This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to the questions: 1. Explanation for using H_2SO_4 instead of HCl for acidification: Hydrochloric acid (HCl) contains chloride ions (Cl^-) which can be oxidized by strong oxidizing agents (e.g., MnO_4^- or Cr_2O_7^2- if present) to chlorine gas (Cl_2). This would interfere with the redox titration by reacting with the titrant or analyte. Sulfuric acid (H_2SO_4) is a non-oxidizing acid, and its sulfate ions (SO_4^2-) are not easily oxidized, making it suitable for acidifying solutions in redox titrations without interfering reactions. 2. Explanation why sodium thiosulphate is not a primary standard: Sodium thiosulphate (Na_2S_2O_3 · 5H_2O) is not a primary standard because its purity and exact water content are variable. It can lose water of crystallization (effloresce) or absorb moisture (deliquesce) from the atmosphere. It is also susceptible to decomposition by atmospheric oxygen, light, bacteria, and in acidic solutions, leading to changes in its concentration over time. 3. Explanation why sodium thiosulphate solution is susceptible to decomposition to give elemental sulphur: Sodium thiosulphate solution is unstable in acidic conditions. When acidified, the thiosulphate ion (S_2O_3^2-) undergoes disproportionation, where sulfur in different oxidation states reacts to form elemental sulfur and sulfur dioxide. The reaction is: S_2O_3^2- (aq) + 2H^+ (aq) → SO_2 (g) + S (s) + H_2O (l) The elemental sulfur appears as a cloudy precipitate, indicating decomposition. (Note: The question mentions "metal sulfide" which is not a typical product of thiosulphate decomposition in acid; elemental sulfur and sulfur dioxide are the common products). 4. Equation for thiosulphate reacting with iodine, producing sodium iodide: The reaction between iodine and thiosulphate ions is: I_2 (aq) + 2S_2O_3^2- (aq) → 2I^- (aq) + S_4O_6^2- (aq) In terms of sodium salts, this is: I_2 (aq) + 2Na_2S_2O_3 (aq) → 2NaI (aq) +