Chemistry - Salt Analysis Question with Solution | TestHub

The question asks to identify the number of cations that first form a precipitate with the given reagent and then form a soluble complex with an excess of the same reagent.

Let's analyze each case:

(i) Bi³⁺ - KI

Bismuth(III) ions react with iodide ions to form a black precipitate of bismuth(III) iodide (BiI₃).

$$B{i}^{3+}+3I^{-}\to Bi{I}_3(s)$$

With excess KI, the precipitate dissolves to form a soluble tetraiodobismuthate(III) complex, which is orange-yellow.

$$Bi{I}_3(s)+I^{-}\to [Bi{I}_4{]}^{-}(aq)$$

This case fits the criteria.

(ii) Cu²⁺ - KCN

Copper(II) ions react with cyanide ions to form a precipitate of copper(II) cyanide (Cu(CN)₂), which quickly decomposes to copper(I) cyanide (CuCN) and cyanogen (C₂N₂).

$$C{u}^{2+}+2C{N}^{-}\to Cu(CN{)}_2(s)$$

$$2Cu(CN{)}_2(s)\to 2CuCN(s)+(CN{)}_2(g)$$

With excess KCN, the CuCN precipitate dissolves to form a soluble tetracyanocuprate(I) complex.

$$CuCN(s)+3C{N}^{-}\to [Cu(CN{)}_4{]}^{3-}(aq)$$

This case fits the criteria.

(iii) Ni²⁺ - NaOH

Nickel(II) ions react with hydroxide ions to form a green precipitate of nickel(II) hydroxide (Ni(OH)₂).

$$N{i}^{2+}+2O{H}^{-}\to Ni(OH{)}_2(s)$$

Nickel(II) hydroxide is INSOLUBLE in excess NaOH. It does not form a soluble complex.

This case does NOT fit the criteria.

(iv) Fe³⁺ - NH₄OH

Iron(III) ions react with ammonium hydroxide to form a reddish-brown precipitate of iron(III) hydroxide (Fe(OH)₃).

$$F{e}^{3+}+3N{H}_{4}OH\to Fe(OH{)}_3(s)+3N{H}_4^{+}(aq)$$

Iron(III) hydroxide is INSOLUBLE in excess NH₄OH. It does not form a soluble complex.

This case does NOT fit the criteria.

(v) Zn²⁺ - NaOH

Zinc(II) ions react with hydroxide ions to form a white gelatinous precipitate of zinc(II) hydroxide (Zn(OH)₂).

$$Z{n}^{2+}+2O{H}^{-}\to Zn(OH{)}_2(s)$$

With excess NaOH, the precipitate dissolves to form a soluble tetrahydroxozincate(II) complex.

$$Zn(OH{)}_2(s)+2O{H}^{-}\to [Zn(OH{)}_4{]}^{2-}(aq)$$

This case fits the criteria.

(vi) Ag⁺ - Na₂S₂O₃

Silver(I) ions react with thiosulfate ions to form a white precipitate of silver thiosulfate (Ag₂S₂O₃).

$$2A{g}^{+}+S_2{O}_3^{2-}\to A{g}_2{S}_2{O}_3(s)$$

With excess Na₂S₂O₃, the precipitate dissolves to form a soluble dithiosulfatoargentate(I) complex.

$$A{g}_2{S}_2{O}_3(s)+3S_2{O}_3^{2-}\to 2[Ag(S_2{O}_3{)}_2{]}^{3-}(aq)$$

This case fits the criteria.

The cations that fit the criteria are Bi³⁺, Cu²⁺, Zn²⁺, and Ag⁺.

Therefore, there are 4 such cations.

The final answer is $\boxed{4}$