Answer is: concentration of the barium ion is 0.245 M.
Chemical reaction: BaF₂ → Ba²⁺ + 2F⁻.
[F⁻] = 1.00·10⁻² M.
Ksp = 2.45·10⁻⁵.
Ksp = [Ba²⁺] · [F⁻]².
[Ba²⁺] = Ksp ÷ [F⁻]².
[Ba²⁺] = 2.45·10⁻⁵ ÷ (1.00·10⁻² M)².
[Ba²⁺] = 0.245 M.
Answer: The concentration of barium ions that must exceed to precipitate the salt is 0.245 M
Explanation:
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is represented as [tex]K_{sp}[/tex]
Barium fluoride is an ionic compound formed by the combination of 1 barium ion and 2 fluoride ions.
The equilibrium reaction for the ionization of barium fluoride follows the equation:
[tex]BaF_2(s)\rightleftharpoons Ba^{2+}(aq.)+2F^-(aq.)[/tex]
The solubility product for the above reaction is:
[tex]K_{sp}=[Ba^{2+}]\times [F^-]^2[/tex]
We are given:
[tex][F^-]=1.00\times 10^{-2}M\\\\K_{sp}=2.45\times 10^{-5}[/tex]
Putting values in above equation, we get:
[tex]2.45\times 10^{-5}=[Ba^{2+}]\times (1.00\times 10^{-2})^2[/tex]
[tex][Ba^{2+}]=\frac{2.45\times 10^{-5}}{(1.00\times 10^{-2})^2}=0.245M[/tex]
Hence, the concentration of barium ions that must exceed to precipitate the salt is 0.245 M
