Step-by-step Solution:

This problem involves a redox titration, specifically an iodometric titration, to determine the concentration of the active ingredient in a household bleach solution, which is sodium hypochlorite ($NaOCl$).

Step 1: Understand the chemical reactions.

1. The bleach solution ($NaOCl$) is treated with an excess of potassium iodide ($KI$) in an acidic medium (acetic acid). The hypochlorite ion ($OCl^-$) oxidizes the iodide ion ($I^-$) to iodine ($I_2$). The oxidation state of chlorine in $OCl^-$ is +1, and it gets reduced to $Cl^-$ (oxidation state -1). $OCl^- + 2I^- + 2H^+ \rightarrow Cl^- + I_2 + H_2O$
2. The liberated iodine ($I_2$) is then titrated with a standard solution of sodium thiosulfate ($Na_2S_2O_3$). In this reaction, iodine is reduced back to iodide ions, and the thiosulfate ion ($S_2O_3^{2-}$) is oxidized to the tetrathionate ion ($S_4O_6^{2-}$). $I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$ Step 2: Apply the law of chemical equivalence.

The principle of iodometry is based on the law of equivalence. The number of equivalents of the initial oxidizing agent ($NaOCl$) is equal to the number of equivalents of the iodine liberated, which in turn is equal to the number of equivalents of the sodium thiosulfate used for titration.

Equivalents of $NaOCl$ = Equivalents of $I_2$ = Equivalents of $Na_2S_2O_3$

We can work with milli-equivalents (mEq) since volumes are given in mL.

Step 3: Calculate the milli-equivalents of $Na_2S_2O_3$ used.

The number of milli-equivalents is calculated as the product of normality (N) and volume (V in mL).

Given:

• Volume of $Na_2S_2O_3$ solution, $V_{thio}$ = 48 mL
• Normality of $Na_2S_2O_3$ solution, $N_{thio}$ = 0.25 N

mEq of $Na_2S_2O_3$ = $N_{thio} \times V_{thio}$ mEq of $Na_2S_2O_3$ = $0.25 \frac{\text{eq}}{L} \times 48 \text{ mL} = 12$ mEq

Step 4: Determine the milli-equivalents of $NaOCl$ in the bleach sample.

Based on the law of equivalence: mEq of $NaOCl$ = mEq of $Na_2S_2O_3$ = 12 mEq

These 12 mEq of $NaOCl$ were present in the initial 25 mL sample of the household bleach solution.

Step 5: Calculate the normality of the bleach solution.

Normality of bleach solution ($N_{bleach}$) = $\frac{\text{mEq of } NaOCl}{\text{Volume of bleach solution (mL)}}$ $N_{bleach} = \frac{12 \text{ mEq}}{25 \text{ mL}} = 0.48$ N

Step 6: Convert the normality of the bleach solution to molarity.

The relationship between molarity (M) and normality (N) is given by: $N = M \times \text{n-factor}$

We need to find the n-factor for $NaOCl$ in the first reaction: $OCl^- + 2I^- + 2H^+ \rightarrow Cl^- + I_2 + H_2O$

In this reaction, the oxidation state of chlorine (Cl) in $OCl^-$ is +1. It is reduced to $Cl^-$, where its oxidation state is -1. The change in oxidation state is $|(+1) - (-1)| = 2$. Therefore, the n-factor for $NaOCl$ is 2.

Now, we can calculate the molarity ($M_{bleach}$): $M_{bleach} = \frac{N_{bleach}}{\text{n-factor}} = \frac{0.48 \text{ N}}{2} = 0.24$ M

Thus, the molarity of the household bleach solution is 0.24 M.