Calculate the millimoles of ${SeO}_{3}^{2 -}$ in solution on the basis of following data:
70 ml of $\frac{M}{60}$ solution of ${KBrO}_{3}$ was added to ${SeO}_{3}^{2 -}$ solution.
The bromine evolved was removed by boiling and excess of ${KBrO}_{3}$ was back titrated with 12.5 mL $\frac{M}{25}$ solution of $NaAs O_{2}$ .
The reactions for the above data is/are given below.
$(1) {SeO}_{3}^{2 -} + {BrO}_{3}^{-} + H^{+} \to {SeO}_{4}^{2 -} + {Br}_{2} + H_{2} O$
$(2) {BrO}_{3}^{-} + {AsO}_{2}^{-} + H_{2} O \to {Br}^{-} + {AsO}_{4}^{3 -} + H^{+}$

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$1.6 \times 10^{- 3}$

1.25

$2.5 \times 10^{- 3}$

zero

answer is C.

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Detailed Solution

$(a) \overset{+ 4}{{SeO}_{3}^{- 2}} + \overset{+ 5}{{BrO}_{3}^{-}} + H^{+} \to \overset{+ 6}{{SeO}_{4}^{+ 2}} + \overset{0}{{Br}_{2}} + H_{2} O$

$(b) \overset{+ 5}{{BrO}_{3}^{-}} + \overset{+ 3}{{AsO}_{2}^{-}} + H_{2} O \to \overset{- 1}{{Br}^{-}} + \overset{+ 5}{{AsO}_{4}^{3 -}} + H^{+}$

In reaction (2)

$gm . eq. of {BrO}_{3}^{-} = gm . eq. of {AsO}_{2}^{-}$

$n_{{BrO}_{3}} \times 6 = n_{{AsO}_{2}} \times 2 = \frac{12.5}{1000} \times \frac{1}{25} \times 2 = 10^{- 3}$

$n_{{BrO}_{3}} = \frac{10^{- 3}}{6}$

In reaction (1)

$moles of {BrO}_{3}^{-} consumed = \frac{70}{1000} \times \frac{1}{60} - \frac{10^{- 3}}{6} = 10^{- 3}$

$gm eq. of {SeO}_{3}^{2 -} = gm . eq. of {BrO}_{3}^{-}$

$n_{{SeO}_{3}^{2 -}} \times 2 = 10^{- 3} \times 5; n_{{SeO}_{3}^{2 -}} = 2.5 \times 10^{- 3}$

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