Question

What concentration of the barium ion, ba2+, must be exceeded to precipitate baf2 from a solution that is 1.00×10−2 m in the fluoride ion, f−? ksp for barium fluoride is 2.45×10−5 . express your answer with the appropriate units?

Answer 1

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 $K_{sp}$

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:

$BaF_2(s)\rightleftharpoons Ba^{2+}(aq.)+2F^-(aq.)$

The solubility product for the above reaction is:

$K_{sp}=[Ba^{2+}]\times [F^-]^2$

We are given:

$[F^-]=1.00\times 10^{-2}M\\\\K_{sp}=2.45\times 10^{-5}$

Putting values in above equation, we get:

$2.45\times 10^{-5}=[Ba^{2+}]\times (1.00\times 10^{-2})^2$

$[Ba^{2+}]=\frac{2.45\times 10^{-5}}{(1.00\times 10^{-2})^2}=0.245M$

Hence, the concentration of barium ions that must exceed to precipitate the salt is 0.245 M

Answer 2

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.