Question

A solution of 10 ml $\frac{M}{10} {FeSO_4}$ was titrated with ${KMnO_4}$ solution in acidic medium. The amount of ${KMnO_4}$ used will be :

• A. 5 ml of 0.1 M
• B. 10 ml of 0.1 M
• C. 10 ml of 0.5 M
• D. 10 ml of 0.02 M

Answer 1

Answer: (D)

10 ml of 0.02 M

Answer 2

Given: The $10\, ml$ of $\frac{ M }{10} FeSO _{4}$ is titrated with $KMnO _{4} .$ To Find: Amount of $KMnO _{4}$ used. Reaction: $MnO _{4}^{-}+5 Fe ^{2+}+8 H ^{+} \rightarrow 5 Fe ^{3+}+ Mn ^{2+}+4 H _{2} O$ Hence, 1 mole of $KMnO _{4}$ is required to titrate 5 moles of $FeSO _{4}$. Moles of $FeSO _{4}$ titrated =Molality $\times$ Volume $=10\, ml \times \frac{M}{10}=1\, m$ mole. The number of moles of $KMnO _{4}$ used $=\left(\frac{\text { Moles of } FeSO _{4}}{5}\right)$ $=\frac{1 mill \text { mole }}{5}=0.2\, m$ mole. So, $0.2\, m$ mole $=10\, ml \times 0.02\, M\, KMnO _{4}$. Therefore, $10\, ml$ of $0.02\, M\, KMO _{4}$ is required to titrate the $FeSO _{4}$ solution.