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Oxalic acid, $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$, reacts with paramagnet ion according to the balanced equation $5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4(\mathrm{aq})+$ $2 \mathrm{MnO}_{4^{-}}(\mathrm{aq}) \rightleftharpoons 2 \mathrm{Mn}^{2+}(\mathrm{aq})+10 \mathrm{CO}_2(\mathrm{~g})+8 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$. The volume in $\mathrm{mL}$ of $0.0162 \mathrm{M} \mathrm{KMnO}_4$ solution required to react with $25.0 \mathrm{~mL}$ of $0.022 \mathrm{M} \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$ solution is :
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Verified Answer
The correct answer is:
$13.6$
In the reaction
\(\begin{aligned}
& \mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{+2} \\
& \text {thus nf }=5 \\
& \mathrm{C}_2^{+3} \mathrm{O}_4^{2-} \rightarrow \mathrm{CO}_2^{+4} \\
& 2 \cdot \mathrm{C}^{+3} \rightarrow \mathrm{C}^{+4} \\
& \mathrm{nf}=2 \times 1=2
\end{aligned}\)
thus by equation
\(\begin{gathered}
\mathrm{KMnO}_4 \mathrm{M}_1 \mathrm{~V}_1 \mathrm{nf}_1=\mathrm{M}_2 \mathrm{~V}_2 \mathrm{nf}_2 \quad \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \\
0.0162 \mathrm{M} \times \mathrm{V}_1 \times 5=0.022 \mathrm{M} \times 25.0 \mathrm{~mL} \times 2 \\
\mathrm{~V}_1=13.58 \mathrm{~mL} \simeq 13.6 \mathrm{~mL}
\end{gathered}\)
\(\begin{aligned}
& \mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{+2} \\
& \text {thus nf }=5 \\
& \mathrm{C}_2^{+3} \mathrm{O}_4^{2-} \rightarrow \mathrm{CO}_2^{+4} \\
& 2 \cdot \mathrm{C}^{+3} \rightarrow \mathrm{C}^{+4} \\
& \mathrm{nf}=2 \times 1=2
\end{aligned}\)
thus by equation
\(\begin{gathered}
\mathrm{KMnO}_4 \mathrm{M}_1 \mathrm{~V}_1 \mathrm{nf}_1=\mathrm{M}_2 \mathrm{~V}_2 \mathrm{nf}_2 \quad \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \\
0.0162 \mathrm{M} \times \mathrm{V}_1 \times 5=0.022 \mathrm{M} \times 25.0 \mathrm{~mL} \times 2 \\
\mathrm{~V}_1=13.58 \mathrm{~mL} \simeq 13.6 \mathrm{~mL}
\end{gathered}\)
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