Consider the sulphides HgS, PbS, CuS, Sb$$_2$$S$$_3$$, As$$_2$$S$$_3$$ and CdS. Number of these sulphides soluble in 50% HNO$$_3$$ is _________.

We have to see, one by one, whether each of the six given sulphides is attacked by $$50\%$$ (that is, roughly one-to-one $$HNO_3 : H_2O$$) nitric acid. In such a medium the acid acts simultaneously as a strong proton source and as an oxidising agent, so a sulphide will dissolve if the metal can pass into an oxidised, soluble species while the sulphur is converted to elemental sulphur or to sulphuric acid. Whenever this happens we say that the sulphide is “soluble” in the acid.

For a simple metallic sulphide $$MS$$, the typical oxidation-dissolution pattern is

$$3\,MS\;+\;8\,HNO_3 \;\longrightarrow\; 3\,M(NO_3)_2 \;+\;2\,NO \;+\;3\,S \;+\;4\,H_2O$$

(If the metal goes to a higher oxidation state the stoichiometry changes, but the guiding idea is the same: the nitrate of the metal is formed and remains in solution.) Using this idea we now examine every sulphide in the list.

1. $$\underline{HgS}$$ - Mercury(II) sulphide (cinnabar) has an extremely compact crystal lattice. Even concentrated $$HNO_3$$ cannot oxidise it; only aqua regia or a cyanide solution succeeds. So

$$HgS\;+\;HNO_3\;(50\%) \;\nrightarrow\; \text{soluble products}$$

Hence $$HgS$$ does not dissolve.

2. $$\underline{PbS}$$ - Lead(II) sulphide is readily oxidised. With $$50\%$$ nitric acid we obtain

$$3\,PbS + 8\,HNO_3 \;\longrightarrow\; 3\,Pb(NO_3)_2 + 2\,NO + 3\,S + 4\,H_2O$$

The salt $$Pb(NO_3)_2$$ is soluble, so the sulphide disappears from the solid phase. Therefore $$PbS$$ is soluble in the medium.

3. $$\underline{CuS}$$ - Copper(II) sulphide behaves exactly like lead sulphide:

$$3\,CuS + 8\,HNO_3 \;\longrightarrow\; 3\,Cu(NO_3)_2 + 2\,NO + 3\,S + 4\,H_2O$$

The nitrate $$Cu(NO_3)_2$$ is freely soluble, so $$CuS$$ dissolves.

4. $$\underline{Sb_2S_3}$$ - When antimony(III) sulphide is treated with $$HNO_3$$ an oxidation sets in, but the antimony produced is not a soluble nitrate; instead an insoluble hydrated oxide, commonly written $$HSbO_3$$, precipitates:

$$Sb_2S_3 + 6\,HNO_3 + 3\,H_2O \;\longrightarrow\; 2\,HSbO_3 \downarrow + 3\,H_2SO_4 + 6\,NO$$

Because the newly formed oxide remains as a solid, the original sulphide has not really passed into solution. Hence $$Sb_2S_3$$ is regarded as insoluble in $$50\%$$ nitric acid.

5. $$\underline{As_2S_3}$$ - Arsenic(III) sulphide, unlike the antimony compound, is completely oxidised to the soluble acid $$H_3AsO_4$$ (arsenic(V) acid). A convenient balanced equation is

$$As_2S_3 + 10\,HNO_3 + 4\,H_2O \;\longrightarrow\; 2\,H_3AsO_4 + 10\,NO_2 + 3\,H_2SO_4$$

Both $$H_3AsO_4$$ and $$H_2SO_4$$ stay in solution, so the solid sulphide disappears completely. Therefore $$As_2S_3$$ is soluble in $$50\%$$ nitric acid.

6. $$\underline{CdS}$$ - Cadmium(II) sulphide is oxidised in exactly the same manner as lead and copper sulphides:

$$3\,CdS + 8\,HNO_3 \;\longrightarrow\; 3\,Cd(NO_3)_2 + 2\,NO + 3\,S + 4\,H_2O$$

The nitrate $$Cd(NO_3)_2$$ is soluble, so $$CdS$$ dissolves.

Collecting the results:

• Insoluble: $$HgS,\; Sb_2S_3$$ (2 compounds)

• Soluble: $$PbS,\; CuS,\; As_2S_3,\; CdS$$ (4 compounds)

So in total, $$4$$ of the listed sulphides dissolve in $$50\%$$ nitric acid.

So, the answer is $$4$$.