To calculate the solubility of BaF2 in moles per litre, we can use the solubility product constant (Ksp). The balanced equation for the dissolution of BaF2 is:
BaF2(s) <=> Ba^2+(aq) + 2F^-(aq)
Let x be the solubility of BaF2 in moles per litre. According to the balanced equation, the concentration of Ba^2+ ions will also be x, and the concentration of F^- ions will be 2x. The Ksp expression for this equilibrium is:
Ksp = [Ba^2+][F^-]^2
Given Ksp = 1.8*10^(-7), we can substitute the concentrations in terms of x:
1.8*10^(-7) = (x)(2x)^2
1.8*10^(-7) = 4x^3
Now, solve for x:
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)
The solubility of BaF2 is approximately 3.56 * 10^(-3) moles per litre.
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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