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Back titrations?

  1. Jan 23, 2007 #1
    1. The problem statement, all variables and given/known data
    Stibnite Sb2S3, is the major ore of the element antimony. A 6.143g sample of the ore was chemically treated to dissolve all the Sb3+ ion in solution. These were then oxidised to antimony(v) by adding 25ml of 0.2017M acidified KMnO4. The excess KMnO4 was titrated against 0.0981M, freshly prepared, acidified FeSO4 solution. This required 23.44ml.

    Calculate the percentage by mass of antimony in the ore sample.

    2. Relevant equations
    Basic chem equations

    3. The attempt at a solution
    I may have trouble understanding the problem. Are they assuming the 6.143g sample of the ore is impure? I.e. it may contain other ores or metals? If not than the percentage by mass of antimony is easily calculated as 121.76g*2/(121.76g*2+3*32.07g) = 0.7168 or 72% by mass of antimony in the pure ore sample. However, that is not the solution of 31.7%

    So from this does it means the ore sample is impure and is actually a mixture.

    If we use the information about the titration. Sb was isolated into its ion in the sample and underwent the half reaction for each ore compound in the sample (2Sb3+ in each compound),
    2Sb3+ -> 2Sb5+ + 4e-

    The KMnO4 underwent the half equation
    MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

    To balance the two half reactions we obtain
    5(2Sb3+ -> 2Sb5+ + 4e-)
    4(MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O)

    So net ionic equation is 10Sb3+ + 4MnO4- ………………… only these two reactants are important.

    To workout how much MnO4- has reduced, we use the titration information

    Some KMnO4 hasn’t reacted. These that hasn’t were reacted with FeSO4. The equation is
    2KMnO4 + FeSO4 -> Fe2+ + 2MnO4- + K2SO4 Is this correct?

    0.02344L*0.0981moles/L = 0.0023 moles of FeSO4 was required so 0.0046 moles of KMnO4 still remained.

    Originally 0.025L*0.2017moles/L = 0.005 moles of KMnO4 were added so only 0.00044moles reacted with antinomy.

    Back to 10Sb3+ + 4MnO4- …………………
    For each mole of MnO4- consumed, 5/2 moles of Sb3+ reacted. So 5/2 * 0.00044moles = 0.0011 moles of Sb3+ was in the sample.

    0.0011moles*121.75g/mole = 0.135g of antimony in the ore sample. This is only 0.135g/6.143g *100 = 2.19%. Far from the claimed answer of 31.7%
  2. jcsd
  3. Jan 25, 2007 #2


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    Research to find the appropriate and balanced equations,

    5 Fe 2+ + MnO4- + 8H+ -------> 5Fe 3+ + Mn 2+ + 4H2O

    show some analysis from here.......
    Last edited: Jan 25, 2007
  4. Jan 27, 2007 #3
    So it shouldn't be 2KMnO4 + FeSO4 -> Fe2+ + 2MnO4- + K2SO4
    but should be 5 Fe 2+ + MnO4- + 8H+ -------> 5Fe 3+ + Mn 2+ + 4H2O

    What was wrong with the reaction I suggested? Is it energetically unfavourable? Sulfate ion is a weak base so wouldn't 'want' to combine with K. Altghough I assume K2SO4 is in aqeous form so should dissociate anyway.

    The reaction you suggested was energetically favourable. And when carried out imply 4.6*10^-4 moles of MnO4- is left for titration. Hence .005-.00046=.00458 moles of MnO4- originally reacted with Sb3+. The reaction is 2MnO4- + 5Sb3+ -> .....
    so 5/2 * .00458 = .01146 moles of Sb3+ was in the original sample which turns out to give a weight of 1.3953g hence a percentage of 1.3953/6.143=22.6% which is still less than the answer of 31.7%
  5. Jan 28, 2007 #4


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    Staff: Mentor

    Perhaps question is badly worded because if you calculate Sb2S3 percentage you get 31.5%...

  6. Jan 28, 2007 #5


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    Show all of your analysis if you want us to correct your steps.
  7. Jan 28, 2007 #6
    You're saying calculate the percentage of Sb2S3 in the sample. So .01146 moles of Sb => .001146/2 moles of Sb2S3 hence .00575(2*121.75 + 32.06*3)=1.946g
    1.946g/6.143=.317 or 31.7% as suggested in the book. It is likely that this is what they meant.
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