Question

In the preparation of $HBr$ or $HI,NaX(X=Br,I)$ is treated with $H_3{PO}_4$ and not by conc. $H_2{SO}_4$ and since:

• A. $H_2{SO}_4$ makes the reaction reversible
• B. $H_2{SO}_4$ oxidises $HX$ to $X_2(Br,I)$
• C. ${Na}_3{PO}_4$ is water soluble
• D. ${Na}_2{SO}_4$ is water soluble

Answer 1

Answer: (B)

$H_2{SO}_4$ oxidises $HX$ to $X_2(Br,I)$

As $H_{2}S{O}_4$ is strong oxidising agent so it will reacts with $HX$ and form halogen gas along with water and sulphur dioxide. Because of this reaction sulphuric acid is not used in this reaction.
$H_{2}S{O}_4+HX\to X_2+H_{2}O+S{O}_2$