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a quantity of 25 ml of a solution containing both Fe2+ and Fe3+ ions is titrated with 23 ml of .02 M KMnO4 solution. as a result all the Fe2+ ions are oxidised to Fe3+ ions. Next the solution is treated with Zn metal to convert all Fe3+ ions to Fe2+ ions. Finally, the solution containing only the Fe2+ ions requires 40ml of the same KMnO4 solution for oxidation to Fe3+. Calculate the molarity of Fe2+ and Fe3+ in the original solution.

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    Write the equation for Fe^2+ and MnO4^-
    The pertinent information is this.
    5Fe^2+ + MnO4^- ==> Mn^2+ + 5Fe^3+

    From the first titration you have
    mols MnO4^- = M x L = ?
    Convert mols MnO4^- to mols Fe^2+ using the coefficients in the balanced equation. That's mols Fe^2+ = 5x mols MnO4^-
    M Fe^2+ = mols Fe^2+/0.025L = ?

    Do the second part as above. This time you have total mols Fe^2+ + Fe^3+. Subtract from this number mols Fe^2+ from the first titration. This give you Fe^3+, then M Fe^3+ = mols/0.025L = ?

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