dilution of hydrogen peroxide

how do u calculate the necessary dilution and how to you calculate the volume of h202 to use?

Reply 3

For a 20 Volume solution, how much would it need to be diluted?

Look, it all depends on what you are going to titrate it with.... or with it!

Lets assume that you are going to titrate a solution of potassium manganate VII whch is about 0.02M

What you want ideally is about 25cm3 of both titrant and solution.

So calculate how many moles of KMnO4 there would be in the 25cm3 aliquot.

Then use the reacting proportions to calculate how many moles of H2O2 this would react with.

Now using the relationship outlined above to ensure that you have about the correct dilution.

Example:

25cm3 of 0.02M KMnO4 contains 0.025 x 0.02 moles = 0.00044 moles of MnO4-

The reaction is:

MnO4- + 8H+ + 5e --> 4H2O + Mn2+
H2O2 ---> 2H+ + O2 + 2e
-----------------------------------
2MnO4- + 16H+ + 10e --> 8H2O + 2Mn2+
5H2O2 ---> 10H+ + 5O2 + 10e
--------------------------------------
2MnO4- + 6H+ + 5H2O2 ---> 8H2O + 2Mn2+ + 5O2
--------------------------------------------------

therefore the ratio of manganate to hydtrogen peroxide is 2:5

so if you have 0.00044 moles of manganate (VII) you need 0.0011 moles H2O2 in your 25cm3 aliquot.

Let's assume that you are going to make up 250 cm3 then you need 0.011 moles H2O2

by the equation: 2H2O2 --> 2H2O + O2

this number of moles of H2O2 would provide 0.011/2 moles of oxygen = 0.0055

This is equivalent to 0.0055 x 24000 cm3 at RTP = 132 cm3

So you need an amount of H2O2 that would release 132 cm3 (approx) in your 250cm3 solution

As you are starting with 100 volume H2O2 then 1 cm3 will release 100 cm3 oxygen

Clearly then you only need 1.32 cm3 in 250 cm3

As this is not an appropriate amount to measure out I would suggest doing a ten fold dilution first of all (of which you would need 13.2 cm3) and then measuring out 15cm3 with a standard pipette and making it up to 250 cm3.

Reply 5

thanks for that it helps a lot

Reply 6

Thank you. But can you just assume 0.02M? I thought you had to work it out from the Moles of H2O2 and the stoichiometric ratio, which gives an answer of around 0.70M?

Oh Dear...

Thank you. But can you just assume 0.02M? I thought you had to work it out from the Moles of H2O2 and the stoichiometric ratio, which gives an answer of around 0.70M?

I didn't assume 0.02M ...

the OP didn't give any information so I just plucked a number from the air, and as 0.02 is often used in KMnO4 titrations that was it!

Reply 8

dont you just use a kmno4 with the known concetration
cant u just ask ur teacher for the one they use in class. because ur not workin out the concentration of kmno4 ur workin out the concentration of h202

Reply 9

snake

dont you just use a kmno4 with the known concetration
cant u just ask ur teacher for the one they use in class. because ur not workin out the concentration of kmno4 ur workin out the concentration of h202

True that your working out the concentration of H2O2. But the "20 Volume" H2O2 has a concentration of approximately 1.76M. There would be no use of using a class concentration, if it is say 0.02M. It would require burettes full of KMnO4 to reach the endpoint if that was the case. So you do need to calculate an approximate concentration of KMnO4 using the equation ratio, to provide a reasonable titre.

Reply 10

on da mark schemes of other papers wiv similar questions itjust says use kmno4 of a known concetration so u dnt need to calculate it

Reply 11

silent ninja

19

Charco you are way too helpful!

This is the OCR Chem planning exercise i take it?
I did the experiment this week. But basically what charco said, 20 vol H2O2 is too strong so you gotta dilute because KMnO4 in a typical lab is only 0.02M.

Reply 12

jamesharrison

You can work out approximately what the concentration of H2O2 is before the titration. Use the moles of O2 gas given off to work out the moles of
H2O2 using the equation, then the moles/volume gives a concentration.

The concentration of KMnO4 can also be predicted from this. Just look at the full equation for the reaction and work out how many moles of H2O2 reacted with KMnO4. Using the moles you can work out an approximate concentration of KMnO4. easy.

Reply 13

When your working out the Concentration of H2O2, after finding the Moles (1.76 Mol), is the Volume 1dm3? Giving an answer of 1.76 moldm-3?

Or is the volume something different...

Reply 14

i calculated the concentration of h202 to be 4.22 moldm3 in the original 20 volume solution i think its rong can some1 check dis.

my mean titre of kmno4 was 25.35
so i did 25.35/1000 x 0.02= moles of kmno4
den i x moles by5/2 to giv moles of h202

because i pipetted 25cm3 i den divded moles by 25cm3 to giv moles in 1 cm3 and den x by 250 to giv moles in 250cm3

i den divided by 3 and x by 1000 to giv me moldm3

i divided by 3 because das wot i originally used in ma dilution of 250cm3 solution

can some help me plz

Reply 15

silent ninja

19

For the approx conc of the original just use the decomposition equation to work out the number of moles.
eg 1dm3 H₂O₂ gives 20 dm3 O₂. Howmany moles of O₂ is that? So then howmany moles of H₂O₂ did it come from in the equation? This is howmany are present in 1dm3 i.e the concentration (roughly).

From your results, your titre looks dodgy.
If we use 1 mole gas occupies 24 dm3 vol at standard room temp & pressure, then the H₂O₂ conc is about 1.67 mol/dm3.

3cm3 contains 3/1000*1.67= 0.005 moles.
You put this many moles in 250cm3= 4x0.005 mol/dm3 = 0.02 mol/dm3 H₂O₂ is the conc of your diluted solution.

Next, 25cm3 of this solution= 25/1000*0.02=0.0005 moles hydrogen peroxide.
This would react with 0.0005/5*2= 0.0002 moles KMnO4.

If you used 0.02 mol/dm3 KMnO4 solution this equates to 0.0002/0.02= 0.01 dm3
which is only 10cm3 of titre. How did you get 25.35? Your must have used 6+cm3 of the original solution?

Reply 16

ok im still confused

first of all how do calculate how much h202 to use and how much to dilute it by if i was using 0.02mol/dm3 KMnO4

Reply 17

help me bod
i fort i jus make up a mean titre and den jus show how to calculate the conc

Reply 18

silent ninja

19

25 cm3 of 0.02 mol/dm3 KMnO4 is 0.0005 moles.
This would react with 0.0005/2*5= 0.00125 moles H₂O₂ from the equation.
So you want 0.00125 moles peroxide in 25 cm3 to give you a decent volume.

20 vol is about 1.67 mol/dm3.
So howmuch of this stuff contains 0.00125 moles?
If you put that much in water and make up to 25cm3, you have your solution. But since you want to make plenty of it, multiply it by 10 and make up to 250cm3 to make your dilute solution.

Reply 19