Which of the following acids is usually used for the preparation of `HBr` ,and `HI` form `NaBr, Kl` respectively?

Options

A. `CH_(3)CO_(2)H` (glacial)
B. `H_(2)SO_(4)` (conc.)
C. `H_(3)PO_(4)` (syrupy)
D. `HF` (anhydrous)

Answer

3

Explanation

Treatment of metal bromindes or iodides with conc.`H_(2)SO_(4)` is not satisfactory for the perparation of `HBr` and `HI` because the `HBr` or `HI` formed are oxidized by `H_(2)SO_(4)` and the protect is contaminated with the respective halogen.
Hydrogen bromide reduces the sulphuric acid to sulphur dioxide while the stronger reducing agent hydrogen iodide converts it mainly into hydrogen sulphide :
`2HBr(g)+H_(2)SO_(4)(l)rarrSO_(2)(g)+2H_(2)O(l)+Br_(2)(l)`
`8HI(g)+H_(2)SO_(4)(l)rarrH_(2)S(g)+4H_(2)O(l)+4I_(2)(s)`
This difficulty can be overcomed by using a nonoxidising acid such as `H_(3)PO_(4)`