Why ZnS can be seen as the solution is made alkaline? Explain with reactions There is an extract in my book which i am not able to get. It says:
Place strip of metallic zinc in aqueous solution of copper nitrate for about one hour. You may notice that the strip becomes coated with reddish metallic copper and the blue colour of the solution disappears. Formation of Zn2+ ions among the products can easily be judged when the blue colour of the solution due to Cu2+ has disappeared. If H2S gas is passed through the colorless solution containing Zn2+ ions, appearance of white zinc sulphide, ZnS can be seen on making solution alkaline with ammonia.
The final solution contains $\ce{Zn^{2+}}$ ions, which reacts with $\ce{H_2S}$ to produce white $\ce{ZnS}$. But you may not avoid that maybe some of the initial $\ce{Cu^{2+}}$ ions still remains in the solution. And these rare $\ce{Cu^{2+}}$ ions do react with $\ce{H_2S}$ to produce Copper sulphide $\ce{CuS}$ which is dark black.
$$\ce{Cu^{2+}  +  H_2S  -&gt;  CuS  +  2 H^+}$$
So the obtained final precipitate is probably a mixture $\ce{ZnS}$ and some $\ce{CuS}$ ; it is probably not perfectly white, but grey or even black. To be able to see the white color of $\ce{ZnS}$, you add some drops of ammonia $\ce{NH3}$ in the dark mixture containing $\ce{ZnS}$. This will dissolve the copper sulphide, producing a solution containing $\ce{[Cu(NH3)4]^{2+}}$.
$$\ce{CuS  +  4 NH3 -&gt; [Cu(NH_3)_4]^{2+}  + S^{2-}}$$
So the black component of the precipitate will disappear, and you will be able to see the white color of insoluble $\ce{ZnS}$.