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pH the Power of H

In neutral so...

Question

In neutral solution, one-mole periodate ion, ${I O}_{4}^{-}$ , reacts with an excess of iodide to produce one mole of iodine; on acidification of the resulting solution, a further three moles of iodine is liberated.

Equations of the reactions which occur under these conditions.

$I O_{4}^{-} + 2 I^{-} + H_{2} O ⟶ I O_{3}^{-} + I_{2} + 2 O H^{-}$
$I O_{3}^{-} + 5 I^{-} + 6 H^{+} ⟶ 3 I_{2} + 3 H_{2} O$

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Solution

In neutral solution, one mole periodate ion, ${I O}_{4}^{-}$ , reacts with excess of iodide to produce one mole of iodine;
$I O_{4}^{-} + 2 I^{-} + H_{2} O ⟶ I O_{3}^{-} + I_{2} + 2 O H^{-}$
On acidification of the resulting solution, a further three moles of iodine are liberated.
$I O_{3}^{-} + 5 I^{-} + 6 H^{+} ⟶ 3 I_{2} + 3 H_{2} O$

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