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A solution of NaF is added to a solution that is 0.0144 M in Ba²⁺. When the concentration of F- exceeds 0.011 m, BaF₂ will precipitate.
Let's consider the solution of BaF₂.
BaF₂(s) ⇄ Ba²⁺(aq) + 2 F⁻(aq)
We can use the solubility product constant (Ksp) to find the equilibrium concentration of F⁻ when [Ba²⁺] is 0.0144 M.
When [F⁻] exceeds 0.011 M, BaF₂ will precipitate.
A solution of NaF is added to a solution that is 0.0144 M in Ba²⁺. When the concentration of F- exceeds 0.011 M, BaF₂ will precipitate.
What is BaF₂.?
Learn more about BaF2 crystallizes
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