For the reaction : $I_{2} + NaOH ⟶ {NaIO}_{3} + NaI + H_{2} O$
Identify the correct statements. (At wt. of Na= 23)

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Reaction is an example of disproportionation

Equivalent weight of $I_{2} = \frac{3}{5} \times (mol . wt. of I_{2})$

Eq. wt of $NaOH$ in the reaction is 6.66

Eq. wt of $NaOH$ in the reaction is 48.

answer is A, B, D.

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Detailed Solution

Reaction is an example of disproportionation.

Equivalent weight of $I_{2} = \frac{3}{5} \times (mol . wt. of I_{2})$

For the reaction,

$3 I_{2} + 6 NaOH ⟶ {NaIO}_{3} + 5 NaI + 3 H_{2} O$

Net $5 e^{-}$ are involved in redox for 6 moles of $NaOH$ .

Hence, eq. weight of $NaOH = \frac{40}{5} = 48$ .

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