Why ZnS can be seen as the solution is made alkaline? Explain with reactions
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 $\ce{ZnS}$.