# Titration help: KMnO4, H2SO4, FeSO4.xH20

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The point of this experiment is to determine the value of x in FeSO4.xH20 by titration.

I started with 3.06g of FeSO4.xH20 then dissolved this in 50cm3 of 1moldm-3 H2SO4. I then made this up to 250cm3 with distilled water.

I then got 25cm3 of that solution, and added about 20cm3 of 1 mol dm-3 H2SO4 in a conical flask and titrated that against 0.01 moldm-3 Potassium Permanganate KMnO4.

Now, I dont understand why more H2SO4 is added after the original 50. I also dont know what the equation for the reaction actually is, so if someone could help me understand the reacion and titration a bit better, I would be grateful.

Thanks for the help.

I started with 3.06g of FeSO4.xH20 then dissolved this in 50cm3 of 1moldm-3 H2SO4. I then made this up to 250cm3 with distilled water.

I then got 25cm3 of that solution, and added about 20cm3 of 1 mol dm-3 H2SO4 in a conical flask and titrated that against 0.01 moldm-3 Potassium Permanganate KMnO4.

Now, I dont understand why more H2SO4 is added after the original 50. I also dont know what the equation for the reaction actually is, so if someone could help me understand the reacion and titration a bit better, I would be grateful.

Thanks for the help.

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#2

The equation should be

10FeSO4 + 2KMnO4 + 8H2SO4 -> 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

You need the number of moles of FeSO4xnH2O, which is equal to the number of moles of FeSO4. You need to determine the volume of KMnO4 needed to react with the solution (the equivalence point is reached when the purple colour won't disappear anymore, as there are no Fe2+ ions left to oxidize). Now you can calculate the number of moles of KMnO4 and then using the equation you can find out the amount of FeSO4 in the solution. As you have diluted the original solution you have to do some additional calculations to determine the amount of Fe2+ ions you started with.

The main formula to use is concentration (mol/dm3) = number of moles (mol) / volume (dm3)

Sorry if this was a bit confusing, I hope this helps a bit.

H2SO4 is needed because acidic environment is needed for the reaction to take place.

Anyway, you can find lots of practical tasks and examples regarding redox titrations using Google and chemistry textbooks should also contain the basics.

10FeSO4 + 2KMnO4 + 8H2SO4 -> 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

You need the number of moles of FeSO4xnH2O, which is equal to the number of moles of FeSO4. You need to determine the volume of KMnO4 needed to react with the solution (the equivalence point is reached when the purple colour won't disappear anymore, as there are no Fe2+ ions left to oxidize). Now you can calculate the number of moles of KMnO4 and then using the equation you can find out the amount of FeSO4 in the solution. As you have diluted the original solution you have to do some additional calculations to determine the amount of Fe2+ ions you started with.

The main formula to use is concentration (mol/dm3) = number of moles (mol) / volume (dm3)

Sorry if this was a bit confusing, I hope this helps a bit.

H2SO4 is needed because acidic environment is needed for the reaction to take place.

Anyway, you can find lots of practical tasks and examples regarding redox titrations using Google and chemistry textbooks should also contain the basics.

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(Original post by

The equation should be

10FeSO4 + 2KMnO4 + 8H2SO4 -> 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

You need the number of moles of FeSO4xnH2O, which is equal to the number of moles of FeSO4. You need to determine the volume of KMnO4 needed to react with the solution (the equivalence point is reached when the purple colour won't disappear anymore, as there are no Fe2+ ions left to oxidize). Now you can calculate the number of moles of KMnO4 and then using the equation you can find out the amount of FeSO4 in the solution. As you have diluted the original solution you have to do some additional calculations to determine the amount of Fe2+ ions you started with.

The main formula to use is concentration (mol/dm3) = number of moles (mol) / volume (dm3)

Sorry if this was a bit confusing, I hope this helps a bit.

H2SO4 is needed because acidic environment is needed for the reaction to take place.

Anyway, you can find lots of practical tasks and examples regarding redox titrations using Google and chemistry textbooks should also contain the basics.

**rag-doll**)The equation should be

10FeSO4 + 2KMnO4 + 8H2SO4 -> 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

You need the number of moles of FeSO4xnH2O, which is equal to the number of moles of FeSO4. You need to determine the volume of KMnO4 needed to react with the solution (the equivalence point is reached when the purple colour won't disappear anymore, as there are no Fe2+ ions left to oxidize). Now you can calculate the number of moles of KMnO4 and then using the equation you can find out the amount of FeSO4 in the solution. As you have diluted the original solution you have to do some additional calculations to determine the amount of Fe2+ ions you started with.

The main formula to use is concentration (mol/dm3) = number of moles (mol) / volume (dm3)

Sorry if this was a bit confusing, I hope this helps a bit.

H2SO4 is needed because acidic environment is needed for the reaction to take place.

Anyway, you can find lots of practical tasks and examples regarding redox titrations using Google and chemistry textbooks should also contain the basics.

But Im still struggling to understand the whole equation. Why do I need to react the FeSO4.xH2O solution with H2SO4? And what does that reaction cause?

Like you say, I can understand that the additional H2SO4 is necessary because the reaction takes place in acidic conditions. But what does the inital reaction between FeSO4 and H2SO4 produce?

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#4

(Original post by

Thanks, that has helped.

But Im still struggling to understand the whole equation. Why do I need to react the FeSO4.xH2O solution with H2SO4? And what does that reaction cause?

Like you say, I can understand that the additional H2SO4 is necessary because the reaction takes place in acidic conditions. But what does the inital reaction between FeSO4 and H2SO4 produce?

**Dr.Hox**)Thanks, that has helped.

But Im still struggling to understand the whole equation. Why do I need to react the FeSO4.xH2O solution with H2SO4? And what does that reaction cause?

