Question

25 ml of household bleach solution was mixed with 30 ml of 0.50M KI and 10 mL of 4N acetic acid. In the
titration of the liberated iodine, 48 ml of 0.25N Na2S2O3 was used to reach the end point. The molarity of the household bleach solution is
(a) 0.48M
(b) 0.96M
(c) 0.24M
(d) 0.024M
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Answer 1

Answer:

Explanation:

Quantity of bleach = 25ml

Acetic Acid = 4N = 10mL

Kl = 0.50M = 30ml

The household bleach solution will react with Kl to liberate iodine.

NaOCl+2Kl+H+(Excess )=l2+NaCl+2K+ +H2O

Iodine will combine with the iodine to produce l3-

l2+l-→l3

This is not titrated with the hypo solution.

l3- +2Na2S2O3→Na2S4O6+3l-

Now NaOCl ≡ l2 ≡ l3 ≡ 2Na2S2O3

Thus, two mol Na2S2O3 is equivalent to one mol NaOCI(1m Na2S2O3,2N Na2,S2O3).

Answer 2

.24

Explanation:

25 mL of household bleach solution was mixed with 30 mL of 0.50 M Kl and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The molarity of the household bleach solution is a) 0.48 M b) 0.96M c) 0.24 M d) 0.024