Question

In basic medium, $Cr{O}_4^{2-}$ oxidises $S_2{O}_3^{2-}$ to form $Cr(OH{)}_4^{-}$ and $S^{}{O}_4^{2-}$. How many mL(nearest integer) of $0.154M$ $Cr{O}_4^{2-}$ are required to react with $40.0\text{mL}$ of $0.246M$ $S_2{O}_3^{2-}$?

$[\text{Hint}:0.0154M=0.154\times 3NCr{O}_4^{2-}\text{and}0.246M=0.246\times 8N{S}_2{O}_3^{2-}]$

Answer 1

The reaction is as follows:

$8Cr{O}_4^{2-}+3S_2{O}_3^{2-}\to 6S{O}_4^{2-}+8Cr(OH{)}_4^{-}$

$\text{The~normality~of}0.154MCr{O}_4^{2-}\text{is}0.154\times 3N.$

$\text{Similarity,~the~normality~of~}0.246M{S}_2{O}_3^{2-}\text{solution~is}0.246\times 8N.$
$N_1{V}_1=N_2{V}_2$
$V\times 0.154\times 3=0.246\times 8\times 40$
$V=170\text{mL.}$