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. A: Define i) A standard solution? A standard solution is a solution of accurately known concentration. It is typically prepared by dissolving a precise mass of a primary standard solute in a precisely known volume of solvent. ii) A disproportionation reaction? A disproportionation reaction is a type of redox reaction in which a single element is simultaneously oxidized and reduced. This means the element acts as both the oxidizing agent and the reducing agent. 1. B: Calculation based on titration i) What is the indicator that can be used to mark the endpoint of this reaction? The indicator used is starch solution. ii) Indicate the colour at the end point The colour at the endpoint is blue-black. iii) Calculate the molarity of the KIO_3 solution Step 1: Calculate the molar mass of KIO_3. M(KIO_3) = K + I + 3 × O = 39 + 127 + 3 × 16 = 214 g/mol Step 2: Calculate the moles of KIO_3. Moles of KIO_3 = MassMolar mass = 0.214 g214 g/mol = 0.001 mol Step 3: Calculate the molarity of the KIO_3 solution. The volume of the solution is 250 ml = 0.250 dm^3. Molarity of KIO_3 = MolesVolume = 0.001 mol0.250 dm^3 = 0.004 mol/dm^3 iv) Write and balance the half equations and hence give the balanced ionic equation for the reaction A) Reduction half-equation The oxidation state of iodine in IO_3^- is +5, and in I_2 it is 0. So, IO_3^- is reduced to I_2. 10e^- + 12H^+ + 2IO_3^-(aq) → I_2(aq) + 6H_2O(l) B) Oxidation half-equation The oxidation state of iodine in I^- is -1, and in I_2 it is 0. So, I^- is oxidized to I_2. 2I^-(aq) → I_2(aq) + 2e^- C) Overall balanced ionic equation To balance the electrons, multiply the oxidation half-equation by 5: 10I^-(aq) → 5I_2(aq) + 10e^- Add the reduction half-equation and the modified oxidation half-equation: (10e^- + 12H^+ + 2IO_3^-) + (10I^-) → (I_2 + 6H_2O) + (5I_2 + 10e^-) 12H^+ + 2IO_3^- + 10I^- → 6I_2 + 6H_2O Divide by 2 to simplify: 6H^+(aq) + IO_3^-( ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._