What is the molarity of the standard solution used for iodimetric titration?

[higherme]

Homework Statement

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.
61.24 mL of iodine solution was required to titrate 0.2802 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?

The Attempt at a Solution

I have to find the concentration of the standard which is iodine. I know that it took 61.24ml of that soln to titrant the analyte As2O3. So, my attempt is to take the mass of As2O3 that was used and convert that to moles using the molar mass.. then this moles will be the same as moles of iodine that reacted. from there i can find concentration...
BUT, what I don't get is the equation given... it started with H2AsO3... is that what the analyte is? or is it a different compound.. since the 2 after the As is missing

 

[chemisttree]

You lost me at hydrazine (H4N2). What does arsenite ion have to do with this?

BTW. Arsenic in As(OH)3 or H2AsO3- is As+3 and Arsenic in As2O3 (O=As-O-As=O) it is +6. Yes, they are different compounds in different oxidation states.

 

[higherme]

I'm confused. But that is what the question is asking... maybe I mis-understood something. I don't know what is reacting with what

 

[higherme]

maybe the hydrazine, is not related to the question, I don't know :S

so is "As2O3 dissolved in a sodium hydroxide" => H2AsO3- ?

 

[chemisttree]

Apparently the arsenous acid is used to standardize the I3- solution. The standardized solution (I3-) is then used to titrate hydrazine, although that part of the procedure is not discussed in this problem.

 

[higherme]

Oh...

so, would this be the way to find the [ ] of the standard:

take the mass of As2O3 given (0.2802g) and divide by its molar mass to get the moles of it.

0.2802g *(1mol As2O3 / 197.839g) = 0.001416 mol As2O3 *2 = 0.00283mol As

0.00283 mol As = moles I3- reacted

0.00283mol I3- / 0.06124 L = 0.046 M I3- <--- that would be the concentration of the standard

is this way right?

 

[chemisttree]

Don't forget that "...As2O3 dissolved in a sodium hydroxide solution adjusted to near neutral pH" means,

As2O3 + 3OH- + H+ ------> 2 H2AsO3-

 

[higherme]

so after i find the moles of As2O3 times 2 = moles of H2AsO3-

0.2802g *(1mol As2O3 / 197.839g) = 0.001416 mol As2O3 *2
= 0.00283mol H2AsO3- = moles I3-0.00283mol I3- / 0.06124 L = 0.046 M I3- <--- that would be the concentration of the standard

like this?