Question

The volume of 0.3 M ferrous ammonium sulphate solution required for the completion of redox titration with 20 mL of 0.1 M potassium dichromate solution is ............. mL.

Hint:

In titration calculations, use the equation \( M_1 V_1 = M_2 V_2 \) to find the volume of one solution required to neutralize the other, given their concentrations and the stoichiometry of the reaction.

Answer 1

Correct Answer: 40

Balanced redox half-reactions in acidic medium:

\[\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O}\]
\[\mathrm{Fe^{2+} \longrightarrow Fe^{3+} + e^-}\]

From these, one \(\mathrm{Cr_2O_7^{2-}}\) (dichromate) accepts 6 electrons, so it oxidizes 6 equivalents of \(\mathrm{Fe^{2+}}\):
\[\mathrm{1; Cr_2O_7^{2-} \longrightarrow 6; Fe^{2+}}\]

Moles of dichromate provided:
\[n_{\mathrm{Cr_2O_7^{2-}}} = C\times V = 0.10\ \mathrm{mol,L^{-1}}\times 0.020\ \mathrm{L} = 0.002\ \mathrm{mol}.\]

Therefore moles of \(\mathrm{Fe^{2+}}\) required:
\[n_{\mathrm{Fe^{2+}}} = 6\times 0.002 = 0.012\ \mathrm{mol}.\]

Given the ferrous ammonium sulfate solution is \(0.30\ \mathrm{M}\), the required volume (V) is
\[V=\frac{n}{C}=\frac{0.012\ \mathrm{mol}}{0.30\ \mathrm{mol,L^{-1}}}=0.040\ \mathrm{L}=40\ \mathrm{mL}.\]

\[\boxed{V=40\ \text{mL}}\]