In the preparation of HBr or HI , NaX (X = Br, I) is treated with H 3 PO 4 and not by concentrated H 2 SO 4 since,
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HBr (or H Ι ) cannot be prepared by heating bromide (iodide) with concentrated H 2 SO 4 because HBr and H Ι are strong reducing agents and reduce H 2 SO 4 to SO 2 and get themselves oxidised to bromine and iodine respectively.
KX + H 2 SO 4 ⎯→ KHSO 4 + HX
H 2 SO 4 + 2HX ⎯→ SO 2 + X 2 + 2H 2 O (X = Br or Ι )
Hence, HBr and H Ι are prepared by heating bromides and iodides respectively with concentrated H 3 PO 4 .
3KBr(K Ι ) + H 3 PO 4 ⎯→ K 3 PO 4 + 3HBr (H Ι )
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