[leena]

A sample of alumina (Al₂O₃) contains iron as an impurity.Explain how you would determine the mass percentage of Al in the sample.You are supplied with aq. NaOH and HCl acid only,as reagents.


Since Al2O3 is an amphoteric oxide ,I thought of adding HCl first to the above mixture,to get the soluble chlorides of Al and Fe.
Al2O3 + 6 HCl  2AlCl3 + 3H20
As for Fe I'm not sure if we get a ferrous or ferric chloride or even a complex,
Fe + 2HCl  FeCl2 + H2
2Fe +6HCl  2FeCl3 + 3H2,
but I think all 3 are soluble.
Next,I thought of adding NaOH, but here too I'm a bit unsure if I should add excess NaOH or little NaOH,
With excess NaOH,Fe²⁺ and Fe³⁺, would precipitate as their hydroxides.
Al³⁺ would form soluble NaAlO2, but I have no idea how to get the mass percentage of Al from a solution of NaAlO2?

But , if I add little NaOH, I might get a precipitate of Al(OH)3,if it was a reversible reaction,but I don't know this for sure and soluble Fe(2+ or 3+?) hydroxide from ironchloride,I think.
But sooo... many assumptions,I'm  so confused...

[plankk]

At first, iron will undergo to Fe²⁺ (to form Fe³⁺ is needed the oxidant).
Secondly, in my opinion you can (or even must) use a scales.

[Borek]

What if you start with dissolution in NaOH?

[leena]

Hi Borek,
I guess,with excess NaOH,we would get NaAlO2 and a precipitate of ferrous hydroxide ,right?

[Borek]

Is Fe soluble in NaOH solution?

[leena]

Hmm..now that you mention it,I'm not so sure,
I just checked my notes I have reactions for Fe2+ and Fe3+,with NaOH,but not for Fe.
Is it soluble?

[leena]

Ok,so I don't think Fe would even react with NaOH,cause I don't think NaOH would oxidise it.
So then we have sodium aluminate solution,Fe  solid and NaOH remaining.
Now do I add HNO3? well,that's the only reagent left so...
HNO3 would neutralise the excess NaOH.But what happens to the Fe and NaAlO2?
Oh... I really don't know.Now I'm even more confused

[Borek]

Filtration perhaps?

[leena]

Ok,so I don't think Fe would even react with NaOH,cause I don't think NaOH would oxidise it.
So then we have sodium aluminate solution,Fe  solid and NaOH remaining.
Now do I add HNO3? well,that's the only reagent left so...
HNO3 would neutralise the excess NaOH.But what happens to the Fe and NaAlO2?
Oh... I really don't know.Now I'm even more confused

I meant HCl.For some reason I can't seem to edit my post.The modifying button's missing!

Filtration perhaps?

So when I filter,I get sodium aluminate and NaOH in the filtrate.Then I add HCl to this filtrate.The excess NaOH would get neutralised and hopefully the amount of HCl used won't affect the NaAlO2 present.And then I can weigh the solution of NaAlO2 and find the mass of Al present.
Does this sound okay?
Thanks

[Borek]

Another hint: back titration.

[leena]

okay,so I backtitrate 25cm³ of the filtrate with HCl in the burette.Then at the endpoint,I can find the volume of unused NaOH,then... I can find the volume of (the known concentration of) NaOH that reacted with Al2O3.
Then with a little bit of stoichiometry,I can find the number of moles of Al2O3 and therefore the mass of Al.
Now does it sound ok?
Pleeease say yes

[leena]

But ,wait.To find the endpoint we have to use an indicator.
So doesn't an indicator also count as a reagent?

[Stephen]

indicator is neutral reagent....so it's not necessary to write it as reagent.

[Borek]

I can't see any way of determining amounts without either indicator or balance.

So far so good. You have to deal with Fe now.

[leena]

Can we just weigh the sample of Al₂O₃ and Fe at first ,without adding any reagents,using a chemical balance?
So then we would have the mass percent of Al=(mass of Al/total mass of initial sample)*100 percent,
or do we have to find the mass of Fe seperately?by drying and weighing the residue of Fe,then,by adding both the masses of Al2O3 and Fe,we can find the mass percent