A quantity of 25.0 mL of a solution containing both Fe^2* and Fe^3+ ions is titr

Question

A quantity of 25.0 mL of a solution containing both Fe^2* and Fe^3+ ions is titrated with 23.0 mL of 0.0200 M KMnO_4 in dilute acidic solution. As a result, all of the Fe^2* ions are converted to Fe^3+ ions. Next, the solution is treated with zinc metal to convert all of the Fe^3+ ions to Fe^2+ ions. Finally, the solution containing only Fe^2+ ions requires 40.0 mL If the same KMnO_4 solution. Calculate the molar concentration of the Fe^2+ and Fe^3+ ions in the original solution. The net ionic equation is: MnO_4^- + 5Fe^2+ + 8H^+ rightarrow Mn^2+ + 5Fe^3+ + 4H_2O

Explanation / Answer

First of all calculate the concentration of Fe^2+ at the last step of the reaction.

M1V1 = M2V2

where M1 and V1 is the concentration and volume of iron solution. M2 and V2 is the concentration and volume of the KMnO4 solution used in the last step. At this stage the volume of the iron solution = 25 mL + 23mL = 48 mL. 23 mL is due to the first step of the process.

M1 = M2V2/V1 = 0.02M *40mL/(25 +23)mL = 0.017 M

This is the total concentration = 0.017 M

So, the total concentration of iron in the original solution = 0.017 *48/25 = 0.032 M

Now, consider the first step of the process where only Fe^2+ is oxidised.

Molarity of Fe^2+ = 23mL *0.02/25 = 0.0184 M

Molar concentration of Fe^2+ = 0.0184 M

Molar concentration of Fe^3+ = 0.032-0.0184 = 0.0136 M

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