The volume $V$ of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ (molar mass $294 \mathrm{~g} \mathrm{~mol}^{-1}$ ) solution (mass concentration $0.005 \mathrm{~g}$ per $\mathrm{mL}$ ) is equivalent to $35.0 \mathrm{~mL}$ of $0.02 \mathrm{M} \mathrm{KMnO}_{4}$ (molar mass $158 \mathrm{~g}…

Question

The volume $V$ of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ (molar mass $294 \mathrm{~g} \mathrm{~mol}^{-1}$ ) solution (mass concentration $0.005 \mathrm{~g}$ per $\mathrm{mL}$ ) is equivalent to $35.0 \mathrm{~mL}$ of $0.02 \mathrm{M} \mathrm{KMnO}_{4}$ (molar mass $158 \mathrm{~g} \mathrm{~mol}^{-1}$ ), when these solutions are used in the titration in acid solution. The value of $V$ would be

Aicharya · Accepted Answer

The reactions are
$$
\frac{1}{6} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\frac{14}{6} \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \frac{1}{3} \mathrm{Cr}^{3+}+\frac{7}{6} \mathrm{H}_{2} \mathrm{O}
$$
and
$\frac{1}{5} \mathrm{MnO}_{4}^{-}+\frac{8}{5} \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \frac{1}{5} \mathrm{Mn}^{2+}+\frac{4}{5} \mathrm{H}_{2} \mathrm{O}$
Thus, $\quad n\left(\frac{1}{6} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}ight) \equiv n\left(\frac{1}{5} \mathrm{MnO}_{4}^{-}ight)$i.e. $6 n\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}ight)=5 n\left(\mathrm{MnO}_{4}^{-}ight)$
Now
$$
\begin{array}{ll}
\text { Now } & n\left(\mathrm{KMnO}_{4}ight)=V M=(0.035 \mathrm{~L})\left(0.02 \mathrm{~mol} \mathrm{~L}^{-1}ight)=0.0007 \mathrm{~mol} \\
& n\left(\mathrm{~K}_{2} \mathrm{CrO}_{7}ight)=V M=V\left(\frac{0.005 \mathrm{~g} \mathrm{~mL}^{-1}}{294 \mathrm{~g} \mathrm{~mol}^{-1}}ight)=V\left(\frac{0.005}{294} \mathrm{~mol} \mathrm{~mL}^{-1}ight) \\
\text { Thus } \quad & 6 V\left(\frac{0.005}{294} \mathrm{~mol} \mathrm{~mL}^{-1}ight)=5(0.0007 \mathrm{~mol}) \quad \text { or } \quad V=\left(\frac{5}{6}ight)\left(\frac{0.0007 \times 294}{0.005}ight) \mathrm{mL}=34.3 \mathrm{~mL}
\end{array}
$$

The volume $V$ of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ (molar mass $294 \mathrm{~g}…

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