Question

$0.1M$ $KMn{O}_4$ 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: $S_2{O}_3^{2-}+Mn{O}_4^{-}+H_{2}O\to Mn{O}_{2(s)}+S{O}_4^{2-}+O{H}^{-}$

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

Answer 1

Answer: (A)

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

'n' factor of $N{a}_2{S}_2{O}_3=8$

'n' factor of $KMn{O}_4=3$

m.eq of $N{a}_2{S}_2{O}_3=$ m.eq of $KMn{O}_4$

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

$V=26.7mL$