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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The image you've sent describes a quantitative analysis experiment (titration), not qualitative analysis. Assuming you're asking for the formulas relevant to the experiment shown in the image, here they are: Formulas for the Quantitative Analysis Experiment 1. Preparation of Standard Potassium Iodate (KIO₃) Solution (Procedure 1): Molar Mass of KIO₃: To calculate the molar mass, you'd sum the atomic masses of its constituent elements: M_KIO_3 = M_K + M_I + (3 × M_O) Using standard atomic masses: M_KIO_3 = 39.10 g/mol + 126.90 g/mol + (3 × 16.00 g/mol) = 214.00 g/mol Moles of KIO₃: Moles of KIO_3 = Mass of KIO_3 (from Table 1)Molar Mass of KIO_3 Concentration of KIO₃ solution: The solution is made up to 250 cm³. C_KIO_3 = Moles of KIO_3Volume of solution (in dm^3) Volume in dm^3 = 250 cm^31000 cm^3/dm^3 = 0.250 dm^3 2. Standardization of Sodium Thiosulphate (Na₂S₂O₃) Solution (Procedure 2): This involves two sequential redox reactions: Reaction 1: Iodate reacts with Iodide in acidic medium to produce Iodine. IO_3^-(aq) + 5I^-(aq) + 6H^+(aq) → 3I_2(aq) + 3H_2O(l) Reaction 2: Iodine is then titrated with Thiosulphate. I_2(aq) + 2S_2O_3^2-(aq) → 2I^-(aq) + S_4O_6^2-(aq) Stoichiometry: From Reaction 1, 1 mole of IO₃⁻ produces 3 moles of I₂. From Reaction 2, 1 mole of I₂ reacts with 2 moles of S₂O₃²⁻. Therefore, 1 mole of IO₃⁻ reacts with 3 × 2 = 6 moles of S₂O₃²⁻. Moles of IO_3^-1 = Moles of S_2O_3^2-6 Calculation of Concentration of Na₂S₂O₃: Moles of S_2O_3^2- = 6 × Moles of IO_3^- C_Na_2S_2O_3 × V_Na_2S_2O_3 = 6 × C_KIO_3 × V_KIO_3 Where V_KIO_3 is the volume of KIO₃ solution used (25 cm³ = 0.025 dm³) and V_Na_2S_2O_3 is the mean titre from Table 2. C_Na_2S_2O_3 = 6 × C_KIO_3 × V_KIO_3V_Na_2S_2O_3 3. Determination of Concentration of Copper (II) Sulphate (CuSO₄) Solution (Procedure 3): This also involves two sequential redox reactions: Reaction 1: Copper(II) ions react with Iodide to produce Iodine. 2Cu^2+(aq) + 4I^-(aq) → 2CuI(s) + I_2(aq) Reaction 2: Iodine is then titrated with Thiosulphate. I_2(aq) + 2S_2O_3^2-(aq) → 2I^-(aq) + S_4O_6^2-(aq) Stoichiometry: From Reaction 1, 2 moles of Cu²⁺ produce 1 mole of I₂. From Reaction 2, 1 mole of I₂ reacts with 2 moles of S₂O₃²⁻. Therefore, 2 moles of Cu²⁺ react with 2 moles of S₂O₃²⁻, which simplifies to a 1:1 molar ratio. Moles of Cu^2+1 =