The value of solubility product constant (Ksp) = 4.4 × 10−9
For lead(II) iodide, the solubility product is written following the dissociation :
PbI2 (s) = Pb2+ (aq) + 2 I- (aq)
So, Ksp = [Pb2+][I-]2
(Ksp is the product of the concentration of the ions raised to their stoichiometric coefficients)
[…] denotes concentration in molarity.
Molar solubility = number of moles of solute ( PbI2) which will dissolve in 1000 mL of solution.
Let molar solubility of lead(II) iodide = s M.
So, the concentration of Pb2+ = s M , and I - = 2s M
(since 1 mole of the salt gives 1 Pb2+ ions and 2 I - ions).
Putting the value of concentration of Pb2+ and I - in the equation for Ksp and solving for s :
Ksp = 4.4 × 10−9= (s)(2s)2
Or, 4.4 × 10−9= 4s3
Or, s = 1.032 x 10-3 M (this is the molar solubility)
So, in 1000 mL of solution, 1.032 x 10-3 moles of lead(II) iodide will dissolve.
Converting moles to mass :
Mass ( in g) = Number of moles x molar mass
Or, mass ( in g) = 1.032 x 10-3 moles x 461 g/mol = 0.476 g = 476 mg ( 1g = 1000 mg)
Thus, in the 1000 mL solution, 476 mg of lead (II) iodide will dissolve .
The remaining mg of lead iodide ( 800 mg – 476 mg = 324 mg) will precipitate.
So, the last option is correct.
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