Calculate the concentration of hydrogen peroxide

[Me!!]

can anone help me with these questions?

a) How many moles of hydrogen peroxide react with each mole of permanganate in the titrationmoles of MnO4- per mole of H2O2

b) Calculate the concentration of hydrogen peroxide in the 10 mL aliquot, taken from the reaction if the titration with 0.0176 M permanganate solution requires 32.8 mL to reach the end-point

c) If the concentration of hydrogen peroxide in the 10 mL aliquot is 0.0568, calculate ln([H2O2])

your help will be greatly appreciated!

 

[chem_tr]

Hello,

I will try to solve the problems.

a) How many moles of hydrogen peroxide react with each mole of permanganate in the titrationmoles of MnO4- per mole of H2O2

My answer: As you know, peroxide reacts with permanganate to yield elemental oxygene and manganese(IV) oxide:

2Mn(7+) + 6e(-) --> 2Mn(4+)
3O2(2-) --> 3O2 + 6e(-)

2KMnO4 + 3H2O2 --> 2MnO2 + 3O2 + 2KOH + 2H2O

Thus, 2 moles of permanganate is required to oxidize 3 moles of peroxide.

b) Calculate the concentration of hydrogen peroxide in the 10 mL aliquot, taken from the reaction if the titration with 0.0176 M permanganate solution requires 32.8 mL to reach the end-point

My answer: I like to express the milimoles by directly multiplying the molarity and titer; so we obtain 0.0176*32.8=0,57728 milimoles of permanganate. We will multiply this value with 1.5 (as seen from the mole ratio between permanganate and peroxide in the reaction above) to learn the amount of reacted peroxide: 0.86592 milimoles of peroxide in 10 mL solution. This gives a 0,086592 M solution.

c) If the concentration of hydrogen peroxide in the 10 mL aliquot is 0.0568, calculate ln([H2O2])

My answer: I assume that [H2O2] is expressed as molar concentration, so we may directly calculate the ln value from this: in this way, we obtain approximately -2.868.

I may be wrong, but I have not found any errors.

Regards
chem_tr

 

[ravenprp]

a) In a titration, the mole ratio between the reactants can be used to calculate the concentration of one substance from the known concentration of the other. In this case, the mole ratio between H2O2 and MnO4- is 1:5. This means that for every 1 mole of H2O2, 5 moles of MnO4- are needed for complete reaction.

b) To calculate the concentration of H2O2 in the 10 mL aliquot, we first need to determine the number of moles of MnO4- used in the titration. This can be calculated by multiplying the volume of MnO4- used (32.8 mL) by its concentration (0.0176 M):

n(MnO4-) = (0.0328 L)(0.0176 mol/L) = 0.0005776 mol

Since the mole ratio between H2O2 and MnO4- is 1:5, the number of moles of H2O2 in the aliquot is 1/5 of the moles of MnO4- used:

n(H2O2) = 0.0005776 mol/5 = 0.00011552 mol

Now, we can use the volume of the aliquot (10 mL) to calculate the concentration of H2O2:

[H2O2] = n(H2O2)/V(aliquot) = 0.00011552 mol/0.010 L = 0.011552 M

c) To calculate ln([H2O2]), we first need to convert the concentration of H2O2 from M to mol/L:

[H2O2] = 0.0568 mol/L

Now, we can use the natural log function to calculate ln([H2O2]):

ln([H2O2]) = ln(0.0568 mol/L) = -2.868