0.1 M solution of KI reacts with excess of H | vedclass

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(A) The balanced chemical equation is $5I^{-} + IO_3^{-} + 6H^{+} \rightarrow 3I_2 + 3H_2O$.Moles of $KI$ in $200 \ mL$ of $0.1 \ M$ solution $= 0.1 \

Vedclass · Accepted Answer

$A$ and $D$ only

Vedclass · Answer

(A) The balanced chemical equation is $5I^{-} + IO_3^{-} + 6H^{+} ightarrow 3I_2 + 3H_2O$.Moles of $KI$ in $200 \ mL$ of $0.1 \ M$ solution $= 0.1 	imes 0.2 = 0.02 \ mol$.From the stoichiometry,$5 \ mol$ of $I^{-}$ reacts with $1 \ mol$ of $IO_3^{-}$.So,$0.02 \ mol$ of $I^{-}$ reacts with $0.02 / 5 = 0.004 \ mol$ of $KIO_3$. Statement $(A)$ is correct.$5 \ mol$ of $I^{-}$ reacts with $6 \ mol$ of $H^{+}$. So,$0.02 \ mol$ of $I^{-}$ reacts with $(6/5) 	imes 0.02 = 0.024 \ mol$ of $H^{+}$. Statement $(B)$ is incorrect.$5 \ mol$ of $I^{-}$ produces $3 \ mol$ of $I_2$. So,$0.05 \ mol$ of $I^{-}$ (in $0.5 \ L$) produces $(3/5) 	imes 0.05 = 0.03 \ mol$ of $I_2$. Statement $(C)$ is incorrect.In the reaction $IO_3^{-} + 6H^{+} + 5e^{-} ightarrow \frac{1}{2}I_2 + 3H_2O$,the change in oxidation state of $I$ in $IO_3^{-}$ $(+5)$ to $I_2$ $(0)$ is $5$. Thus,the n-factor is $5$. Equivalent weight $= \frac{	ext{Molecular weight}}{5}$. Statement $(D)$ is correct.

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