Answer to Question #197408 in General Chemistry for Felix

Question

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

Answer 1

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 K_sp = x(2x)²=4x³

K_sp=4(6.5×10^-4)³ =1.0985 ×10^-9

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