl2_2 in pure water is approximately

  • A

    1.6×10−51.6 \times 10^{-5} M

  • B

    Ksp/43≈1.6×10−2\sqrt[3]{K_{sp}/4} \approx 1.6 \times 10^{-2} M

    check_circle
  • C

    Ksp/2K_{sp}/2

  • D

    Ksp≈4.0×10−3\sqrt{K_{sp}} \approx 4.0 \times 10^{-3} M

Explanation

PbCl2_2 ⇌ Pb2+^{2+} + 2 Cl−^-. Let s = molar solubility. Ksp=s(2s)2=4s3K_{sp} = s(2s)^2 = 4s^3. s=1.6×10−5/43=4×10−63=1.59×10−2s = \sqrt[3]{1.6\times10^{-5}/4} = \sqrt[3]{4\times10^{-6}} = 1.59\times10^{-2} M.

Want 10 more like this — adaptive to your weak spots?

Practice 10 like thisarrow_forwardPredict my AP score

Related questions

auto_awesomeAll Toolsquestion_answerHomework Q&Amenu_bookAP & SAT PreptimelineLearn Mode