[alo159]

Anhydrous liquid hydrazine is used as a rocket propellant. It is a highly hazardous, toxic, carcinogenic substance.
The %-purity of a sample, presumed to be neat anhydrous hydrazine, is to be determined by iodimetric titration.
First, the titrant iodine (+potassium iodide) solution (I₂ + I⁻ <-> I₃⁻) was standardized.
61.26 mL of iodine solution was required to titrate 0.3944 g of As₂O₃ dissolved in a sodium hydroxide solution adjusted to near neutral pH.
H₂AsO₃−> + I₃^- + 4H₂O-> HAsO₄²⁻ + 3I⁻ + 3H₃O⁺
What is the molarity of the standard solution?
I got the molarity to be 0.06508 M, but then the second part, I can't seem to get it
The second part is this:
Next, a 1.7556-g sample of neat hydrazine is carefully dissolved in water and diluted to 2000.0 mL.
A 50.00-mL aliquot of this solution is withdrawn and titrated with 34.36 mL of the standard iodine solution.

N₂H₄ + 4H₂O + 2I₃⁻ -> N₂ + 6I⁻ + 4H₃O⁺

What is the percentage purity of the hydrazine?
any help would be good

[Borek]

I got the molarity to be 0.06508 M

That's OK

but then the second part I can't seem to get it

What's the problem? Show how you have tried and tell us why do you think it is wrong.

[alo159]

I tried converting the mass of hydrazine, which is given, into mole. Then, I
used that value to find the concentration of hydrazine after it was diluted
to 2000mL. Then,the question says that 50mL of the aliquot is withdrawn and
titrated with 34.36 mL of the standard iodine solution. So, I took the
concentration I found after it was diluted and times it by 0.05L to get my
mole and I divided that number by 0.03436L to get my concentration. Finally,
I divided the concentration of hyrazine after the dilution by the
concentration of hydrazine after the titration...but it is not right..am I
doing something wrong. Just in case the explanation is kind of vague, I will
also show the calculation:
1.7556g(1mol/32.045g)=0.05479mol of hydrazine Note:32.045 is the molar mass of hydrazine
0.05479mol/2L= 0.02740M(after the dilution)
0.02740mol/L(0.05L)=0.00137mol(after 50mL was taken out)
0.00137mol/0.03436L=0.03987M(after the titration)

[Borek]

You have mixed things... Calculate amount of hydrazine that was titrated (you know number of moles of iodine, you know reaction equation - that's simple stoichiometry now). This was in 50 mL sample, convert to total amount of hydrazine (you took 1/40 of total), convert total to mass and calculate %.

[alo159]

hey, thanks for the help. i got the answer....