Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Please Help Me this is Tough

  1. Oct 8, 2006 #1
    How do I find x in FeC2O4.xH2O. I am told that this is a hydrated salt of 1.75g and was dissolved in acid and made up to 250ml. A 25mL sample of this solution required 29.15mL of standardised 0.0200mol/L KMnO4 solution for complete oxidation.

    I worked out n(MnO4-) = cv = 5.83 x 10^-4 moles
    and n(C2O4^2-) = 1.4575 x 10^-3 moles
    and c(C2O4^2-) = n/v = 0.0583 molL-1
    I just don't know how to find x
    Last edited: Oct 8, 2006
  2. jcsd
  3. Oct 8, 2006 #2
    Well your titration gives you the quantity of Fe Ions in solution (the dissociated Iron). You can take that concentration and find the total mass in the solution, and subtracting that mass from the Hydrous compund mass you can find the mass of H2O present in the hydrated salt. Then you need to divide that mass by the molecular mass of water to find your x.

    Hope thats clear....
    Last edited: Oct 8, 2006
  4. Oct 8, 2006 #3
    Thanks for helping
    I still don't really understand. I am not very good with chemistry
  5. Oct 8, 2006 #4
    Ok. You have done all the hard work, you just need to make the last step. You take a fixed mass of the salt (Hydrated) and disolve it in the acid solution.

    [tex]V\ =\ 250ml \ =\ 0.25dm^{3}[/tex]

    [tex]m\ =\ 1.75g\ = \ 1+\frac{3}{4}=\frac{7}{4}g[/tex]

    [tex]0.25dm^{3}[/tex] is [tex]\frac{1}{4}dm^{3}[/tex]

    Thefore your concentration is simply:

    [tex][FeC_{2}O_{4}.xH_{2}O]\ = \ 7g\ dm^{3}[/tex]

    You take [tex]25ml\ = 25\times10^{-3}\ dm^{3} = 0.025dm^{3}[/tex]

    And titrate it with Potasium Permanganate [tex]KMnO_{4}[/tex], For which the half-reaction is:

    [tex]MnO_{4}^{-}_{(aq)}\ + \ 8H^{+}_{(aq)} \ + \ 5e^{-} \ \rightarrow \ Mn^{2+}_{(aq)} \ + \ 4H_{2}O_{(l)}[/tex]

    The indicator changes color as it goes from Mn(VII) to Mn(II) and Fe is oxidised from Fe(II) to Fe(II).

    [tex]Fe^{3+}_{(aq)}\ \rightarrow \ Fe^{3+}_{(aq)}\ +\ e^{-} [/tex]

    If you balence the half reactions you get a 5:1 ratio of [tex]KMnO_{4}:FeC_{2}O_{4}[/tex]

    This means the quantiy of [tex][Fe^{2+}_{(aq)}][/tex] is 5 times the amount of titrant used.
    Last edited: Oct 8, 2006
  6. Oct 8, 2006 #5
    Blah, see the other post on titration......this post was taking forever to latex...
  7. Oct 8, 2006 #6
    Thanks heaps!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook