Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry. 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 Na 2 S 2 O 3 was used to reach the end point. The molarity of the household bleach solution is

Question

Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry. 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 Na 2 S 2 O 3 was used to reach the end point. The molarity of the household bleach solution is (A) 0.48 M (B) 0.96 M (C) 0.24 M (D) 0.024 M

Tardigrade Admin · Accepted Answer

Consider the titration reaction CaOCl 2+2 KI arrow I 2+ Ca ( OH )2+ KCl According to this reaction, 25 mL of CaOCl 2, reacts with 30 mL of 0.50 M KI I 2+2 Na 2 S 2 O 3 arrow Na 2 S 4 O 6+2 NaI Given that 48 mL of 0.25 N Na 2 S 2 O 3 was used to reach the end point. So, the number of moles of I 2 produced =48 × 0.25 / 2=6. According to the reaction, Number of millimoles of bleaching powder = Number of moles of I2 = (1 / 2) × Number of moles of Na 2 S 2 O 3=6 So, the molarity of CaOCl 2 is ( text Number of moles of bleaching powder / text Volume of solution ) =6 ( text m mol /25 mL )=0.24 M