Asked by Anonymous
Consider the two equilibria:
BaF2=Ba2+ + 2F- Ksp=1.7x10^(-6)
F- + H20=HF + OH- Kb=2.9x10^(-11)
Determine the solubility of BaF2 at pH=5?
What I'm mostly confused about is how to determine if F- is dominant or if you have to take HF into account while doing your calculations.
BaF2=Ba2+ + 2F- Ksp=1.7x10^(-6)
F- + H20=HF + OH- Kb=2.9x10^(-11)
Determine the solubility of BaF2 at pH=5?
What I'm mostly confused about is how to determine if F- is dominant or if you have to take HF into account while doing your calculations.
Answers
Answered by
DrBob222
You must take HF into consideration during the calculation.
The whole point of these problems is to show you that BaF2 has a certain solubility in neutral solution. As the solution becomes more and more acid the formation of HF becomes more (because HF is a weak acid); that shifts the solubility equilibrium to the right and makes BaF2 more soluble in acid solutions.
The whole point of these problems is to show you that BaF2 has a certain solubility in neutral solution. As the solution becomes more and more acid the formation of HF becomes more (because HF is a weak acid); that shifts the solubility equilibrium to the right and makes BaF2 more soluble in acid solutions.
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