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Equilibrium

Question
CBSEENCH11006610

Calculate the solubility of lead iodide in water at 298 K. (Ksp of PbI2 = 7·1×10–9).

Solution
Let the solubility of lead iodide in water be S mol L–1.
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i.e. 1 mole of PbI2 in the solution gives 1 mole of Pb2+ ions and 2 moles of I- ions. 
 Hence in the solution open square brackets Pb to the power of 2 plus end exponent close square brackets space equals space straight S space mol space straight L to the power of negative 1 end exponent
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Applying the law of solubility product, 
  space space space space space space straight K subscript sp space equals space open square brackets Pb to the power of 2 plus end exponent close square brackets space open square brackets straight I to the power of minus close square brackets squared space equals space straight S left parenthesis 2 straight S right parenthesis squared space equals space 4 straight S cubed
or space space space space straight S space equals space open parentheses straight K subscript sp over 4 close parentheses to the power of 1 divided by 3 end exponent space equals space open parentheses fraction numerator 7.1 space cross times space 10 to the power of negative 9 end exponent over denominator 4 end fraction close parentheses to the power of 1 divided by 3 end exponent
space space space space space space space space space space space space equals space 1.21 space cross times space 10 to the power of negative 3 end exponent
Hence solubility of PBI2 in water is 1·21 × 10–3 mol L–1.