Both Fe2+ and Cd2+ are precipitated by sulfide. However, they can be quantitatively separated through pH adjustment. You are given a solution containing 0.086 M Fe2+ and 0.062 M Cd2+. Calculate the [H+] range for a saturated H2S solution (ie, [H2S]=0.10 M) in which at least 99.99% of the least soluble ion can be precipitated while the other remains entirely in solution. The relevent equilibria are:
FeS(s) ¡ê Fe2+ + S2‾ Ksp = 4.9 10-18
CdS(s) ¡ê Cd2+ + S2‾ Ksp = 3.6 10-29
H2S(aq) ¡ê 2 H+ + S2‾ Ka12 = 1.3 10-20
What is the lower limit [H+] of the separation range at which the most soluble ion will begin to precipitate?
What is the upper limit [H+] of the separation range at which ¡Ã99.99% of the least soluble ion will remain precipitated?
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To answer these questions, we need to consider the solubility product constants (Ksp) and the acid dissociation constant (Ka) of the relevant equilibria.
1. Calculate the [H+] range for the saturated H2S solution at which the most soluble ion (Cd2+) will begin to precipitate:
We can start by writing the equilibrium equation for the precipitation of CdS:
CdS(s) ⇌ Cd2+ + S2‾
The solubility product constant for CdS is given as Ksp = 3.6 x 10^(-29). Since [Cd2+] is 0.062 M, we can assume that [S2‾] is also 0.062 M (since it is in stoichiometric ratio).
The concentration of H2S ([H2S]) is given as 0.10 M. The equation for the dissociation of H2S is: H2S(aq) ⇌ 2 H+ + S2‾. The acid dissociation constant (Ka) for this reaction is given as Ka12 = 1.3 x 10^(-20).
To determine the lower limit [H+] at which CdS begins to precipitate, we need to find the [H+] at which the concentration of S2‾ equals the solubility product constant (Ksp).
Using the equilibrium expression: Ksp = [Cd2+][S2‾], we can substitute the known values: 3.6 x 10^(-29) = (0.062)[H+]^2
Rearranging the equation, we get: [H+]^2 = (3.6 x 10^(-29))/(0.062) = 5.8 x 10^(-28)
Taking the square root of both sides, we find: [H+] = 7.6 x 10^(-14)
Therefore, the lower limit [H+] of the separation range (at which Cd2+ begins to precipitate) is approximately 7.6 x 10^(-14) M.
2. Calculate the upper limit [H+] at which at least 99.99% of the least soluble ion (Fe2+) will remain precipitated:
Using similar reasoning, we can calculate the concentration of Fe2+ that will precipitate based on the solubility product constant (Ksp) for FeS.
The equilibrium equation for the precipitation of FeS is: FeS(s) ⇌ Fe2+ + S2‾
The solubility product constant for FeS is given as Ksp = 4.9 x 10^(-18). Since [Fe2+] is 0.086 M, we can assume that [S2‾] is also 0.086 M (since it is in stoichiometric ratio).
Again, using the equilibrium expression: Ksp = [Fe2+][S2‾], we can substitute the known values: 4.9 x 10^(-18) = (0.086)[H+]^2
Rearranging the equation, we get: [H+]^2 = (4.9 x 10^(-18))/(0.086) = 5.7 x 10^(-17)
Taking the square root of both sides, we find: [H+] = 7.5 x 10^(-9)
Therefore, the upper limit [H+] of the separation range (at which at least 99.99% of Fe2+ remains precipitated) is approximately 7.5 x 10^(-9) M.
In summary:
- The lower limit [H+] is approximately 7.6 x 10^(-14) M.
- The upper limit [H+] is approximately 7.5 x 10^(-9) M.