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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Welcome back TINATTY — missed you this week. Part (a): This part describes the experimental setup for the titration. Solution D (0.100 mol/dm^3 NaOH) is the standard solution, and Solution E (dilute HCl) is the analyte of unknown concentration. Solution E is placed in the burette and titrated against 25.0 cm^3 portions of Solution D using methyl orange indicator. Part (b): To tabulate results and calculate the average volume of acid used, experimental data is required. Since no titration data is provided in the image, we will assume an average titre volume for the acid (Solution E) to demonstrate the subsequent calculations. Assumption: Let the average volume of acid (Solution E) used be 20.00 cm^3. Part (c): Step 1: Write the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl). NaOH (aq) + HCl (aq) NaCl (aq) + H_2O (l) Part (d): Step 1: Identify the known values and the assumed average titre. Concentration of base, C_B = C_D = 0.100 mol/dm^3 Volume of base, V_B = V_D = 25.0 cm^3 Volume of acid, V_A = V_E = 20.00 cm^3 (assumed from part b) From the balanced equation, the mole ratio of acid to base is n_A : n_B = 1:1. Step 2: Use the titration formula to calculate the concentration of the acid (C_A). (C_A V_A)/(n_A) = (C_B V_B)/(n_B) C_A = (C_B V_B n_A)/(V_A n_B) C_A = (0.100 mol/dm^3) × (25.0 cm^3) × 1(20.00 cm^3) × 1 C_A = (2.50)/(20.00) mol/dm^3 C_A = 0.125 mol/dm^3 The concentration of E in mol/dm^3 is 0.125 mol/dm^3. Step 3: Calculate the molar mass of HCl. Molar mass of H = 1.008 g/mol Molar mass of Cl = 35.453 g/mol Molar mass of HCl = 1.008 + 35.453 = 36.461 g/mol Step 4: Convert the concentration from mol/dm^3 to g/dm^3. Concentration in g/dm^3 = Concentration in mol/dm^3 × Molar mass Concentration in g/dm^3 = 0.125 mol/dm^3 × 36.461 g/mol Concentration in g/dm^3 = 4.557625 g/dm^3 The concentration of E in g/dm^3 is 4.56 g/dm^3 (to 3 significant figures). Part (e): Step 1: Determine the mass of pure HCl in 1.0 dm^3 of solution E. From part (d), the concentration of pure HCl is 4.557625 g/dm^3. So, in 1.0 dm^3 of solution, the mass of pure HCl is 4.557625 g. Step 2: Calculate the percentage purity. Given that the impure acid used to prepare the solution was 4.0 g of HCl in 1.0 dm^3. Percentage purity = Mass of pure HClMass of impure HCl × 100\% Percentage purity = 4.557625 g4.0 g × 100\% Percentage purity = 113.940625\% This result (over 100%) indicates that the assumed average titre volume or the given mass of impure HCl might be inconsistent with a real-world scenario where purity cannot exceed 100%. However, based on the numbers provided and calculated, this is the mathematical result. The percentage purity of the acid is 113.9\% (to 1 decimal place). Part (f): (i) Dilution factor Step 1: Identify the initial and final volumes. Initial volume, V_1 = 100 cm^3 Final volume, V_2 = 500 cm^3 Step 2: Calculate the dilution factor. Dilution factor = Final volumeInitial volume Dilution factor = 500 cm^3100 cm^3 Dilution factor = 5 The dilution factor is 5. (ii) Concentration of the new solution in mol/dm^3 Step 1: Use the dilution formula C_1 V_1 = C_2 V_2. Initial concentration, C_1 = 0.125 mol/dm^3 (from part d) Initial volume, V_1 = 100 cm^3 Final volume, V_2 = 500 cm^3 Final concentration, C_2 = ? Step 2: Calculate the new concentration (C_2). C_2 = (C_1 V_1)/(V_2) C_2 = (0.125 mol/dm^3) × (100 cm^3)500 cm^3 C_2 = (12.5)/(500) mol/dm^3 C_2 = 0.025 mol/dm^3 The concentration of the new solution is 0.025 mol/dm^3$. Send me the next one 📸