How Much Na2S2O3 Is Needed for Titration in This Redox Reaction?

[Maharg]

Homework Statement

Bleaching powder, Ca(OCl)Cl, reacts with iodide in acidic medium according to the equation:

OCl– + 2 I– + 2 H3O– ---> I2 + Cl– + 3 H2O

How many millilitres of 0.0645 M Na2S2O3 are required to titrate the iodine liberated from a 0.6028 g sample of bleaching powder containing 10.50 % (w/w) Cl ?

Homework Equations

I2 + 2S203 --> 2I- + S4O6

The Attempt at a Solution

I have already submitted an answer for this assignment, and this was one of only 2 questions I got wrong, we can do a second submission so I went to figure this one out.

I think this is where I went wrong.

I Figured out the Cl- moles by: g Cl/0.6028 g = 0.01050

0.063294 g Cl / 35.453 g/mol = 1.78529 mmol

I think I may have done that wrong, but from that.
From mol Cl- I used 1:1 ratio of original reaction to get mol I2

Mol I2 in second reaction are 1/2 mol S203

mol S2O3 = 3.57058641 mmol

Then calculated mL needed.

V = 3.57058641 mmol/0.0645 M
= 55.4 mL

Any suggestions?

 

[Borek]

None

I got 55.3 mL, close enough.

 

[Maharg]

Hm.. that's odd then. The computer usually gives part marks when we're close.

 

[Borek]

Hm, could be we are both wrong. It is 10.5% of Ca(OCl)Cl. Part of the chlorine is not taking part in the iodine oxidation. I have concentrated on your solution, instead of reading the question

 

[Maharg]

Would I just divide it by 2? So only 1/2(1.785 mmol Cl) take place in the reaction?

 

[Maharg]

Ah, possibly I'm going about it the wrong way, can I instead just avoid the w/w of Cl?

0.6028 g / MW Cl-/ MW Ca(OCl)Cl = mol Cl-

Does anyone know, I'm really lost in the solution.