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$0.1M$ $KMnO_4$ is used for the following titration $S_2O_3^{2-}+KMnO_4^-+H_2O\rightarrow MnO_2(s)+SO_4^{2-}+OH^-$.What volume of the solution in mL will be required to react with 0.158 of $Na_2S_2O_3?

$\begin{array}{1 1}26.66mL\\25.55mL\\23.66mL\\12.66mL\end{array} $

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The balanced equation is
$3S_2O_3^{2-}+8MnO_4^-+H_2O\rightarrow 8MnO_2(s)+6SO_4^{2-}+2OH^-$
$3Na_2S_2O_3+8KMnO_4+H_2O\rightarrow 8MnO_2+3Na_2SO_4+2KOH+3K_2SO_4$
$3\times 158$ g $Na_2S_2O_3$ reacts with =$8\times 158g$ $KMnO_4$
$\therefore 0.158g$ $Na_2S_2O_3$ react with =$\large\frac{8\times 158\times 0.158}{3\times 158}$ g $KMnO_4$
$\Rightarrow 0.4213g$ $KMnO_4$
0.1M $KMnO_4$ contains $0.1\times 158$ =15.8g/L
$15.8g$ of $KMnO_4$ is present in =1000mL
$\therefore 0.4213g$ of $KMnO_4$ is present in = $\large\frac{1000\times 0.4213}{15.8}$
$\Rightarrow 26.66mL$
answered Aug 7, 2014 by sreemathi.v

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