We are given the equilibrium constant (Ksp) for the dissolution of BaF2:
BaF2 (s) <=> Ba^2+ (aq) + 2F^- (aq)
Ksp = [Ba^2+] * [F^-]^2
We are asked to find the solubility of BaF2 in moles per liter, which means we need to find the concentration of Ba^2+ ions in the solution.
Let x be the concentration of Ba^2+ ions in moles per liter. Since there are two moles of F^- ions for every mole of Ba^2+ ions, the concentration of F^- ions is 2x.
Now, we can substitute these concentrations into the Ksp equation:
Ksp = (x) * (2x)^2 = 4x^3
We are given Ksp = 1.8 * 10^(-7), so we can solve for x:
(1.8 * 10^(-7)) = 4x^3
x^3 = (1.8 * 10^(-7)) / 4
x^3 = 4.5 * 10^(-8)
x = (4.5 * 10^(-8))^(1/3)
x ≈ 3.56 * 10^(-3)
Therefore, the solubility of BaF2 in moles per liter is approximately 3.56 * 10^(-3) M.
The equilibrium for the saturated solution of BaF^2 is 1.8*10^-7. Calculate the solubility of BaF^2 in moles per litre?
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