Question

A salt unknown containing ferrous ion was dissolved and diluted to 250.0 mL. A 25.00 mL aliquot of the ferrous ion solution was titrated with 0.01450 M potassium permanganate solution, and the mean of three acceptable, corrected titrations volumes was 14.43 mL. Calculate the mass of iron in the 250.0 mL solution.

Give your answer to 4 places after the decimal. Do not enter units.

Answer 1

Question #72909, Chemistry / Other / Completed

A salt unknown containing ferrous ion was dissolved and diluted to 250.0 mL. A 25.00 mL aliquot of the ferrous ion solution was titrated with 0.01450 M potassium permanganate solution, and the mean of three acceptable, corrected titrations volumes was 14.43 mL. Calculate the mass of iron in the 250.0 mL solution.

Give your answer to 4 places after the decimal. Do not enter units.

Solution

$$C_a = \frac{C_t V_t M}{V_a}$$

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.

$$\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5\mathrm{Fe^{2+}} \longrightarrow 5\mathrm{Fe^{3+}} + \mathrm{Mn^{2+}} + 4\mathrm{H_2O}$$

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 $\mathrm{Fe^{2+}}$:

$$\mathrm{C_a} = \mathrm{C_t V_t M} / \mathrm{V_a} = 0.01450 \mathrm{M} \cdot 14.43 \mathrm{mL} \cdot 5 / 25 \mathrm{mL} = 0.041847 \mathrm{M}$$ $$\mathrm{n} \left(\mathrm{Fe^{2+}}\right) = 0.041847 \mathrm{M} \cdot 0.250 \mathrm{L} = 0.01046175 \mathrm{mol} - \text{in } 250 \mathrm{mL} \text{ of the solution}$$ $$\mathrm{m} \left(\mathrm{Fe^{2+}}\right) = \mathrm{n} \cdot \mathrm{M} = 0.01046175 \mathrm{mol} \cdot 55.85 \mathrm{g/mol} = 0.5843 \mathrm{g}$$

Answer: 0.5843.

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