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 your questions: 8.1.1 Define a base in terms of the Lowry-Brønsted theory. A base is a substance that acts as a proton acceptor. 8.1.2 Write down the NAME or FORMULA of the substance that acts as an ampholyte in the above equations. Give a reason for the answer. Substance: HSO_4^- (Hydrogen sulfate ion) Reason: In equation II, HSO_4^- donates a proton to H_2O (acting as an acid). In the reverse of equation I, HSO_4^- accepts a proton from H_3O^+ to form H_2SO_4 (acting as a base). 8.2.1 Calculate the pH of this nitric acid solution. Given: Concentration of HNO_3 = 0.096 mol·dm^-3. HNO_3 is a strong acid, so it dissociates completely: HNO_3(aq) + H_2O(l) → H_3O^+(aq) + NO_3^-(aq) Step 1: Determine the concentration of hydronium ions. Since HNO_3 is a strong acid, [H_3O^+] = [HNO_3]. [H_3O^+] = 0.096 mol·dm^-3 Step 2: Calculate the pH. pH = -[H_3O^+] pH = -(0.096) pH ≈ 1.02 8.2.2 Calculate the concentration of the EXCESS HNO_3 in the new solution. Balanced equation: 2HNO_3(aq) + Na_2CO_3(aq) → 2NaNO_3(aq) + CO_2(g) + H_2O(l) Given: Volume of HNO_3 (V_A) = 50 cm^3 = 0.050 dm^3 Concentration of HNO_3 (C_A) = 0.096 mol·dm^-3 Volume of Na_2CO_3 (V_B) = 25 cm^3 = 0.025 dm^3 Concentration of Na_2CO_3 (C_B) = 0.075 mol·dm^-3 Step 1: Calculate the initial moles of HNO_3. n(HNO_3)_initial = C_A × V_A n(HNO_3)_initial = (0.096 mol·dm^-3) × (0.050 dm^3) n(HNO_3)_initial = 0.0048 mol Step 2: Calculate the initial moles of Na_2CO_3. n(Na_2CO_3)_initial = C_B × V_B n(Na_2CO_3)_initial = (0.075 mol·dm^-3) × (0.025 dm^3) n(Na_2CO_3)_initial = 0.001875 mol Step 3: Determine the moles of HNO_3 that react with Na_2CO_3. From the balanced equation, the mole ratio HNO_3 : Na_2CO_3 is 2:1. n(HNO_3)_reacted = 2 × n(Na_2CO_3)_initial n(HNO_3)_reacted = 2 × 0.001875 mol n(HNO_3)_reacted = 0.00375 mol Since n(HNO_3)_initial (0.0048 mol) > n(HNO_3)_reacted (0.00375 mol), HNO_3 is in excess. Step 4: Calculate the moles of excess HNO_3. n(HNO_3)_excess = n(HNO_3)_initial - n(HNO_3)_reacted n(HNO_3)_excess = 0.0048 mol - 0.00375 mol n(HNO_3)_excess = 0.00105 mol Step 5: Calculate the total volume of the new solution. V_total = V_A + V_B V_total = 0.050 dm^3 + 0.025 dm^3 V_total = 0.075 dm^3 Step 6: Calculate the concentration of the excess HNO_3. C(HNO_3)_excess = n(HNO_3)_excessV_total C(HNO_3)_excess = 0.00105 mol0.075 dm^3 C(HNO_3)_excess = 0.014 mol·dm^-3 3 done, 2 left today. You're making progress.