Calculate the millimoles of SeO_3^{2-} in sol | vedclass

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(C) Reaction $(I): \ SeO_3^{2-} + BrO_3^{-} + H^{+} 	o SeO_4^{2-} + Br_2 + H_2O$ (n-factor for $SeO_3^{2-} = 2$,for $BrO_3^{-} = 5$)Reaction $(II): \

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$2.5 	imes 10^{-3}$

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(C) Reaction $(I): \ SeO_3^{2-} + BrO_3^{-} + H^{+} 	o SeO_4^{2-} + Br_2 + H_2O$ (n-factor for $SeO_3^{2-} = 2$,for $BrO_3^{-} = 5$)Reaction $(II): \ BrO_3^{-} + AsO_2^{-} + H_2O 	o Br^{-} + AsO_4^{3-} + H^{+}$ (n-factor for $BrO_3^{-} = 6$,for $AsO_2^{-} = 2$)Step $1$: Calculate milliequivalents of $AsO_2^{-}$ used in back titration:$meq \ of \ AsO_2^{-} = 12.5 \ mL 	imes (\frac{1}{25} \ M) 	imes 2 = 1 \ meq$Step $2$: Calculate millimoles of excess $BrO_3^{-}$:$meq \ of \ BrO_3^{-} = meq \ of \ AsO_2^{-} = 1 \ meq$$mmol \ of \ BrO_3^{-} = \frac{1}{6} \ mmol$Step $3$: Calculate millimoles of $BrO_3^{-}$ consumed in reaction $(I)$:Total $mmol \ of \ BrO_3^{-} = 70 \ mL 	imes \frac{1}{60} \ M = \frac{7}{6} \ mmol$$mmol \ of \ BrO_3^{-} \ consumed = \frac{7}{6} - \frac{1}{6} = 1 \ mmol$Step $4$: Calculate millimoles of $SeO_3^{2-}$:$meq \ of \ SeO_3^{2-} = meq \ of \ BrO_3^{-} \ consumed$$mmol \ of \ SeO_3^{2-} 	imes 2 = 1 \ mmol 	imes 5$$mmol \ of \ SeO_3^{2-} = 2.5 \ mmol = 2.5 	imes 10^{-3} \ mol$

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