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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The table for burette readings is empty. To complete the calculations, I will assume an average titre value for the acid. Assumption: Let the average titre value (volume of acid used) be 22.50 cm^3. Given Information: Solution B: Contains 0.02 mol of sodium hydroxide (NaOH). Volume of solution B (stock) = 250 cm³. Solution A: Tetraoxosulphate (IV) acid (H_2SO_4). Volume of solution B used in titration (V_b) = 25 cm³. Average titre value (volume of acid A used, V_a) = 22.50 cm³ (assumed). Step 1: Calculate the concentration of the base (Solution B). Moles of NaOH = 0.02 mol Volume of stock solution B = 250 cm³ = 0.250 dm³ M_b = Moles of NaOHVolume of solution B (dm^3) = 0.02 mol0.250 dm^3 = 0.08 mol/dm^3 So, the molar concentration of the base (M_b) is 0.08 mol/dm^3. Step 2: Write the balanced chemical equation for the reaction. The reaction is between sulfuric acid (H_2SO_4) and sodium hydroxide (NaOH). H_2SO_4 (aq) + 2NaOH (aq) → Na_2SO_4 (aq) + 2H_2O (l) From the balanced equation: Moles of acid (n_a) = 1 Moles of base (n_b) = 2 --- i) The molar concentration of the acid. Step 1: Convert volumes to dm³. V_a = 22.50 cm^3 = 0.02250 dm^3 V_b = 25 cm^3 = 0.025 dm^3 Step 2: Use the titration formula. (M_aV_a)/(n_a) = (M_bV_b)/(n_b) Rearrange to solve for M_a: M_a = (M_bV_b n_a)/(V_a n_b) Substitute the known values: M_a = (0.08 mol/dm^3) × (0.025 dm^3) × 1(0.02250 dm^3) × 2 M_a = (0.002)/(0.045) M_a = 0.04444... mol/dm^3 Rounding to three significant figures: M_a = 0.0444 mol/dm^3 ii) The concentration mass of the acid. Step 1: Calculate the molar mass of H_2SO_4. Atomic masses: H = 1.01 g/mol, S = 32.07 g/mol, O = 16.00 g/mol MM_H_2SO_4 = (2 × 1.01) + 32.07 + (4 × 16.00) MM_H_2SO_4 = 2.02 + 32.07 + 64.00 = 98.09 g/mol Step 2: Calculate the mass concentration. Mass Concentration = M_a × MM_H_2SO_4 Mass Concentration = 0.04444 mol/dm^3 × 98.09 g/mol Mass Concentration = 4.359 g/dm^3 Rounding to three significant figures: Mass Concentration = 4.36 g/dm^3 iii) From a balanced chemical equation, determine the mole ratio of the base. Step 1: Write the balanced chemical equation. H_2SO_4 (aq) + 2NaOH (aq) → Na_2SO_4 (aq) + 2H_2O (l) Step 2: Determine the mole ratio. The stoichiometric coefficient for the base (NaOH) is 2. The mole ratio of acid to base (H_2SO_4 : NaOH) is 1:2. The mole ratio of the base (NaOH) in the reaction is 2. iv) Sketch the titration curve of the acid-base titration against the pH. This is a titration of a strong base (NaOH) with a strong acid (H_2SO_4). The base is initially in the conical flask, and the acid is added from the burette. Characteristics of the curve: Initial pH: High (basic), as the solution starts with NaOH. Initial pH of 0.08 M NaOH: pOH = -(0.08) = 1.10, so pH = 14 - 1.10 = 12.90. Equivalence point: At pH 7.0, because it's a strong acid-strong base titration. Shape: The pH decreases gradually, then sharply drops around the equivalence point (pH 7), and then continues to decrease gradually as excess acid is added. Final pH: Low (acidic), as excess H_2SO_4 is added. Here is a sketch of the titration curve: [scale=0.8] [->] (0,0) -- (10,0) node[below] Volume of Acid (cm^3); [->] (0,0) -- (0,10) node