Calculating PPM Iron: Iron Titration

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SUMMARY

This discussion focuses on calculating the parts per million (ppm) of iron in a water sample using redox titration techniques. A 100.0 mL water sample was treated with 25 mL of 0.002107 M potassium dichromate, converting iron to ferric ion and dichromate to Cr3+. The excess dichromate was back titrated with 7.47 mL of 0.00979 M Fe2+. The final calculations involve determining the millimoles of potassium dichromate and applying stoichiometry to find the ppm of iron in the original solution.

PREREQUISITES
  • Understanding of redox chemistry and electron balancing
  • Knowledge of titration techniques and calculations
  • Familiarity with molarity and ppm calculations
  • Basic chemical reaction equations involving Fe2+ and Cr6+
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  • Study the stoichiometry of redox reactions in detail
  • Explore the concept of ppm calculations in various contexts
  • Investigate the properties and reactions of potassium dichromate
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Chemistry students, laboratory technicians, and professionals involved in water quality analysis or environmental chemistry will benefit from this discussion.

tipton12
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Please Help! I have no idea where to start with this one!

A 100.0mL sample of water was treated to convert any iron present to ferrous ion. Addition of 25mL of 0.002107M potassium dichromate converted the iron to ferric ion and the dichromate to Cr3+. Excess dichromate was then back titrated with 7.47 mL of standard 0.00979M Fe2+. Calculate the ppm iron in the original solution.
 
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Hello, it is easy one, so I don't find it useful to solve it for you.

You know that ppm=\frac {mg}{L} for liquids, and ppm=\frac {mg}{kg} for solids. This means that you'll express your result in milligrams in one liter of solution.

Find how many millimoles are there in 25 mL of 0,002107 M K_2Cr_2O_7, but note that you'll have to use redox chemistry to balance the electrons. I'll help as well as the other members if you stuck here. Subtract the back titration millimoles from this, and the remaining is yours.
 
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ok, i haven't done redox in 4 years and i remember how to do the balancing, but i am having a lot of trouble setting up the reactant side of the equation. am i wrong to include H2O in the reactant side? I know this is basic, but I still am having trouble.
 
Well, since redox chemistry is a bit tough, I'll help you.

Fe \rightarrow Fe^{3+}~+~3e^-
Cr^{6+}~+~3e^-\rightarrow Cr^{3+}
Then you find this, by remembering that dichromate is actually Cr_2O_7^{2-}:
2~Fe~+~(Cr^{6+})_2\rightarrow 2~Fe^{3+}~+~2~Cr^{3+}

The medium is aqueous, and potassium dichromate is a basic salt, so it is very likely that iron(III) hydroxide precipitates. You are right to include water in the reactant side:

Fe^{3+}~+~3H_2O \rightarrow Fe(OH)_3\downarrow~+~3H^+

So, six moles of water per one mole of dichromate is reacted.

About the back-titration redox scheme, I'd write like this:
6~Fe^{2+} \rightarrow 6~Fe^{3+}~+~6~e^-
(Cr^{6+})_2~+~6~e^- \rightarrow 2Cr^{3+}

As a conclusion, the back-titration redox reaction should be like that:
6~Fe^{2+}~+~(Cr^{6+})_2 \rightarrow 6~Fe^{3+}~+~2Cr^{3+}

Note that this approach is not very logical as only trace amounts of elemental iron may be present in water; so you'll convert the redox equations for just Fe2+, as the present Fe3+ in water will not be affected so not detected at all.
 
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I have said in my last thread that the present Fe3+ will not be affected and detected at all, but this is wrong, I'm afraid. It is apparent that almost all of the iron present will be converted to Fe3+ and precipitated in a somewhat basic medium. Please review all the reactions, and give some feedback.
 
Sorry, I didn't get your last messege in time b/c I didn't check it again before class. I went ahead and got the final balanced equation, then I converted the dichromate to mmoles as well as the Fe2+, then I used the mmoles of the dichromate to due a molar ratio, which I think I got 6:1, then I took that answer and subtracted from the back titration. I think I got it right but it hasn't been returned to me yet, but most of the other students got the same answer. Thanks for your help!
 

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