What is the maximum concentration of Ca2+ ions in a solution containing 0.010 M of F- ions?
Is it 3.9 x 10^-7 M?
CaF2 ==> Ca^2+ + 2F^-
Ksp = (Ca^2+)(F^-)^2
You know F and K, solve for Ca.
Well, well, it seems like we have some chemistry going on here! If we have a solution that contains 0.010 M of F- ions, we can determine the maximum concentration of Ca2+ ions by looking at the solubility product constant (Ksp) of calcium fluoride (CaF2).
I'd tell you a chemistry joke, but all the good ones argon. Anyway, the Ksp expression for calcium fluoride is written as:
Ksp = [Ca2+][F-]^2
Since we're given the concentration of F- ions (0.010 M), we can substitute that value into the equation and solve for [Ca2+]. Remember, the Ksp value for calcium fluoride is 3.9 x 10^-11.
Once we solve for [Ca2+], we find the maximum concentration of Ca2+ ions in the solution. So, let's crunch those numbers (or elements) and calculate it!
To determine the maximum concentration of Ca2+ ions in a solution containing 0.010 M of F- ions, we need to consider the solubility product constant (Ksp) of calcium fluoride (CaF2).
The balanced equation for the dissociation of CaF2 is:
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
The Ksp expression for this reaction is:
Ksp = [Ca2+] * [F-]^2
Given that the concentration of F- ions is 0.010 M, let's assume that the solubility of CaF2 is "x" M. Thus, the concentration of Ca2+ ions will also be "x" M.
Therefore, the Ksp expression becomes:
Ksp = (x) * (0.010)^2
Since the Ksp value for CaF2 is 3.9 x 10^-11, we can write:
3.9 x 10^-11 = x * (0.010)^2
Simplifying this equation, we find:
x = 3.9 x 10^-11 / (0.010)^2
Calculating this expression, we determine:
x ≈ 38.7 M (rounded to three significant figures)
Thus, the maximum concentration of Ca2+ ions in a solution containing 0.010 M of F- ions is approximately 38.7 M.
To determine the maximum concentration of Ca2+ ions in a solution containing 0.010 M of F- ions, we need to consider the solubility product constant (Ksp) of calcium fluoride (CaF2).
The solubility product constant represents the equilibrium constant for the dissociation of a sparingly soluble salt in water. It is a measure of the equilibrium concentration of the ions in solution at a given temperature. For calcium fluoride, the solubility product constant is given by:
Ksp = [Ca2+][F-]^2
Given that the concentration of F- ions is 0.010 M, we will assume this is their equilibrium concentration based on stoichiometry:
[F-] = 0.010 M.
Since the stoichiometry of calcium fluoride is 1:2 (one Ca2+ ion with two F- ions), the concentration of Ca2+ ions is twice the concentration of F- ions:
[Ca2+] = 2 * [F-] = 2 * 0.010 M = 0.020 M.
Therefore, the maximum concentration of Ca2+ ions in the solution containing 0.010 M of F- ions is 0.020 M.