25 m L of household bleach solution was mixed with 30 m L of 0.50 M K I and 10 m L of 4 N acetic acid. In the titration of the liberated iodine, 48 m L of 0.25 N N a 2 S 2 O 3 was used to reach the end point. The molarity of the household bleach solution is

Step-by-step solution

H 2 O 2 Bleach 25 ml + 2 I - 0 . 5M 30 ml + 2 H + ( from C H 3 COOH ) 4 N ( 10 ml ) → I 2 + 2 H 2 O ...(i) I 2 + 2N a 2 S 2 O 3 0 . 25 N ( 48 ml ) → N a 2 S 4 O 6 + 2NaI ...(ii) m. mol of I 2 = 1 2 (m. Moles of N a 2 S 2 O 3 ) = 1 2 × ( 0 . 25 × 48 ) = 6 m . mol Using equation (i) 1 m. Mol of I 2 ≡ 1 m. Mol of H 2 O 2 m.mol of H 2 O 2 = 6 m.mol Molarity of H 2 O 2 = 6 25 = 0.24 M