Answer to Question #197408 in General Chemistry for Felix

Question #197408

Into 2 liter of water 0,6 g of lead iodide is dissolved and formed saturated solution. Calculate KSP of this salt, if molecular mass of PbI2 is 461 g/mole.


1
Expert's answer
2021-05-24T07:20:09-0400

PbI2 = Pb2+ + 2I-

x 2x

Given,

Solubility of PbI2 = (.6/2)=0.3g/L

Molar Mass of PbI2 = 461g/mol

[PbI2]= (.3/461)mol/L = 6.5 ×10-4

Solubility product Ksp = x(2x)2=4x3

Ksp=4(6.5×10-4)3 =1.0985 ×10-9

Hence the solubility product of PbI2 is 1.0985×10-9 mol2/lit2

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