SUMMARY
The molar solubility of barium fluoride (BaF2) in water at 25 degrees Celsius is calculated using its solubility product constant (Ksp), which is 1.0 x 10^-16. The dissolution of BaF2 can be represented by the equation BaF2 (s) ⇌ Ba2+ (aq) + 2F- (aq). In pure water, the molar solubility is determined to be approximately 1.0 x 10^-8 M. When BaF2 is dissolved in a 0.15 M sodium fluoride (NaF) solution, the presence of F- ions decreases the molar solubility of BaF2 due to the common ion effect, resulting in a significantly lower solubility compared to that in pure water.
PREREQUISITES
- Understanding of solubility product constant (Ksp)
- Knowledge of chemical equilibrium principles
- Familiarity with the common ion effect
- Basic skills in stoichiometry and concentration calculations
NEXT STEPS
- Study the common ion effect in detail
- Learn how to calculate molar solubility using Ksp values
- Explore the concept of chemical equilibrium in aqueous solutions
- Investigate the solubility of other salts in the presence of common ions
USEFUL FOR
Chemistry students, educators, and professionals involved in analytical chemistry or environmental science who are interested in solubility principles and their applications in real-world scenarios.