Calculate the molar solubility of MgF2 in 0.08M MgCl2 at 25 degrees C.
do I use the Ksp of MfF2 and plug it in the equation?
Yes.
from the equation:
Ksp=(.08M)(F^-)^2
there are two unknown how do we solve it
you have to find the Ksp in the back of your chemistry book.
the value 3.0x10^-5 was wrong for me?
To calculate the molar solubility of MgF2 in 0.08M MgCl2 at 25 degrees C, we will use the concept of the common ion effect.
The common ion effect states that the solubility of a salt is reduced when a common ion is present in the solution. In this case, MgCl2 contains the common ion Mg2+, which can reduce the solubility of MgF2 in the same solution.
To find the molar solubility of MgF2, we need to determine the concentration of Mg2+ ion in the 0.08M MgCl2 solution.
Since MgCl2 dissociates into Mg2+ and 2Cl- ions in water, the concentration of Mg2+ ion is twice the initial concentration of the MgCl2 solution, which is 0.08M.
Therefore, the concentration of the Mg2+ ion is 2 * 0.08M = 0.16M.
Next, we need to know the solubility product constant, Ksp, for MgF2 at 25 degrees C. The Ksp value for MgF2 is 1.5 x 10^-11.
The solubility product expression for MgF2 is given by:
Ksp = [Mg2+][F-]^2
Let's assume the molar solubility of MgF2 is "x".
So, the concentration of F- ions is also "x".
Therefore, we can write:
Ksp = (0.16 - x)(x)^2
To simplify the equation and neglect the value of "x" in comparison to 0.16 (due to the common ion effect), we can approximate the equation as follows:
Ksp ≈ (0.16)(x)^2
Now, we can solve for "x" by rearranging the equation:
x^2 = Ksp / (0.16)
Substituting the value of Ksp (1.5 x 10^-11), we get:
x^2 = (1.5 x 10^-11) / (0.16)
Taking the square root of both sides, we find:
x ≈ 2.22 x 10^-6
Therefore, the molar solubility of MgF2 in 0.08M MgCl2 at 25 degrees C is approximately 2.22 x 10^-6 M.
MgF2 ==> Mg^+2 + 2F^-
Ksp = (Mg^+2)(F^-)^2
Substitute 0.08 M for (Mg^+2) and solve for (F^-).