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..
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House hold bleach solution will react with Kl to liberate iodine.
Bleach solution will react with Kl to liberate iodine.
NaOCl+2Kl+H+(Excess)=l2+NaCl+2K+ +H2O
Iodine will combine with iodine to produce l3-
l2+l-→l3
This is not titrated with hypo solution.
l3- +2Na2S2O3→Na2S4O6+3l-
Now NaOCl≡l2≡l3-≡2Na2S2O3
So,2 mol Na2S2O3 EQUVALENT TO 1 mol NaOCI(1m Na2S2O3,2N Na2,S2O3)
Bleach solution will react with Kl to liberate iodine.
NaOCl+2Kl+H+(Excess)=l2+NaCl+2K+ +H2O
Iodine will combine with iodine to produce l3-
l2+l-→l3
This is not titrated with hypo solution.
l3- +2Na2S2O3→Na2S4O6+3l-
Now NaOCl≡l2≡l3-≡2Na2S2O3
So,2 mol Na2S2O3 EQUVALENT TO 1 mol NaOCI(1m Na2S2O3,2N Na2,S2O3)
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