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Question

PbI2 has a solubility product of 8.0×10-9. Lead Iodide has a solubility in 0.1molar Lead Nitrate is x×10-6mol/L. What is the value of X? (Round off to the nearest integer)


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Solution

Step 1:

  • Given,
  • KspPbI2=8×10-9
  • PbNO32(s)Pb2+(s)+2NO3-(aq)0.1M0.1M0.2M
  • PbI2(s)Pb2+(aq)+2I-(aq)(s)(2s)
  • Ksp=Pb2+I-2
  • Pb2+=s+0.10.1

Step 2:

  • Ksp=8×10-9=Pb2+I-2
  • 8×10-9=0.12s28×10-8=4s2s2=8×10-910-1×4s2=2×10-8s=10-2×2×10-410-2s=2×10-4=141×10-6M=141

Therefore, is the value of X is 141


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