91Ó°ÊÓ

Bleaching powder, also known as calcium hypochlorite, is commonly used as a disinfectant and in the bleaching of fabrics. When it reacts with potassium iodide (\(\mathrm{KI}\)), in the presence of an acidic solution such as diluted sulfuric acid (dil. \(\mathrm{H}_2\mathrm{SO}_4\)), chlorine gas (\(\mathrm{Cl}_2\)) is liberated. This is a crucial process in the qualitative analysis of bleaching powder. The liberated chlorine reacts with \(\mathrm{KI}\) to produce iodine (\(\mathrm{I}_2\)), as shown in the reaction:

• \(\mathrm{Cl}_2 + 2\mathrm{KI} \rightarrow 2\mathrm{KCl} + \mathrm{I}_2\)

This reaction serves a dual purpose: it confirms the presence of chlorine in the bleaching powder and quantitatively measures it by subsequent steps in the process. By liberating iodine, the reaction showcases a redox process where chlorine acts as an oxidizing agent. This step is essential for analytical purposes as it sets the stage for the titration process that follows.

To solve problems in chemistry, calculating moles is a foundational skill. In this exercise, moles help us transition from the reaction of iodine to the chlorination level in bleaching powder. The conversion from volume and concentration to moles is critical:

• The amount of iodine liberated is determined using sodium thiosulfate solution (Naâ‚‚Sâ‚‚O₃), commonly known as hypo, for titration.

Given the solution used is \(\frac{N}{10}\), with a volume of \(50~\mathrm{mL}\), we find the moles of \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\) using the formula:\[\text{Moles of Na}_2\mathrm{S}_2\mathrm{O}_3 = \frac{N \times V}{1000} = \frac{0.1 \times 50}{1000} = 0.005\, \text{moles}\]As per reaction stoichiometry, 1 mole of \(\mathrm{I}_2\) needs 2 moles of \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\). Thus, moles of \(\mathrm{I}_2\) is half:\[\text{Moles of } \mathrm{I}_2 = \frac{0.005}{2} = 0.0025 \text{ moles}\]Accurate calculation of moles helps us in tracking the quantity of substances through a chemical reaction.

Titration is a common laboratory method used to determine the concentration of a solute in solution. In this context, iodometry is used, involving the titration of iodine with a thiosulfate solution.Once iodine is liberated, it is titrated against sodium thiosulfate (\(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\)), which acts as a titrant. The reaction proceeds as:

• \(\mathrm{I}_2 + 2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \rightarrow 2\mathrm{NaI} + \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6\)

By measuring how much thiosulfate is required to completely react with the \(\mathrm{I}_2\) formed, one can back-calculate the amount of \(\mathrm{I}_2\) initially present. The endpoint is usually detected with a starch indicator, which forms a blue-black complex with iodine and fades when iodine is completely consumed.Titration allows for precise quantification of the liberated iodine, which in turn, indicates the amount of active chlorine available in the bleaching powder.

Stoichiometry is the calculation of reactants and products in chemical reactions and is based on the balanced chemical equations. It helps us understand the proportions of substances consumed and produced in a reaction.In these reactions:

• 1 mole of \(\mathrm{Cl}_2\) liberates 1 mole of \(\mathrm{I}_2\), demonstrating a \(1:1\) stoichiometric ratio.
• Similarly, 1 mole of \(\mathrm{I}_2\) reacts with 2 moles of \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\) in a \(1:2\) ratio.

These ratios are crucial for determining the amounts of reactants and products involved. Successfully applying stoichiometry in this problem allows the precise calculation of \(\mathrm{Cl}_2\) liberated, as it’s directly related to moles of \(\mathrm{I}_2\).This methodology leads to determining the concentration of chlorine in the bleaching powder sample, ensuring consistency in quantitative chemistry analyses.