My chemistry questions

[david10d]

Hey guys, I'm going to post some questions that either I don't understand completely or need assurance about.

I'll start with this one:

To standardise a solution of KMnO4 of approximate concentration 0.2M, an analyst took 25.00ml of the solution and made it up to 250.00ml in a volumetric flask. He then measured out 1.117g of oxalic acid crystals, and made it up to 250.00ml in another volumetric flask, using sulfuric acid of approximate concentration 0.2M. Next, 20.00ml aliquots  of the oxalic acid were titrated against the diluted KMnO4 solution The end point was detected at the first permanent pink colour. The mean titre was found to be 18.95ml.

Equation: 2MnO4-(aq) + 5H2C2O4(aq) + 6H+(aq) --> 2Mn+2(aq) + 10CO2(g) + 8H2O(l)

a) Calculate the original concentration of the KMnO4 solution. (The concentration of the diluted solution is 0.02096M. Where's the dilution factor for KMnO4?)
b) Explain why the addition of sulfuric acid was necessary.


Thanks.

[luken93]

Immediately you can see the following things:

n(KMnO4) = .025L x ~.2M = 0.005mol in 250mL flask

n(H2C2O4) = (1.117/90) = 0.0124 mol
c(H2C2O4) = 0.0124 / .25 = 0.0496 mol/L

20ml aliquots contain 0.0496 x .02 = 0.000992mol of H2C2O4

a) n(KMnO4) = (2/5) x n(H2C2O4) = 0.0003968 mol of KMnO4
c(KMnO4) = 0.0003968 / .01895 = 0.020934M

[david10d]

The mol ratio is incorrect, it's supposed to be 2/5. Once you've corrected that then that's the concentration of the diluted solution

I also reread my answer in the OP. It should be 0.02096M, sorry for any confusion.

[Conflagaration]

Gonna assume your equation is:

2MnO4-(aq) + 5H2C2O4(aq) + 6H+(aq) --> 2Mn+2(aq) + 10CO2(g) + 8H2O(l)

Sulfuric acid supplies H+ ions on the left hand side of the equation, so the stable product of water can be formed from hydrogen ions and the oxygen freed from the permanganate as the manganese is reduced.

Actual chemistry reason? Probably something to do with O(2-) being a more unstable product, so the redox reaction will not spontaneously occur. Energetically unfavourable and all that.

[david10d]

BUMP

We conducted a redox titration experiment a few days ago with KMnO4 against H2O2. The standard solution was 1M of KMnO4 and the 7g of H2O2 was topped up to 250ml volumetric flask. We then poured a 20.00ml aliquots of H2O2 in a beaker and began titrating.

Anyways we had gotten 3 concordant results with the average titre being 26.96ml.

The equation of the experiment is:

5H2O2(aq) + 2MnO4-(aq) + 6H+(aq)  >  2Mn2+(aq) + 5O2(g) + 8H2O(l)

The first question asks for the mass of H2O2 in each 20.00ml aliquot. My answer was 2.2916 grams.

Now the next question asks 'Use this result to calculate the total mass of H2O2 in the volumetric flask.'

watdo?

[Water]

To find the mole of H2O2 in the volumetric flask , which is primarily to find the mass of H2O2.


We know that the n(H2O2) in 20.00ml aliquot =     Z


We now want to find the C(H2O2) = n(H2O2) in aliquot divided by 20.00ml.    

This would give us the concentration so that we can find the n(H2O2 in Volumetric flask)

n(H2O2) in volumetric flask = C(H2O2) x .250L


In short, you could do

n(H2O2) in volumetric flask = n(H2O2) in aliquit x .250/.020





From the n(H2O2) in volumetric flask, you can find the mass


Basically, all we are doing throughout the process is using the formula n = cv

[Dr.Lecter]

V(MnO4-)=26.96
n(KMnO4)=0.01998 x (26.96/1000)= 0.0005386608
n(H2O2)= 5/2 x 0.0005386608 = 0.001346652
mass(H2O2)= 0.001346652 x (M(H2O2)=34)= 0.045786168

mass(H202)= (250/20) x 0.045786168 = 0.5723271

[david10d]

Edit2:

It actually doesn't work out - I had forgotten to * by the molar mass.

I'm seriously doubting the accuracy of the results...

Edit3: A friend has told me (Dr. Lecter) that the concentration on the bottle of KMnO4 WAS NOT 1M, but 0.01998M. What a major waste of time..


I DUN GOOFED

[Dr.Lecter]

bl sir