[brnlow]

Its been a while since I've done any chem homework, but I could use some help with a titration problem concerning titration of H2O2 with KMnO4 in H2SO4.  

I ran across the following:  "... each mL 0.1N KMnO4 is equivalent to 1.7005 mg H2O2"

For the life of me I cant figure out where this comes from.  I searched for the reaction and found

5 h2o2 + 2 kmno4 + 3 h2so4 -> k2so4 + 2 mnso4 + 8 h2o + 5 o2

if this is correct then,
mol h2o2 = 2/5 mol kmno4 consumed

for kmn04, 0.1 N = 0.1 M (correct?) so each mL = 1 e-4 mol kmno4
mol h2o2 = 2/5 * 1 e-4 = 4 e-5 mol h2o2

(4 e-5 mol h2o2)(34 g/mol) = 1.36 mg h2o2

Where am I going wrong, or is 1.7005 a misprint?

Help, its driving me NUTS!

[sdekivit]

MnO4(-) gives 5 equivalents of electrons so 0,1 N is 0,02 M KMnO4

so now take 1 mL KMnO4 and calculate like you did, will lead to the answer 1,7008 mg H2O2

(normality is amount of equivalents per L)

[brnlow]

Thanks, I obviously have a weak hold at best on Normality!  I can only calculate N for acids (number of protons liberated).  

How do you get 5 from KMnO4?

KMnO4 (s) -> K+ (aq) + MnO4- (aq) (correct?)

Since Mno4 - can "neutralize"  1 H+ I thought the Normality would be 1

Please explain in detail.  

Thanks

 

[macdonda]

Normality = equivalents of solute / L of solution

For redox reactions, the # equivalents is the number of electrons given or recieved per mol of species (in this case MnO4)

[sdekivit]

Thanks, I obviously have a weak hold at best on Normality!  I can only calculate N for acids (number of protons liberated).  

How do you get 5 from KMnO4?

KMnO4 (s) -> K+ (aq) + MnO4- (aq) (correct?)

Since Mno4 - can "neutralize"  1 H+ I thought the Normality would be 1

Please explain in detail.  

Thanks

 

normality for redoxreaction: see macdonda's reply.

The halfreaction for MnO4(-) can be found in a table for standards electrode potentials:

MnO4(-) + 8 H(+) + 5e(-) --> Mn(2+) + 4 H2O

[brnlow]

Thanks,

Making more sense now!

[Borek]

The halfreaction for MnO4(-) can be found in a table for standards electrode potentials:

MnO4(-) + 8 H(+) + 5e(-) --> Mn(2+) + 4 H2O

That's for acidic conditions only, so it holds in this case, but permanganate can also react accepting only one electron:

MnO₄^- + e^- -> MnO₄^2-

[sdekivit]

That's for acidic conditions only, so it holds in this case, but permanganate can also react accepting only one electron:

MnO₄^- + e^- -> MnO₄^2-

we also have:

MnO4(-) + 2 H2O + 3e(-) --> MnO2(s) + 4 OH(-)

it's nice that you can follow the reaction due to the color of the different Mn-solutions:

the acidic reaction will go from purple to colorless. The reaction with water will go from purple to colorless with the formation of a brown/black precipitate. And the reaction above will go from purple to green.

[brnlow]

That's for acidic conditions only, so it holds in this case, but permanganate can also react accepting only one electron:

MnO₄^- + e^- -> MnO₄^2-

Does this mean in this that the N of the soution changes?  If so:

 WTF does it mean when a  bottle of KMnO4 labeled as 0.1 N  by the manufacturer?  Damn I hate it when % (wt/v) or M isn't listed on the label.

[Borek]

Does this mean in this that the N of the soution changes?

Yes. Read comment here:

http://www.chembuddy.com/?left=concentration&right=normality

WTF does it mean when a  bottle of KMnO4 labeled as 0.1 N  by the manufacturer?  Damn I hate it when % (wt/v) or M isn't listed on the label.

Probably that it is 0.1N for 5 electron reaction (so 0.02M).

And you better don't use % wt/v, as it doesn't mean anything either Only w/w has any reasonable meaning (read What's wrong with % and Mass-volume percentage in the above lectures).

[brnlow]

Probably that it is 0.1N for 5 electron reaction (so 0.02M).

Thanks, I think thats what I'm looking for.  I suppose there is a reason for N, but I simply like to plug and chug with M.  So if I go back to my original post I should be OK using the following?:

[KMnO4] = 0.1 N = 0.02 M

mol H2O2 in sample = 2/5(mol KMnO4) used in the titration

I really appreciate all the help here! You guys are great.

[Borek]

mol H2O2 in sample = 2/5(mol KMnO4) used in the titration

Seems OK.