when Ksp = [A2+] [S2-]
when A is the metal: Fe, Ni, Pb, and Cu
When we have [S2-] = 0.1 m and we have Ksp for each metal So by substitution in Ksp formula we can get [A2+] for each metal and compare its value with solution concentration 0.01 M, when we have a concentration more than 0.01 M So there are no sulfides precipitates
- [Fe2+] = Ksp/[S2-]
by substitution with Fe2+ Ksp value:
= 6x10^2 / 0.1
= 6x10^3 M
when [Fe2+] > 0.01 M ∴ no precipitate
- [Ni2+] = Ksp /[S2-]
by sustitution with Ni Ksp value :
= 8x10^-1 / 0.1
= 8 M
When [Ni2+] > 0.01 M ∴ no precipitate
-[Pb2+] = Ksp / [S2-]
by substitution with Pb Ksp value:
= 6x10^-7 / 0.1
= 6 x 10^-6 M
when [Pb2+] < 0.01 M ∴PbS will be precipited
-[Cu2+] = Ksp / [S2-]
by substitution with Cu2+ Ksp value:
= 6x10^-16 / 0.1
= 6x10^-15 M
when [Cu2+] < 0.01 M ∴ CuS will be precipited
∴The sulfides precipitates are CuS & PbS