Question

$0 .1 M$ $KMnO_{4}$ is used for following titration. What volume of the solution in $mL$ will be required to react with $0 .158 $ of $N a_{2}S_{2}O_{3}$ ?

Not balanced: $S_{2}O_{3}^{2 -}+MnO_{4}^{-}+H_{2}O \rightarrow MnO_{2 \left(\right. s \left.\right)}+SO_{4}^{2 -}+OH^{-}$

• A. 26.7 mL
• B. 50 mL
• C. 65 mL
• D. 75 mL

Answer 1

Answer: (A)

$\stackrel{+2}{\mathrm{~S}_2} \mathrm{O}_3^{2-}+\mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\stackrel{+6}{\mathrm{SO}_4^{2-}}+\mathrm{OH}^{-}$

'n' factor of $Na_{2}S_{2}O_{3}=8$

'n' factor of $KMnO_{4}=3$

m.eq of $Na_{2}S_{2}O_{3}=$ m.eq of $KMnO_{4}$

$\frac{0.158}{158} \times 8 \times 1000 = 0.1 \times \text{V} \times 3$

$V=26.7 \, mL$