NEETp-Block elementsHard
Question
In the preparation of HBr or HI , NaX (X = Br, I) is treated with H3PO4 and not by concentrated H2SO4 since,
Options
A.H2SO4 makes the reaction reversible
B.H2SO4 oxidises HX to X2 (Br2, I2)
C.Na2SO4 is water soluble and Na3PO4 is water insoluble
D.Na3PO4 is water insoluble and Na2SO4 is water soluble
Solution
HBr (or HI) cannot be prepared by heating bromide (iodide) with concentrated H2SO4 because HBr and HI are strong reducing agents and reduce H2SO4 to SO2 and get themselves oxidised to bromine and iodine respectively.
KX + H2SO4 → KHSO4 + HX
H2SO4 + 2HX → SO2 + X2 + 2H2O (X = Br or I)
Hence, HBr and HI are prepared by heating bromides and iodides respectively with concentrated H3PO4.
3KBr(KI) + H3PO4 → K3PO4 + 3HBr (HI)
KX + H2SO4 → KHSO4 + HX
H2SO4 + 2HX → SO2 + X2 + 2H2O (X = Br or I)
Hence, HBr and HI are prepared by heating bromides and iodides respectively with concentrated H3PO4.
3KBr(KI) + H3PO4 → K3PO4 + 3HBr (HI)
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