Why can't HBr and HI be prepared by heating Conc. H 2SO 4 with bromide and iodide respectively ? Write down correct reactions to prepare them. [5] How would you test the presence of Br^ ions in its…

Why $\mathrm{HBr}$ and $\mathrm{HI}$ cannot be prepared by heating concentrated $\mathrm{H}{}_2\mathrm{SO}{}_4$ with bromide and iodide salts

Concentrated sulfuric acid ($\mathrm{H}{}_2\mathrm{SO}{}_4$) is a strong oxidizing agent, especially when hot. When it reacts with bromide ($\mathrm{Br}{}^{-}$) and iodide ($\mathrm{I}{}^{-}$) ions, it does not produce $\mathrm{HBr}$ or $\mathrm{HI}$ but instead oxidizes them to bromine ($\mathrm{Br}{}_2$) and iodine ($\mathrm{I}{}_2$), respectively. This happens because $\mathrm{H}{}_2\mathrm{SO}{}_4$ is reduced to $\mathrm{SO}{}_2$ (sulfur dioxide) in the process.

H₂SO₄ (conc.)+2KI+heathot conc. H₂SO₄SO₂+I₂+K₂SO₄+H₂O

H₂SO₄ (conc.)+2KBr+heathot conc. H₂SO₄SO₂+Br₂+K₂SO₄+H₂O

Incorrect Reactions (Oxidation occurs instead of simple acid-base reaction)

1. With bromide ($\mathrm{Br}{}^{-}$): $$2\mathrm{NaBr}+2\mathrm{H}{}_2\mathrm{SO}{}_4(\mathrm{conc}\cdot )\to \mathrm{Br}{}_2+\mathrm{SO}{}_2+2\mathrm{H}{}_2\mathrm{O}+\mathrm{Na}{}_2\mathrm{SO}{}_4$$

   • Here, $\mathrm{Br}{}^{-}$ is oxidized to $\mathrm{Br}{}_2$ (reddish-brown gas), and $\mathrm{H}{}_2\mathrm{SO}{}_4$ is reduced to $\mathrm{SO}{}_2$.

2. With iodide ($\mathrm{I}{}^{-}$): $$8\mathrm{NaI}+9\mathrm{H}{}_2\mathrm{SO}{}_4(\mathrm{conc}\cdot )\to 4\mathrm{I}{}_2+\mathrm{H}{}_2\mathrm{S}+4\mathrm{H}{}_2\mathrm{O}+4\mathrm{Na}{}_2\mathrm{SO}{}_4$$

   • Here, $\mathrm{I}{}^{-}$ is oxidized to $\mathrm{I}{}_2$ (purple vapor), and $\mathrm{H}{}_2\mathrm{SO}{}_4$ is reduced to $\mathrm{H}{}_2\mathrm{S}$ (hydrogen sulfide gas).

Correct Methods to Prepare $\mathrm{HBr}$ and $\mathrm{HI}$

Since $\mathrm{H}{}_2\mathrm{SO}{}_4$ oxidizes $\mathrm{Br}{}^{-}$ and $\mathrm{I}{}^{-}$, a non-oxidizing acid like phosphoric acid ($\mathrm{H}{}_3\mathrm{PO}{}_4$) is used instead.

NaBr+H₃PO₄ (conc.)HBr+NaH₂PO₄

1. Preparation of $\mathrm{HBr}$: $$\mathrm{NaBr}+\mathrm{H}{}_3\mathrm{PO}{}_4\to \mathrm{HBr}+\mathrm{NaH}{}_2\mathrm{PO}{}_4$$

   • Sodium bromide reacts with phosphoric acid to give hydrobromic acid ($\mathrm{HBr}$).

2. Preparation of $\mathrm{HI}$: $$\mathrm{NaI}+\mathrm{H}{}_3\mathrm{PO}{}_4\to \mathrm{HI}+\mathrm{NaH}{}_2\mathrm{PO}{}_4$$

   • Sodium iodide reacts with phosphoric acid to give hydroiodic acid ($\mathrm{HI}$).

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Testing for $\mathrm{Br}{}^{-}$ ions in aqueous solution

1. Reaction with Chlorine Water ($\mathrm{Cl}{}_2$)

• When chlorine water is added to a solution containing $\mathrm{Br}{}^{-}$ ions, bromine ($\mathrm{Br}{}_2$) is liberated (orange-brown color). $$2\mathrm{Br}{}^{-}+\mathrm{Cl}{}_2\to \mathrm{Br}{}_2+2\mathrm{Cl}{}^{-}$$
• Observation: Orange-brown color of $\mathrm{Br}{}_2$ appears.

Br^-+Cl₂Br₂+2Cl^-

2. Reaction with Silver Nitrate ($\mathrm{AgNO}{}_3$)

• When $\mathrm{AgNO}{}_3$ solution is added to a solution containing $\mathrm{Br}{}^{-}$ ions, a pale yellow precipitate of silver bromide ($\mathrm{AgBr}$) forms. $$\mathrm{Br}{}^{-}+\mathrm{AgNO}{}_3\to \mathrm{AgBr}(\mathrm{s})+\mathrm{NO}{}_3{}^{-}$$
• Observation: Pale yellow precipitate (insoluble in dilute nitric acid).

Br^-+AgNO₃AgBr (s)+NO₃^-

3. Confirmation Test (Optional)

• The pale yellow precipitate ($\mathrm{AgBr}$) dissolves in ammonia ($\mathrm{NH}{}_3$) solution, forming a colorless complex: $$\mathrm{AgBr}+2\mathrm{NH}{}_3\to [\mathrm{Ag}(\mathrm{NH}{}_3){}_2]{}^{+}+\mathrm{Br}{}^{-}$$
• Observation: Precipitate dissolves in excess ammonia.

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AgBr (s)+2NH₃[Ag(NH₃)₂]⁺+Br^-