# A solution with a pH=6.00 is 4.0E-3M in each of the metal cations Mn2+, Fe2+, Co2+, Ni2+, and Zn2+ and 0.10M in H2S. Under these conditions, which of the metal cations will be precipitated as a sulfide? Hint: Calculate IP.

General Equilibrium MS(s) + 2H3O+(aq) = M2+(aq) + 2H2O(l) + H2S(aq); Kspa

Metal Sulfide Kspa

MnS 3E10

FeS 6E2

CoS 3

NiS 8E-1

ZnS 3E-2

a.) Mn2+ and Fe2+

b.) Zn2+ only

c.) Zn2+ and Ni2+

d.) none of the above

I know the answer is d, from the back of my book, I just don't know how to solve it. Thanks a lot

sorry about this second post,

something was wrong with my computer

I don't know the meaning of IP and Kspa but I would do the problem this way.

H2S ==> 2H^+ + S^-2

k1k2 = (H^+)^2(S^-2)/(H2S).

You know pH, convert that to (H^+). You know (H2S). Use the above k1k2 expression to solve for sulfide.

Then, use that sulfide to calculate the (M^+2)(S^-2)= ?? and compare that number with Ksp (which may not be the same as Kspa since I don't know the meaning of Kspa. If the ?? number is higher than Ksp, there will be a ppt. If not, no ppt. Post your work if you get stuck and please let me know the meaning of IP and Kspa.

## To solve this problem, we need to calculate the ion product (IP) for each metal sulfide and compare it to their respective solubility product constants (Kspa). The IP is calculated by multiplying the concentrations of the metal cation and sulfide ion.

First, let's calculate the concentration of the sulfide ion in the solution. We are given that the solution has a pH of 6.00. Since the pH is negative logarithm of the hydrogen ion concentration ([H+]), we can find it using:

[H+] = 10^(-pH)

[H+] = 10^(-6.00)

[H+] = 1.00 x 10^(-6) M

In the reaction: H2S(aq) ⇌ 2H+(aq) + S^(-2)(aq)

We know the concentration of H2S is 0.10 M.

Now, let's calculate the IP for each metal sulfide. Remember, IP = [M^2+][S^(-2)].

For MnS:

[Mn^2+] = 4.0 x 10^(-3) M

[S^(-2)] = [H+]^2 = (1.00 x 10^(-6))^2 = 1.00 x 10^(-12) M

IP for MnS = (4.0 x 10^(-3))(1.00 x 10^(-12)) = 4.0 x 10^(-15)

For FeS:

[Fe^2+] = 4.0 x 10^(-3) M

[S^(-2)] = [H+]^2 = (1.00 x 10^(-6))^2 = 1.00 x 10^(-12) M

IP for FeS = (4.0 x 10^(-3))(1.00 x 10^(-12)) = 4.0 x 10^(-15)

For CoS:

[Co^2+] = 4.0 x 10^(-3) M

[S^(-2)] = [H+]^2 = (1.00 x 10^(-6))^2 = 1.00 x 10^(-12) M

IP for CoS = (4.0 x 10^(-3))(1.00 x 10^(-12)) = 4.0 x 10^(-15)

For NiS:

[Ni^2+] = 4.0 x 10^(-3) M

[S^(-2)] = [H+]^2 = (1.00 x 10^(-6))^2 = 1.00 x 10^(-12) M

IP for NiS = (4.0 x 10^(-3))(1.00 x 10^(-12)) = 4.0 x 10^(-15)

For ZnS:

[Zn^2+] = 4.0 x 10^(-3) M

[S^(-2)] = [H+]^2 = (1.00 x 10^(-6))^2 = 1.00 x 10^(-12) M

IP for ZnS = (4.0 x 10^(-3))(1.00 x 10^(-12)) = 4.0 x 10^(-15)

Now, let's compare the IPs to the Kspa values for each metal sulfide:

MnS: IP = 4.0 x 10^(-15), Kspa = 3 x 10^10, IP < Kspa

FeS: IP = 4.0 x 10^(-15), Kspa = 6 x 10^2, IP < Kspa

CoS: IP = 4.0 x 10^(-15), Kspa = 3, IP < Kspa

NiS: IP = 4.0 x 10^(-15), Kspa = 8 x 10^(-1), IP < Kspa

ZnS: IP = 4.0 x 10^(-15), Kspa = 3 x 10^(-2), IP < Kspa

From the comparison, we can see that the IP for each metal sulfide is much smaller than their respective Kspa values. This indicates that the ion product is less than the solubility product for each compound. As a result, none of the metal cations (Mn2+, Fe2+, Co2+, Ni2+, and Zn2+) will be precipitated as a sulfide under these conditions.

Therefore, the correct answer is d) none of the above.