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's the solution for Question 1: Given average titre (volume of acid A used) = 12.30 cm^3 Part (b)(i): Calculate the concentration of B in mol dm^-3. Step 1: Calculate the molar mass of NaOH. Molar mass of NaOH = Na + O + H = 23.0 + 16.0 + 1.00 = 40.0 g/mol Step 2: Calculate the concentration of solution B (NaOH). Solution B contains 1 g of NaOH in 250 cm^3. Moles of NaOH = massmolar mass = 1 g40.0 g/mol = 0.025 mol Volume of solution B = 250 cm^3 = 0.250 dm^3 Concentration of B = molesvolume = 0.025 mol0.250 dm^3 = 0.10 mol dm^-3 Part (b)(ii): Calculate the concentration of A in mol dm^-3. Step 3: Calculate the concentration of solution A (H_2Y) using titration data. The balanced chemical equation is: H_2Y + 2NaOH Na_2Y + 2H_2O From the equation, the mole ratio of H_2Y to NaOH is 1:2. Given: V_A = 12.30 cm^3 = 0.01230 dm^3 V_B = 25 cm^3 = 0.025 dm^3 C_B = 0.10 mol dm^-3 (from Step 2) Using the titration formula (C_A V_A)/(n_A) = (C_B V_B)/(n_B): C_A = (n_A × C_B × V_B)/(n_B × V_A) = 1 × 0.10 mol dm^-3 × 0.025 dm^32 × 0.01230 dm^3 C_A = (0.0025)/(0.0246) mol dm^-3 ≈ 0.1016 mol dm^-3 Part (b)(iii): Calculate the number of hydrogen ions in 1dm^3 of A. Step 4: Calculate the number of hydrogen ions in 1dm^3 of solution A. Concentration of A (C_A) = 0.1016 mol dm^-3. In 1 dm^3 of solution A, moles of H_2Y = 0.1016 mol. Since H_2Y is a diprotic acid, each molecule produces 2 H^+ ions. Moles of H^+ ions in 1 dm^3 = 2 × 0.1016 mol = 0.2032 mol Using Avogadro's number (6.02 × 10^23 ions/mol): Number of H^+ ions = 0.2032 mol × 6.02 × 10^23 ions/mol ≈ 1.22 × 10^23 ions Part (b)(iv): Calculate the volume of distilled water that should be added to 25cm^3 of solution B in order to dilute it twice.