Like you say, I can understand that the additional H2SO4 is necessary because the reaction takes place in acidic conditions. But what does the inital reaction between FeSO4 and H2SO4 produce?

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#5

The H

^{+}ions are needed for the redox reaction of Fe^{2+}with the oxidising agent.
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(Original post by

--Useful stuff--

**rag-doll**)--Useful stuff--

After the titration, I got two concordant titres of 19.35 and 19.40 cm3 of KMnO4 which took the titration to its end point.

So with a titre of (19.35+19.40) / 2 = 19.375, the moles of MnO4 is 1.9375X10-4.

So no. of moles of Fe2+ is (1.9375X10-4) X 5 = 9.6875X10-4

But thats the no. of moles of 25cm of FeSO4.xH20, but I made 250cm3. So total no. of moles of FeSO4.xH20 is 9.6875X10-3.

From here though, I dont know how to work out what x is.

Help please?

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#7

(Original post by

Hi, just need your help again.

After the titration, I got two concordant titres of 19.35 and 19.40 cm3 of KMnO4 which took the titration to its end point.

So with a titre of (19.35+19.40) / 2 = 19.375, the moles of MnO4 is 1.9375X10-4.

So no. of moles of Fe2+ is (1.9375X10-4) X 5 = 9.6875X10-4

But thats the no. of moles of 25cm of FeSO4.xH20, but I made 250cm3. So total no. of moles of FeSO4.xH20 is 9.6875X10-3.

From here though, I dont know how to work out what x is.

Help please?

**Dr.Hox**)Hi, just need your help again.

After the titration, I got two concordant titres of 19.35 and 19.40 cm3 of KMnO4 which took the titration to its end point.

So with a titre of (19.35+19.40) / 2 = 19.375, the moles of MnO4 is 1.9375X10-4.

So no. of moles of Fe2+ is (1.9375X10-4) X 5 = 9.6875X10-4

But thats the no. of moles of 25cm of FeSO4.xH20, but I made 250cm3. So total no. of moles of FeSO4.xH20 is 9.6875X10-3.

From here though, I dont know how to work out what x is.

Help please?

You know both the mass of the substance and the no. of moles. Now you can calculate its molar mass. M[FeSO4*nH2O]=m/n

Now calculate M[FeSO4]

M[FeSO4] + n*M[H2O] = M[FeSO4*nH2O]

From there you should be able to find n.

I hope I didn't mess up anything, as I didn't solve it on paper beforehand

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(Original post by

The total no. of moles of FeSO4 equals that of FeSO4*nH2O.

You know both the mass of the substance and the no. of moles. Now you can calculate its molar mass. M[FeSO4*nH2O]=m/n

Now calculate M[FeSO4]

M[FeSO4] + n*M[H2O] = M[FeSO4*nH2O]

From there you should be able to find n.

I hope I didn't mess up anything, as I didn't solve it on paper beforehand

**rag-doll**)The total no. of moles of FeSO4 equals that of FeSO4*nH2O.

You know both the mass of the substance and the no. of moles. Now you can calculate its molar mass. M[FeSO4*nH2O]=m/n

Now calculate M[FeSO4]

M[FeSO4] + n*M[H2O] = M[FeSO4*nH2O]

From there you should be able to find n.

I hope I didn't mess up anything, as I didn't solve it on paper beforehand

Mass of FeSO4.xH20

------------------------- = Mr of FeSO4.xH20

No. of moles of FeSO4.xH2O

So it was 3.06/(9.6875X10-3) = 315.871

Mr. of FeSO4 = 151.9

Mr of xH20 = 315.871 - 151.9 = 163.971

163.971/Mr. of H2O = Value of x

Which was then 163.971/18 = 9.11

Is this right?

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#9

(Original post by

Hi, I didnt quite understand all of that, but this is what I did from that, is it what you meant?

Mass of FeSO4.xH20

------------------------- = Mr of FeSO4.xH20

No. of moles of FeSO4.xH2O

So it was 3.06/(9.6875X10-3) = 315.871

Mr. of FeSO4 = 151.9

Mr of xH20 = 315.871 - 151.9 = 163.971

163.971/Mr. of H2O = Value of x

Which was then 163.971/18 = 9.11

Is this right?

**Dr.Hox**)Hi, I didnt quite understand all of that, but this is what I did from that, is it what you meant?

Mass of FeSO4.xH20

------------------------- = Mr of FeSO4.xH20

No. of moles of FeSO4.xH2O

So it was 3.06/(9.6875X10-3) = 315.871

Mr. of FeSO4 = 151.9

Mr of xH20 = 315.871 - 151.9 = 163.971

163.971/Mr. of H2O = Value of x

Which was then 163.971/18 = 9.11

Is this right?

FeSO4*9H2O is a bit weird, though, as it is usually FeSO4*7H2O. But I guess it isn't of that much importance.

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(Original post by

Seems to be.

FeSO4*9H2O is a bit weird, though, as it is usually FeSO4*7H2O. But I guess it isn't of that much importance.

**rag-doll**)Seems to be.

FeSO4*9H2O is a bit weird, though, as it is usually FeSO4*7H2O. But I guess it isn't of that much importance.

Hopefully the actual value of x doesnt matter as much as the method used to get it.

Thanks a lot for your help

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#13

would i be wrong if i were to exclude the H2SO4 from the general equation because it jst serves as a medium?

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#14

(Original post by

would i be wrong if i were to exclude the H2SO4 from the general equation because it jst serves as a medium?

**Boddhi**)would i be wrong if i were to exclude the H2SO4 from the general equation because it jst serves as a medium?

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#15

**Boddhi**)

would i be wrong if i were to exclude the H2SO4 from the general equation because it jst serves as a medium?

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