Question

Most sulfide compounds of the transition metals are insoluble in water. Many of these metal sulfides have striking and characteristic colors by which we can identify them. Therefore, in the analysis of mixtures of metal ions, it is very common to precipitate the metal ions by using dihydrogen sulfate (commonly called hydrogen sulfide), H2S. Suppose you had a mixture of Fe2 , Cr3 , and Ni2 . Complete the net ionic equations for the precipitation of these metal ions by the use of H2S. (Type your answers using the format Fe2 for Fe2 .)

Answer 1

Answer:

$Fe^{2+}(aq)+H_2S(aq)\rightarrow FeS (s)+2H^+ (aq)$

$2Cr^{3+}(aq)+3H_2S(aq)\rightarrow Cr_2S_3 (s)+6H^+ (aq)$

$Ni^{2+}(aq)+H_2S(aq)\rightarrow NiS (s)+2H^+ (aq)$

Explanation:

We're given the following ions:

$Fe^{2+}, Cr^{3+}, Ni^{2+}$

Hydrogen sulfide is a weak acid, so it only ionizes to ions to a very low extent. This means we would expect to see it in a molecular form in a net ionic equation rather than a dissociated form (hydrogen cations and sulfide anions). In each of these net ionic equations, we expect the three cations to displace hydrogen from hydrogen sulfide and form three precipitates.

Firstly, iron(II) displaces hydrogen forming iron(II) sulfide and acidic conditions:

$Fe^{2+}(aq)+H_2S(aq)\rightarrow FeS (s)+2H^+ (aq)$

Secondly, chromium(III) cation displaces hydrogen forming chromium(III) sulfide:

$2Cr^{3+}(aq)+3H_2S(aq)\rightarrow Cr_2S_3 (s)+6H^+ (aq)$

Thirdly, nickel(II) cation displaces hydrogen forming nickel sulfide:

$Ni^{2+}(aq)+H_2S(aq)\rightarrow NiS (s)+2H^+ (aq)$