What is the Ksp for PbI2 at 25°C?

[Soaring Crane]

Calculate Ksp for PbI2 at 25 C based on the following data:
SPECIES------------deltaG_f kJ/mol
Pb (2+) (aq) -------(-)24.4
I- (aq)-------------(-)51.6
PbI2 (s)------------(-) 173.6



a. 4 x 10-31
b. 8 x 10-18
c. 9 x 10-9
d. 5 x 10-5

PbI2 <-> Pb (2+) + 2I-

Use G_f = sigma prod. - sigma reactants

[2(-51.6) - 24.4]kJ - [-173.6]kJ

G_f = 46 kJ = 4.6E3 J

ln K = -G/RT
K = e^-(4.6E3/(8.314*298.15 K) = 8.72E-9

Thanks.

 

[siddharth]

Your approach to solve this problem looks right.

46 kJ = 4.6E3 J

That looks like a mathematical error.

 

[Blueshift5]

Based on the data provided, the equilibrium constant (K) for the reaction between PbI2 and its dissociation products, Pb (2+) and I-, is calculated to be 8.72E-9 at 25 degrees Celsius. This indicates that the reaction strongly favors the formation of PbI2 at this temperature. This value can also be used to calculate the solubility product constant (Ksp) for PbI2 by using the formula Ksp = [Pb (2+)]*[I-]^2, where [Pb (2+)] and [I-] are the concentrations of the respective ions in solution. Using the data provided, the Ksp for PbI2 at 25 degrees Celsius is calculated to be approximately 9 x 10^-9. This value is extremely small, indicating that PbI2 is not very soluble in water at this temperature and that the solid compound is more stable than its dissociation products in solution.