Question

A 3.7122 g sample of an unknown salt containing ferrous ion was dissolved and diluted to 250.0 mL. Repeat 25.00 mL samples of the total solution were titrated with 0.01860 M KMnO4 solution, and the mean of three accepted corrected titration volumes was 19.00 mL. Calculate the %w/w of iron in the original sample mass.

State your answer to 2 places after the decimal place. Do not enter units.

Answer 1

Question #72910, Chemistry / Other / Completed

A 3.7122 g sample of an unknown salt containing ferrous ion was dissolved and diluted to 250.0 mL. Repeat 25.00 mL samples of the total solution were titrated with 0.01860 M KMnO4 solution, and the mean of three accepted corrected titration volumes was 19.00 mL. Calculate the %w/w of iron in the original sample mass.

State your answer to 2 places after the decimal place. Do not enter units.

Solution

$$\mathrm{C}_a = \frac{\mathrm{C}_t \mathrm{V}_t \mathrm{M}}{\mathrm{V}_a}$$

where Ca is the concentration of the analyte, typically in molarity; Ct is the concentration of the titrant, typically in molarity; Vt is the volume of the titrant used, typically in liters; M is the mole ratio of the analyte and reactant from the balanced chemical equation; and Va is the volume of the analyte used, typically in liters.

$$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{Fe}^{2+} \longrightarrow 5\mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$$

The quantity of $\mathrm{Fe}^{2+}$ ions are 5 times more than that of $\mathrm{KMnO}_4$ by the reaction. So M = 5

The concentration of $\mathrm{Fe}^{2+}$:

$$\mathrm{C}_a = \mathrm{C}_t \mathrm{V}_t \mathrm{M} / \mathrm{V}_a = 0.01860 \, \mathrm{M} \cdot 19.00 \, \mathrm{mL} \cdot 5 / 25 \, \mathrm{mL} = 0.07068 \, \mathrm{M}$$ $$n \, (\mathrm{Fe}^{2+}) = 0.07068 \, \mathrm{M} \cdot 0.250 \, \mathrm{L} = 0.01767 \, \mathrm{mol} - \text{in} \, 250 \, \mathrm{mL} \text{ of the solution}$$ $$m \, (\mathrm{Fe}^{2+}) = n \cdot M = 0.01767 \, \mathrm{mol} \cdot 55.85 \, \mathrm{g/mol} = 0.9868695 \, \mathrm{g}$$ $$\% \mathrm{w/w} \, (\mathrm{Fe}^{2+}) = 0.9868695 \, \mathrm{g} \cdot 100\% / 3.7122 \, \mathrm{g} = 26.58\%$$

Answer: 26.58.

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