25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na2​S2​O3​ was used to reach the endpoint. The molarity of the household bleach solution is :

Answers (1)

Household bleach + 2KI→I+ products

I+ 2Na2S2O3→Na2S4O+ 2NaI

Amount of Na2S2O3 used = VM=(48×10-3 L)(0.25 mol L-1) = 12×10-3 mol

Amount of I2 generated = (12×10-3 mol ) / 2 =6×10-3 mol

Assuming 1 mol of household bleach produces 1 mol I2, we will have 

Amount of household bleach in 25 mL solution = 6×10-3 mol

Molarity household bleach = n/V = 6×10-3 mol / 25×10-3 L = 0.24M

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