25ml of household bleach solution was mixed with 30ml of 0.5M KI and 10ml of 4N acetic acid. In the titration of the liberated iodine, 48ml of 0.25N hypo was used to reach the end point. The molarity of the household bleach solution is??

ans: 0.24MI want to know how to solve it..

house hold bleach solution is sodium hypochloride (NaOCl)
 
Bleach solution will react with KI to liberate iodine
 
NaOCl + 2KI + H⁺ (excess) = I₂ + NaCl + 2K⁺  + H₂O
 
Iodine will combine with iodide to produce I₃^-
 
I₂ + I^-  →  I₃^-
 
This is not titrated with hypo solution.
 
I₃^- + 2 Na₂S₂O₃ → Na₂S₄O₆ + 3 I^- 
 
Now NaOCl ≡ I₂ ≡ I₃^- ≡ 2 Na₂S₂O₃ 
 
So, 2 mol Na₂S₂O₃  equvalent to 1 mol NaOCl
 
(1M Na₂S₂O₃ = 2N Na₂S₂O₃ )
 
Now calculate the amount of bleach solution.