What concentration of the barium ion,Ba^2+ ,must be exceeded to precipitate BaF2 from a solution that is 1.00*10^-2M in the fluoride ion,F-? Ksp for barium fluoride is 2.45*10^-5.
To determine the concentration of the barium ion (Ba²⁺) needed to precipitate BaF₂ from a solution that is 1.00 × 10⁻² M in fluoride ion (F⁻), we can use the solubility product constant (Ksp) for barium fluoride.
The balanced chemical equation for the precipitation reaction is:
BaF₂ (s) ↔ Ba²⁺ (aq) + 2F⁻ (aq)
According to the Ksp expression:
Ksp = [Ba²⁺] × [F⁻]²
We know the value of Ksp = 2.45 × 10⁻⁵, and the concentration of F⁻ is 1.00 × 10⁻² M.
Since the stoichiometry of the reaction shows that the concentration of Ba²⁺ is equal to the concentration of F⁻, we assume the concentration of Ba²⁺ is also x M.
Therefore, the Ksp expression can be rearranged to solve for x:
Ksp = x × (1.00 × 10⁻²)²
Substituting the known values:
2.45 × 10⁻⁵ = x × (1.00 × 10⁻²)²
Simplifying:
2.45 × 10⁻⁵ = x × 1.00 × 10⁻⁴
To solve for x (concentration of Ba²⁺), divide both sides by 1.00 × 10⁻⁴:
x = (2.45 × 10⁻⁵) / (1.00 × 10⁻⁴) = 0.245
Therefore, the concentration of the barium ion (Ba²⁺) that must be exceeded to precipitate BaF₂ from a solution that is 1.00 × 10⁻² M in fluoride ion (F⁻) is 0.245 M.
To determine the concentration of the barium ion, Ba^2+, required to precipitate BaF2 from a solution that is 1.00*10^-2M in fluoride ion, F-, we need to use the solubility product constant, Ksp, for barium fluoride.
The balanced equation for the dissociation of barium fluoride is:
BaF2(s) ⇌ Ba^2+(aq) + 2F^-(aq)
The Ksp expression for barium fluoride is:
Ksp = [Ba^2+][F^-]^2
Given:
Ksp = 2.45*10^-5
[F^-] = 1.00*10^-2 M
Let's substitute the given values into the Ksp expression and solve for [Ba^2+].
2.45*10^-5 = [Ba^2+][1.00*10^-2]^2
Simplifying the equation:
2.45*10^-5 = [Ba^2+][1.00*10^-4]
Now, solve for [Ba^2+]:
[Ba^2+] = (2.45*10^-5)/(1.00*10^-4)
Calculating the value:
[Ba^2+] = 0.00245 or 2.45*10^-3 M
Therefore, the concentration of the barium ion, Ba^2+, must be at least 2.45*10^-3 M or 0.00245 M to exceed the solubility product constant and precipitate BaF2 from the solution.