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 #72911, 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

$$\mathbf{C}_a = \frac{\mathbf{C}_t \mathbf{V}_t \mathbf{M}}{\mathbf{V}_a}$$ $$\mathbf{MnO}_4^- + 8\mathbf{H}^+ + 5\mathbf{Fe}^{2+} \longrightarrow 5\mathbf{Fe}^{3+} + \mathbf{Mn}^{2+} + 4\mathbf{H}_2\mathbf{O}$$ $$M = 5$$

The concentration of $\mathrm{Fe}^{2+}$: $C_a = C_t V_t M / V_a = 0.01450 \, \mathrm{M} \cdot 14.43 \, \mathrm{mL} \cdot 5 / 25 \, \mathrm{mL} = 0.041847 \, \mathrm{M}$

n ($\mathrm{Fe}^{2+}$) = 0.041847 M · 0.250 L = 0.01046175 mol – in 250 mL of the solution

m ($\mathrm{Fe}^{2+}$) = n · M = 0.01046175 mol · 55.85 g/mol = 0.5843 g

Answer: 0.5843.

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