25 mL of household bleach solution was mixed with 30 mL of 0.50 M and 10 mL of 4 N acetic acid. In t — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

This is an iodometric titration. Equivalents of Bleach = Equivalents of liberated $I_2$ = Equivalents of $Na_2S_2O_3$. The n-factor of $NaClO$ (where Cl goes from +1 to -1) is 2. $N_{bleach} \times V_{bleach} = N_{hypo} \times V_{hypo} \Rightarrow N_{bleach} \times 25 = 48 \times 0.25 \Rightarrow N_{bleach} \times 25 = 12 \Rightarrow N_{bleach} = 0.48 N$. Since $N = M \times n$, $M = 0.48 / 2 = 0.24 M$.