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jacksonmeg
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in an experiment I dissolved a steel sample in nitric acid (turned yellow, Fe3+?), boiled it, added ammonium persulphate and boiled it again (remained yellow) then boiled it again after adding water, phorphoric( I think it means phosphoric?) acid and potassium iodate(VII). After adding the acid the solution turned colourless straight away and idk why, any help?
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charco
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(Original post by jacksonmeg)
in an experiment I dissolved a steel sample in nitric acid (turned yellow, Fe3+?), boiled it, added ammonium persulphate and boiled it again (remained yellow) then boiled it again after adding water, phorphoric( I think it means phosphoric?) acid and potassium iodate(VII). After adding the acid the solution turned colourless straight away and idk why, any help?
This sounds like the analysis of manganese in steel samples.

The ammonium persulphate in phosphoric acid oxidises the manganese dissolved by the nitric acid to manganate(VII). The phosphoric acid prevents precipitation of manganese dioxide and interference by tungsten in the steel.
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jacksonmeg
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(Original post by charco)
This sounds like the analysis of manganese in steel samples.

The ammonium persulphate in phosphoric acid oxidises the manganese dissolved by the nitric acid to manganate(VII). The phosphoric acid prevents precipitation of manganese dioxide and interference by tungsten in the steel.
yes it is the analysis of manganese in steel samples

i'm just confused cause we have to write reactions to explain the colour changes, but I have no explaination as to why it went yellow -> colourless when I added the phosphoric acid (this was before I added the iodate)
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charco
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(Original post by jacksonmeg)
yes it is the analysis of manganese in steel samples

i'm just confused cause we have to write reactions to explain the colour changes, but I have no explaination as to why it went yellow -> colourless when I added the phosphoric acid (this was before I added the manganese)
Phosphoric acid reacts with iron(III) forming a colourless complex.
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jacksonmeg
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(Original post by charco)
Phosphoric acid reacts with iron(III) forming a colourless complex.
ahhh ok thanks

would you mind checking my reaction?

step 1
3NO3- + 6H+ +Fe -> 3H2O +3NO2 + Fe3+
(colourless to yellow)

for the colourless ->purple, it is a reaction between the IO4- and Mn2+?
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jacksonmeg
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(Original post by charco)
Phosphoric acid reacts with iron(III) forming a colourless complex.
2Mn2+ + 5IO4- + 3H2O -> 2MnO4- + 5IO3- + 6H+ maybe?
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charco
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(Original post by jacksonmeg)
ahhh ok thanks

would you mind checking my reaction?

step 1
3NO3- + 6H+ +Fe -> 3H2O +3NO2 + Fe3+
(colourless to yellow)
In dilute solution nitrogen(II) oxide is preferentially formed (initially)

3Mn + 2NO3- + 8H+ --> 3Mn2+ + 2NO + 4H2O

(Original post by jacksonmeg)

for the colourless ->purple, it is a reaction between the IO4- and Mn2+?
Yes, it's oxidation of manganese(II) to manganate(VII)
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jacksonmeg
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(Original post by charco)
In dilute solution nitrogen(II) oxide is preferentially formed (initially)

3Mn + 2NO3- + 8H+ --> 3Mn2+ + 2NO + 4H2O



Yes, it's oxidation of manganese(II) to manganate(VII)
the reaction I put up was reaction given to us but I dont rlly see how its relevant :/
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charco
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(Original post by jacksonmeg)
the reaction I put up was reaction given to us but I dont rlly see how its relevant :/
If concentrated nitric acid is used then iron will react as shown.

It's only relevant in that you are dissolving all of the components of steel and then you are masking the iron absorbance (assuming that you are using colorimetry) by complexing with phosphoric acid.
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jacksonmeg
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(Original post by charco)
If concentrated nitric acid is used then iron will react as shown.

It's only relevant in that you are dissolving all of the components of steel and then you are masking the iron absorbance (assuming that you are using colorimetry) by complexing with phosphoric acid.
we used a spectrophotometer and then compared the absorbance at 525.0 nm which was the peak of the spectrum obtained in the first part of the experiment ( we made up solutions of diffierent concentrations of MnO4-)

to calculate % of Mn in the steel sample, should I just use the mass of KIO4 used, convert to moles and use mole ratios of the equation I made, then convert mols of Mn+ to mass, mass/sample mass?
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jacksonmeg
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i'm guessing we should be using the absorbances though...
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charco
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(Original post by jacksonmeg)
i'm guessing we should be using the absorbances though...
Clearly there is no point in using a colorimeter unless you do something with the data!!

You should have made a calibration curve using the standard solutions of manganate(VII). You simply read off the concentration from the absorbance of your sample.

As moles of manganate(VII) = moles of manganese in sample the rest is easy.
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jacksonmeg
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(Original post by charco)
Clearly there is no point in using a colorimeter unless you do something with the data!!

You should have made a calibration curve using the standard solutions of manganate(VII). You simply read off the concentration from the absorbance of your sample.

As moles of manganate(VII) = moles of manganese in sample the rest is easy.
I plotted the calibration curve and measured the absorbance of my sample and got the concentration from it, did some calculations and got over 100% manganese :/
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jacksonmeg
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(Original post by charco)
Clearly there is no point in using a colorimeter unless you do something with the data!!

You should have made a calibration curve using the standard solutions of manganate(VII). You simply read off the concentration from the absorbance of your sample.

As moles of manganate(VII) = moles of manganese in sample the rest is easy.
I got 0.315 absorbance which correlates to 6.45 mg dm^-3 of Mn
then I converted to mol dm^-3 by dividing by 54.938045 which is atomic mass of Mn, then used M=C*V, M=(6.45/54.938045)*(250/1000) = 0.029351244 moles, moles * atomic mass = mass, 0.029351244 * 54.938045 = 1.6125g, not sure what I did wrong
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charco
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(Original post by jacksonmeg)
I got 0.315 absorbance which correlates to 6.45 mg dm^-3 of Mn
then I converted to mol dm^-3 by dividing by 54.938045 which is atomic mass of Mn, then used M=C*V, M=(6.45/54.938045)*(250/1000) = 0.029351244 moles, moles * atomic mass = mass, 0.029351244 * 54.938045 = 1.6125g, not sure what I did wrong
6.45 mg dm-3 = 6.25 x 10-3 g = 1.174 x 10-4 mol dm-3

but you don't need to do all of that.

You should make your calibration curve using solutions of different concentration and then read off your sample value directly into concentration...

You have not given me any data to work with.

1. The mass of the steel sample
2. Any dilutions and solutions (volumes) made up after digestion with nitric acid.
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jacksonmeg
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(Original post by charco)
6.45 mg dm-3 = 6.25 x 10-3 g = 1.174 x 10-4 mol dm-3

but you don't need to do all of that.

You should make your calibration curve using solutions of different concentration and then read off your sample value directly into concentration...

You have not given me any data to work with.

1. The mass of the steel sample
2. Any dilutions and solutions (volumes) made up after digestion with nitric acid.
I figured it out :P I plotted the calibration curve and just drew a line across from the absorbence down to the concentration and worked backwards to find the mass and % mass
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