A solution is $0.30\mathrm{M}$ in $\mathrm{HCl}$, saturated with $\mathrm{H}{}_2\mathrm{S}$ ($0.10\mathrm{M}$), and contains each of $\mathrm{Cd}{}^{2+}$, $\mathrm{Cu}{}^{2+}$, $\mathrm{Zn}{}^{2+}$ and $\mathrm{Ni}{}^{2+}$ at $0.010\mathrm{M}$. Use $K=[\mathrm{H}{}^{+}]{}^2[\mathrm{S}{}^{2-}]/[\mathrm{H}{}_2\mathrm{S}]=1.0\times 10^{-21}$ and neglect the $\mathrm{H}{}^{+}$ from $\mathrm{H}{}_2\mathrm{S}$. Given $K_{sp}$: $\mathrm{CdS}$ $8.0\times 10^{-27}$, $\mathrm{CuS}$ $6.0\times 10^{-36}$, $\mathrm{ZnS}$ $3.0\times 10^{-22}$, $\mathrm{NiS}$ $3.0\times 10^{-19}$. Which sulphides precipitate?

1. A$\mathrm{CuS}$
2. B$\mathrm{ZnS}$
3. C$\mathrm{CdS}$
4. D$\mathrm{NiS}$