Both \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\) can be used to titrate \(\mathrm{Fe}^{2+}(\mathrm{aq})\) to \(\mathrm{Fe}^{3+}(\mathrm{aq}) .\) Suppose you have available as titrants two solutions: \(0.1000 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and \(0.1000 \mathrm{M} \mathrm{MnO}_{4}^{-}(\mathrm{aq})\). (a) For which solution would the greater volume of titrant be required for the titration of a particular sample of \(\mathrm{Fe}^{2+}(\text { aq }) ?\) Explain. (b) How many \(\mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{MnO}_{4}^{-}(\) aq) would be required for a titration if the same titration requires \(24.50 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) ?\)

Step by step solution

01

Analyze the Stoichiometric Relationship

The first step is to write down the reduction half equations for \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} and \(\mathrm{MnO}_{4}^{-}\), and the oxidation half equation for \(\mathrm{Fe}^{2+}\). Then combine the appropriate reduction and oxidation half equations to obtain the full redox equations. For \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), the full redox equation is \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{Fe}^{2+} + 14H^{+} \rightarrow 2Cr^{3+} + 6\mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}\), and for \(\mathrm{MnO}_{4}^{-}\), the equation is \( \mathrm{MnO}_{4}^{-} + 5\mathrm{Fe}^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5\mathrm{Fe}^{3+} + 4 \mathrm{H}_{2} \mathrm{O}\).

02

Determine the Volume of Titrant

We will now explore the stoichiometric ratio between the titrant and the sample, which is \(\mathrm{Fe}^{2+}\) ions. For part (a), comparing the stoichiometric ratios, one can see that the same amount of \(\mathrm{Fe}^{2+}\) ions requires more moles of \(\mathrm{MnO}_{4}^{-}\) than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) since 5 moles of \(\mathrm{Fe}^{2+}\) ions react with 1 mole \(\mathrm{MnO}_{4}^{-}\), while 6 moles of \(\mathrm{Fe}^{2+}\) ions react with 1 mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\). Therefore, the solution of \(\mathrm{MnO}_{4}^{-}\) would require greater volume for the titration.

03

Convert Volume of one Titrant to Another

For part (b), since the molarity of the two solutions is the same (0.1000 M), the required volume of \(\mathrm{MnO}_{4}^{-}\) will be proportional to the stoichiometric ratio of the two species. The stoichiometric ratio of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to \(\mathrm{MnO}_{4}^{-}\) is 6:5 (from step 1). Therefore, the required volume of \(\mathrm{MnO}_{4}^{-}\) = \(24.50 mL* \frac{5}{6} = 20.42 mL\).

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