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 calculations for part (b). Part (b): Calculate the concentration (mol/dm³) of the dichromate ions, Cr₂O₇²⁻, in solution M, using the titre t₂. Step 1: Calculate the moles of Na₂S₂O₃ used in the titration (t₂). The mean titre (t₂) = 30.55 cm³ = 0.03055 dm³ The concentration of Na₂S₂O₃ (from part a) = 0.3 mol/dm³ Moles of Na_2S_2O_3 = Concentration × Volume = 0.3 mol/dm^3 × 0.03055 dm^3 = 0.009165 mol Step 2: Determine the moles of I₂ that reacted with Na₂S₂O₃. The reaction is: I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq) From the stoichiometry, 1 mole of I₂ reacts with 2 moles of S₂O₃²⁻. Moles of I_2 = (1)/(2) × Moles of Na_2S_2O_3 = (1)/(2) × 0.009165 mol = 0.0045825 mol Step 3: Determine the moles of Cr₂O₇²⁻ that produced the I₂. The reaction is: Cr₂O₇²⁻(aq) + 6I⁻(aq) + 14H⁺(aq) → 3I₂(aq) + 2Cr³⁺(aq) + 7H₂O(l) From the stoichiometry, 1 mole of Cr₂O₇²⁻ produces 3 moles of I₂. Moles of Cr_2O_7^2- = (1)/(3) × Moles of I_2 = (1)/(3) × 0.0045825 mol = 0.0015275 mol Step 4: Calculate the concentration of Cr₂O₇²⁻ in solution M. The volume of solution M used for titration was 25 cm³ = 0.025 dm³. Concentration of Cr_2O_7^2- = MolesVolume = 0.0015275 mol0.025 dm^3 = 0.0611 mol/dm^3 The concentration of dichromate ions, Cr₂O₇²⁻, in solution M is 0.0611 mol/dm^3. 3 done, 2 left today. You're making progress.