Determine whether Cd2+ can be separated from Zn2+ by bubbling H2S through a 0.3 M HCl solution that contains 0.005 M Cd2+ and 0.005 M Zn2+. Ksp for CdS is 8x10^-7 and Ksp for ZnS is 3x10^-2.

K for H2S is k1k2 = (H^+)^2(S^-2)/(H2S)

These problems usually say that the solution is saturated with respect to H2S and that is 0.1 M.
So plug in H^+ from HCl and 0.1M for H2S and calculate (S^-2).
Then calculate Qsp for ZnS and CdS.
Qsp = (Zn^+2)(S^-2)
Qsp = (Cd^+2)(S^-2)
Finally, compare Qsp with Ksp of each.
I should point out that neither Ksp you list is correct. CdS is closer to 10^-28 and ZnS closer to 10^-24

My experience in 0.3M HCl is that both CdS and ZnS ppt.

To determine whether Cd2+ can be separated from Zn2+ by bubbling H2S through a 0.3 M HCl solution, we need to compare the solubility product constants (Ksp) for CdS and ZnS and consider the effect of the HCl concentration.

First, let's write the balanced chemical equations for the precipitation of CdS and ZnS when H2S is added to the solution:

For Cd2+ + H2S --> CdS + 2H+
For Zn2+ + H2S --> ZnS + 2H+

Since both CdS and ZnS are sparingly soluble salts, they will precipitate out of the solution when their concentrations exceed their respective Ksp values. Therefore, we need to calculate the concentration of Cd2+ and Zn2+ ions at which precipitation occurs.

For CdS:
Ksp(CdS) = [Cd2+][S2-] = 8x10^-7

Given that the concentration of Cd2+ is 0.005 M, we can assume that [Cd2+] << [S2-] since the concentration of S2- is determined by the H2S added. Therefore, we can approximate [S2-] to be equal to [H2S].

For ZnS:
Ksp(ZnS) = [Zn2+][S2-] = 3x10^-2

Given that the concentration of Zn2+ is also 0.005 M, we can make the same approximation that [S2-] is equal to [H2S].

Considering that H2S is bubbled through a 0.3 M HCl solution, it will dissociate into H+ and HS-. The HS- ion will then react with H+ to form H2S:

HS- + H+ --> H2S

Since the HCl concentration is much higher than the initial concentration of H2S, we can assume that almost all of the HS- ions react with H+ to form H2S. Therefore, we can approximate [H2S] to be equal to [HS-].

Now, let's calculate the concentrations of H2S and compare them to the Ksp values for CdS and ZnS:

[H2S] = [HS-] = concentration of HCl = 0.3 M

For CdS:
Ksp(CdS) = [Cd2+][HS-] = 8x10^-7

Substituting the known values, we get:
8x10^-7 = (0.005)(0.3) --> [Cd2+] = 5.33x10^-5 M

For ZnS:
Ksp(ZnS) = [Zn2+][HS-] = 3x10^-2

Again, substituting the known values, we have:
3x10^-2 = (0.005)(0.3) --> [Zn2+] = 1.0x10^-3 M

Comparing the calculated concentrations of Cd2+ and Zn2+ with their respective Ksp values, we can conclude that Cd2+ ions will precipitate as CdS, whereas Zn2+ ions will remain in solution since their concentration is lower than the Ksp value for ZnS.

Therefore, Cd2+ can be separated from Zn2+ by bubbling H2S through the 0.3 M HCl solution.