Determine Molarity of Household Bleach Solution by Iodometry

To determine the molarity of the household bleach solution, we utilize the principle of redox titration (iodometry).

Step 1: Chemical Reactions Involved

1. The hypochlorite ion ($\text{ClO}^-$) present in the household bleach oxidizes iodide ions ($\text{I}^-$) to iodine ($\text{I}_2$) in an acidic medium: $\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{Cl}^- + \text{I}_2 + \text{H}_2\text{O}$

2. The liberated iodine ($\text{I}_2$) is then titrated against sodium thiosulfate solution ($\text{Na}_2\text{S}_2\text{O}_3$): $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$

Step 2: Calculating Milliequivalents of Sodium Thiosulfate

The milliequivalents ($\text{meq}$) of $\text{Na}_2\text{S}_2\text{O}_3$ used in the titration are: $\text{meq of Na}_2\text{S}_2\text{O}_3 = \text{Normality} \times \text{Volume in mL}$ $\text{meq of Na}_2\text{S}_2\text{O}_3 = 0.25\text{ N} \times 48\text{ mL} = 12\text{ meq}$

Step 3: Equating Equivalents

According to the law of equivalence: $\text{meq of bleach solution} = \text{meq of liberated I}_2 = \text{meq of Na}_2\text{S}_2\text{O}_3$

Let $N$ be the normality of the bleach solution: $N \times 25\text{ mL} = 12\text{ meq}$ $N = \frac{12}{25} = 0.48\text{ N}$

Step 4: Calculation of Molarity

In the reduction reaction of hypochlorite ion to chloride ion: $\text{ClO}^- + 2\text{e}^- + 2\text{H}^+ \rightarrow \text{Cl}^- + \text{H}_2\text{O}$

The oxidation state of chlorine changes from $+1$ in $\text{ClO}^-$ to $-1$ in $\text{Cl}^-$. Thus, the $n$-factor ($v$) for $\text{ClO}^-$ is $2$.

Using the relationship between Normality ($N$) and Molarity ($M$): $\text{Normality} = \text{Molarity} \times n\text{-factor}$ $0.48 = M \times 2$ $M = \frac{0.48}{2} = 0.24\text{ M}$

Thus, the molarity of the household bleach solution is $0.24\text{ M}$.

Correct Option: C