What is the Ksp for PbI2 at 25°C?

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SUMMARY

The solubility product constant (Ksp) for lead(II) iodide (PbI2) at 25°C is calculated to be approximately 8.72 x 10-9. This value is derived using the Gibbs free energy of formation (G_f) for the species involved: Pb2+ at -24.4 kJ/mol, I- at -51.6 kJ/mol, and PbI2 at -173.6 kJ/mol. The calculation follows the equation G_f = sigma products - sigma reactants, leading to the final Ksp value through the relationship ln K = -G/RT.

PREREQUISITES
  • Understanding of Gibbs free energy and its applications in chemical equilibrium.
  • Familiarity with the concept of solubility product constants (Ksp).
  • Knowledge of thermodynamic equations, specifically ln K = -G/RT.
  • Basic proficiency in using scientific notation for large and small numbers.
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  • Explore the calculation of Ksp for other sparingly soluble salts, such as BaSO4.
  • Learn about the impact of temperature on solubility and Ksp values.
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Soaring Crane
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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.
 
Physics news on Phys.org
Your approach to solve this problem looks right.

46 kJ = 4.6E3 J
That looks like a mathematical error.
 

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