Determine whether Cd^2+ can be separated from Zn^2+ by bubbling H2S through a 0.3 M HCl solution that contains 0.005M Cd^2+ and 0.005 M Zn ^2+. Ksp for CdS is 8*10^-7. Ksp for ZnS is 3*10^-2
I don't have any of your data (and I don't believe the Ksp values either---they are far from correct) but here is what you do. You know H^+ is 0.3M, use that and the concn of a saturated solution of H2S to calculate (S^-2). Then
(Cd^+2)(S^-2) = Qsp
(Zn^+2)(S^-2) = Qsp
and compare with Ksp for each.
Using Ksp values for ZnS and CdS from my quant book, this will separate Zn from Cd (but my quant book Ksp values don't agree with your post).
To determine whether Cd^2+ can be separated from Zn^2+ by bubbling H2S through a 0.3 M HCl solution that contains 0.005 M Cd^2+ and 0.005 M Zn ^2+, we need to compare the solubility product constants (Ksp) of CdS and ZnS with the concentrations of Cd^2+ and Zn^2+ present in the solution.
The solubility product constant (Ksp) is an equilibrium constant that measures the solubility of a sparingly soluble compound in a solution. It can be used to determine the maximum concentration of ions that can exist in a saturated solution.
In this case, the Ksp values for CdS and ZnS are given as:
Ksp for CdS = 8*10^-7
Ksp for ZnS = 3*10^-2
Since H2S is bubbled through a 0.3 M HCl solution, it will react with HCl to form H2S gas and chloride ions (Cl^-). This reaction is not relevant to the precipitation of CdS and ZnS. So, we can disregard the presence of HCl in our analysis.
Now, let's calculate the ion product (IP) for CdS and ZnS in the given solution:
IP for CdS = [Cd^2+][S^2-]
= (0.005 M)(x), where x is the concentration of S^2- ions in the solution
IP for ZnS = [Zn^2+][S^2-]
= (0.005 M)(x), where x is the concentration of S^2- ions in the solution
To determine whether CdS and ZnS will precipitate, we need to compare the IP values with the respective Ksp values.
For CdS:
IP for CdS = (0.005 M)(x)
Ksp for CdS = 8*10^-7
If the IP for CdS is greater than Ksp for CdS (IP > Ksp), then CdS will precipitate and can be separated from Zn^2+.
For ZnS:
IP for ZnS = (0.005 M)(x)
Ksp for ZnS = 3*10^-2
If the IP for ZnS is greater than Ksp for ZnS (IP > Ksp), then ZnS will precipitate and Cd^2+ cannot be separated from Zn^2+.
To determine the values of x for CdS and ZnS, we need to look at their molar ratios in the respective chemical formulas.
CdS: 1 Cd^2+ ion reacts with 1 S^2- ion, so the molar ratio is 1:1.
ZnS: 1 Zn^2+ ion reacts with 1 S^2- ion, so the molar ratio is also 1:1.
Now, let's calculate the IP values for CdS and ZnS using the molar ratio:
IP for CdS = (0.005 M)(x)
IP for ZnS = (0.005 M)(x)
Since the molar ratio is the same for both compounds, we can conclude that the concentration of S^2- ions (x) will be the same for both CdS and ZnS.
To determine which compound will precipitate, we need to compare the IP values with the respective Ksp values:
For CdS:
IP for CdS = (0.005 M)(x)
Ksp for CdS = 8*10^-7
For ZnS:
IP for ZnS = (0.005 M)(x)
Ksp for ZnS = 3*10^-2
Comparing the IP values with the respective Ksp values, we find that the IP values for both CdS and ZnS are much smaller than their respective Ksp values. Therefore, neither CdS nor ZnS will precipitate in the given solution.
In conclusion, Cd^2+ cannot be separated from Zn^2+ by bubbling H2S through a 0.3 M HCl solution that contains 0.005 M Cd^2+ and 0.005 M Zn ^2+ because both CdS and ZnS will remain dissolved in the solution.