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Mole Concept

0.1 M KMnO4 i...

Question

$0.1 M K M n O_{4}$ is used for the following titration. What volume of the solution in $m L$ will be required to react will $0.158 g$ of $N a_{2} S_{2} O_{3}$ ?
${S_{2} O_{3}}^{2 -} + {M n O_{4}}^{-} + H_{2} O \to M n O_{2} (s) + {S O_{4}}^{2 -} + O H^{-}$

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Solution

$S_{2} O_{3}^{2 -} \to 2 S O_{4}^{2 -}$
Change in oxidation number of sulphur per molecule of $S_{2} O_{3}^{2 -} = 2 \times (6 - 2) = 8$
Change in oxidation number of $M n$ per molecule of $M n O_{4}^{-} = 7 - 4 = 3$
No. of moles in $0.158 g$ of $N a_{2} S_{2} O_{3} = \frac{0.158}{158} = 1 \times 10^{- 3}$
No. of equivalents $= 8 \times 10^{- 3}$
Normality of $0.1 M K M n O_{4}$ solution $= 0.1 \times 3 = 0.3$
Let $V m L$ of volume of $K M n O_{4}$ be required; then
$\frac{V}{1000} \times 0.3 = 8 \times 10^{- 3}$
or $V = \frac{8}{0.3} \times 10^{- 3} = 10^{3} = 26.7 m L$ .

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