To determine the molar concentration of sodium hypochlorite ((\text{NaOCl})) in the diluted bleach solution, we’ll follow these steps:
1. Understanding the Reaction
The titration involves sodium thiosulfate ((\text{Na}_2\text{S}_2\text{O}_3)) reacting with sodium hypochlorite ((\text{NaOCl})). The balanced redox reaction is:
From the equation, 1 mole of (\text{ClO}^-) reacts with 2 moles of (\text{S}_2\text{O}_3^{2-}).
2. Calculating Moles of (\text{Na}_2\text{S}_2\text{O}_3)
Given:
-
Volume of \(\text{Na}_2\text{S}_2\text{O}_3\) solution = 27.74 mL** = **0.02774 L*
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Molarity of \(\text{Na}_2\text{S}_2\text{O}_3\) = **0.0197 M**
3. Determining Moles of (\text{NaOCl})
From the balanced equation:
4. Calculating Molarity of (\text{NaOCl})
Given:
- Volume of diluted bleach solution = 10.00 mL = 0.01000 L
Final Answer
The molar concentration of (\text{NaOCl}) in the diluted bleach solution is 0.0273 M.
\[ \begin{align*} \text{Oxidation:} & \quad 3\text{I}^- \rightarrow \text{I}_3^- + 2e^- \\ \text{Reduction:} & \quad \text{OCl}^- + 2\text{H}^+ + 2e^- \rightarrow \text{Cl}^- + \text{H}_2\text{O} \\ \hline \text{Overall:} & \quad \text{OCl}^- + 3\text{I}^- + 2\text{H}^+ \rightarrow \text{I}_3^- + \text{Cl}^- + \text{H}_2\text{O} \end{align*} \] \[ \begin{align*} \text{Reduction:} & \quad \text{IO}_3^- + 6 \text{H}^+ + 6 \text{e}^- \rightarrow \text{I}^- + 3 \text{H}_2\text{O} \\ \text{Oxidation:} & \quad 6 \text{S}_2\text{O}_3^{2-} \rightarrow 3 \text{S}_4\text{O}_6^{2-} + 6 \text{e}^- \\ \hline \text{Net Ionic Equation:} & \quad \text{IO}_3^- + 6 \text{S}_2\text{O}_3^{2-} + 6 \text{H}^+ \rightarrow \text{I}^- + 3 \text{S}_4\text{O}_6^{2-} + 3 \text{H}_2\text{O} \end{align*} \]