An aqueous solution of
KF is add dropwise to an aqueous micture containing 0.25M Ca2+ and 0.75M Mg2+
calcuate the minimum molarity of F that is nessaryvto initate the precipitation of each salt? Which precipitate first?
Ksp of CaF2 :5.4x10^-9
Ksp of MgF2 :3.6x10^-8
To calculate the minimum molarity of F necessary to initiate the precipitation of each salt (CaF2 and MgF2), we need to compare the solubility product constant (Ksp) to the concentrations of Ca2+ and Mg2+ ions.
To initiate the precipitation, the product of the concentrations of the reacting ions (Ca2+ and F- in the case of CaF2, and Mg2+ and F- in the case of MgF2) should exceed their respective Ksp values.
Let's start with CaF2:
CaF2(s) <--> Ca2+(aq) + 2F-(aq)
The Ksp expression for CaF2 is:
Ksp = [Ca2+][F-]^2 = 5.4x10^-9
Since we know the concentration of Ca2+ is 0.25 M, let's assume the initial concentration of F- is x M.
Now, let's look at MgF2:
MgF2(s) <--> Mg2+(aq) + 2F-(aq)
The Ksp expression for MgF2 is:
Ksp = [Mg2+][F-]^2 = 3.6x10^-8
Again, assuming the initial concentration of F- is x M, we have the following constraints based on the solubility product principle:
For CaF2:
Ksp = [Ca2+][F-]^2 = (0.25)(x)^2 = 5.4x10^-9
Solving for x:
x^2 = (5.4x10^-9) / (0.25)
x^2 = 2.16x10^-8
x ≈ 1.469x10^-4 M
For MgF2:
Ksp = [Mg2+][F-]^2 = (0.75)(x)^2 = 3.6x10^-8
Solving for x:
x^2 = (3.6x10^-8) / (0.75)
x^2 = 4.8x10^-8
x ≈ 2.19x10^-4 M
Therefore, the minimum molarity of F- necessary to initiate the precipitation of CaF2 is approximately 1.469x10^-4 M, and the minimum molarity of F- necessary to initiate the precipitation of MgF2 is approximately 2.19x10^-4 M.
To determine which salt precipitates first, we need to compare their respective ion product (IP) values.
For CaF2, IP = [Ca2+][F-]^2 = (0.25)(1.469x10^-4)^2 ≈ 5.37x10^-9
For MgF2, IP = [Mg2+][F-]^2 = (0.75)(2.19x10^-4)^2 ≈ 3.62x10^-7
Comparing the IP values, we can see that the IP for CaF2 is smaller compared to the IP for MgF2. Therefore, CaF2 will precipitate first before MgF2 in this given scenario.