Question

Determine the volume of $\mathrm{M} / 8 \mathrm{KMnO}{4}$ solution (in mL) that is required to react completely with $25.0$ $\mathrm{cm}^{3}$ of $\mathrm{M} / 4 \mathrm{FeSO}{4}$ solution in acidic medium.

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Answer

Problem Statement: Determine the volume of a $\mathrm{M}/8 \mathrm{KMnO}_4$ solution (in mL) that is required to completely react with $25.0 , \mathrm{cm}^3$ of an $\mathrm{M}/4 \mathrm{FeSO}_4$ solution in an acidic medium.

:

The balanced ionic equation for the reaction is: $$ \mathrm{MnO}_4^- + 5 \mathrm{Fe}^{2+} + 8 \mathrm{H}^+ \rightarrow \mathrm{Mn}^{2+} + 5 \mathrm{Fe}^{3+} + 4 \mathrm{H}_2\mathrm{O} $$

From the balanced ionic equation, it is evident that 1 mole of $\mathrm{KMnO}_4$ reacts with 5 moles of $\mathrm{FeSO}_4$.

Using the law of chemical equivalence, we get: $$ \mathrm{M}_1 \mathrm{V}_1 \mathrm{n}_1 (\mathrm{KMnO}_4) = \mathrm{M}_2 \mathrm{V}_2 \mathrm{n}_2 (\mathrm{FeSO}_4) $$ where $\mathrm{n}$ represents the equivalence factor.

Given:

  • Molarity of $\mathrm{KMnO}_4$ = $\frac{1}{8} , \mathrm{M}$
  • Volume of $\mathrm{KMnO}_4$ solution = $\mathrm{V}_1$
  • $n_1$ for $\mathrm{KMnO}_4$ = 5
  • Molarity of $\mathrm{FeSO}_4$ = $\frac{1}{4} , \mathrm{M}$
  • Volume of $\mathrm{FeSO}_4$ solution = $25 , \mathrm{cm}^3$
  • $n_2$ for $\mathrm{FeSO}_4$ = 1

Substitute the values into the equation: $$ \frac{1}{8} \times \mathrm{V}_1 \times 5 = \frac{1}{4} \times 25 \times 1 $$

Solve for $\mathrm{V}_1$: $$ \frac{5}{8} \mathrm{V}_1 = \frac{25}{4} $$ $$ \mathrm{V}_1 = \frac{25}{4} \times \frac{8}{5} $$ $$ \mathrm{V}_1 = 10 , \mathrm{cm}^3 $$

Therefore, the volume of the $\mathrm{M}/8 \mathrm{KMnO}_4$ solution required is 10 mL.


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