A 50.00 -mL sample of solution containing \(\mathrm{Fe}^{2+}\) ions is titrated with a 0.0216 \(\mathrm{M} \mathrm{KMnO}_{4}\) solution. It required 20.62 \(\mathrm{mL}\) of \(\mathrm{KMnO}_{4}\) solution to oxidize all the \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\) ions by the reaction $$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \stackrel{\text { Acidic }}{\longrightarrow} \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q) \text{(Unbalanced)} $$ a. What was the concentration of \(\mathrm{Fe}^{2+}\) ions in the sample solution? b. What volume of 0.0150\(M \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution would it take to do the same titration? The reaction is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Fe}^{2+}(a q) \stackrel{\mathrm{Acidic}}{\longrightarrow} \mathrm{Cr}^{3+}(a q) +\mathrm{Fe}^{3+}(a q) \text {(Unbalanced)} $$

First, we need to balance both the given reactions before we proceed further. For the first reaction, \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)\) Balanced reaction: \(5\,\mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_4^-(a q)+8\,\mathrm{H}^+(a q)\longrightarrow\mathrm{Mn}^{2+}(a q)+5\,\mathrm{Fe}^{3+}(a q)+4\,\mathrm{H}_2\mathrm{O}(l)\) For the second reaction, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Fe}^{3+}(a q)\) Balanced reaction: \(6\,\mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_2\mathrm{O}_7^{2-}(a q)+14\,\mathrm{H}^+(a q)\longrightarrow 2\,\mathrm{Cr}^{3+}(a q)+6\,\mathrm{Fe}^{3+}(a q)+7\,\mathrm{H}_2\mathrm{O}(l)\) Now that both reactions are balanced, we can proceed with solving the exercise.

We're given that a 50.00 -mL sample of solution containing \(\mathrm{Fe}^{2+}\) ions was titrated with a 0.0216 \(\mathrm{M} \mathrm{KMnO}_{4}\) solution, requiring 20.62 \(\mathrm{mL}\) of the \(\mathrm{KMnO}_{4}\) solution. First, we will find the moles of \(\mathrm{KMnO}_{4}\) used in the titration using its molarity and volume. Moles of \(\mathrm{KMnO}_{4}\) = Molarity \(\times\) Volume in Liters Moles of \(\mathrm{KMnO}_{4}\) = \(0.0216\,\mathrm{M}\times(20.62\,\mathrm{mL}\times(1\,\mathrm{L}/1000\,\mathrm{mL}))=4.45472\times10^{-4}\,\mathrm{mol}\) Now let's use stoichiometry to find the moles of \(\mathrm{Fe}^{2+}\). From the balanced equation, we have 1 mol of \(\mathrm{MnO}_4^-\) reacting with 5 mol of \(\mathrm{Fe}^{2+}\). Moles of \(\mathrm{Fe}^{2+}\) = Moles of \(\mathrm{KMnO}_{4}\) × \(\frac{5\,\mathrm{mol}\,\mathrm{Fe}^{2+}}{1\,\mathrm{mol}\,\mathrm{KMnO}_4}\) Moles of \(\mathrm{Fe}^{2+}\) = \(4.45472\times10^{-4}\,\mathrm{mol}\times5=2.22736\times10^{-3}\,\mathrm{mol}\) Now let's calculate the concentration of \(\mathrm{Fe}^{2+}\) ions in the solution by dividing the moles found by the volume of the sample in Liters. Concentration of \(\mathrm{Fe}^{2+}\) ions = \(\frac{\text{Moles of }\,\mathrm{Fe}^{2+}}{\text{Volume of the Sample in Liters}}\)= \(\frac{2.22736\times10^{-3}\,\mathrm{mol}}{(50\,\mathrm{mL}\times(1\,\mathrm{L}/1000\,\mathrm{mL})}\) Concentration of \(\mathrm{Fe}^{2+}\) ions = 0.04454 \(\mathrm{M}\) a. The concentration of \(\mathrm{Fe}^{2+}\) ions in the sample solution is 0.04454 \(\mathrm{M}\).

We are given a solution with a molarity of 0.0150\(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) and asked to find the volume required for the same titration. From the balanced reaction of the second equation, we have 1 mol of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) reacting with 6 mol of \(\mathrm{Fe}^{2+}\). Using stoichiometry, we can find the moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) required for the titration. Moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) = Moles of \(\mathrm{Fe}^{2+}\) × \(\frac{1\,\mathrm{mol}\,\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7}{6\,\mathrm{mol}\,\mathrm{Fe}^{2+}}\) Moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) = \(2.22736\times10^{-3}\,\mathrm{mol}\times\frac{1}{6}=3.71226\times10^{-4}\,\mathrm{mol}\) Now let's calculate the volume of 0.0150 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution required for the titration. Volume of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution = \(\frac{\text{Moles of }\,\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\text{Molarity of }\,\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\,\text{solution}}\)= \(\frac{3.71226\times10^{-4}\,\mathrm{mol}}{0.015\,\text{M}}\) Volume of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution = 0.02475 \(\mathrm{L}\) (or 24.75\(\,\mathrm{mL}\)) b. The volume of 0.0150\(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution required for the same titration is 24.75\(\,\mathrm{mL}\).