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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You're on a roll — My apologies for the visibility issue. Let's re-do the titration calculations with clearer formatting. Titration of 25.0 cm^3 of solution A with 0.100 mol/dm^3 hydrochloric acid. Step 1: Record the burette readings and calculate the average volume of HCl used. | Titration | 1st | 2nd | 3rd | | :--------------- | :-------- | :-------- | :-------- | | Final reading | 24.50 cm^3 | 48.20 cm^3 | 47.00 cm^3 | | Initial reading | 0.00 cm^3 | 24.50 cm^3 | 48.20 cm^3 | | Volume used | 24.50 cm^3 | 23.70 cm^3 | 18.80 cm^3 | The consistent readings are 23.70 cm^3 and 18.80 cm^3. Average volume of HCl used = (23.70 cm^3 + 18.80 cm^3) / 2 Average volume of HCl used = 42.50 cm^3 / 2 Average volume of HCl used = 21.25 cm^3 Step 2: Calculate the number of moles of hydrochloric acid (HCl) used. Formula: Moles = Concentration × Volume (in dm³) Volume of HCl = 21.25 cm^3 = 21.25 / 1000 dm^3 = 0.02125 dm^3 Concentration of HCl = 0.100 mol/dm^3 Moles of HCl = 0.100 mol/dm^3 × 0.02125 dm^3 Moles of HCl = 0.002125 mol Step 3: Write the balanced chemical equation for the reaction. Assuming solution A is sodium carbonate (Na_2CO_3), the reaction with hydrochloric acid is: Na_2CO_3(aq) + 2HCl(aq) → 2NaCl(aq) + H_2O(l) + CO_2(g) Step 4: Calculate the number of moles of solution A in 25.0 cm^3. From the balanced equation, the mole ratio of Na_2CO_3 to HCl is 1:2. Moles of Na_2CO_3 = Moles of HCl / 2 Moles of Na_2CO_3 = 0.002125 mol / 2 Moles of Na_2CO_3 = 0.0010625 mol Step 5: Calculate the concentration of solution A in mol/dm^3. Formula: Concentration = Moles / Volume (in dm³) Volume of solution A = 25.0 cm^3 = 25.0 / 1000 dm^3 = 0.0250 dm^3 Concentration of Na_2CO_3 = 0.0010625 mol / 0.0250 dm^3 Concentration of Na_2CO_3 = 0.0425 mol/dm^3 Step 6: Calculate the concentration of solution A in g/dm^3. First, calculate the molar mass of Na_2CO_3: Molar mass of Na_2CO_3 = (2 × Na) + (1 × C) + (3 × O) Molar mass of Na_2CO_3 = (2 × 22.99) + 12.01 + (3 × 16.00) Molar mass of Na_2CO_3 = 45.98 + 12.01 + 48.00 = 105.99 g/mol Formula: Concentration (g/dm³) = Concentration (mol/dm³) × Molar mass Concentration of Na_2CO_3 = 0.0425 mol/dm^3 × 105.99 g/mol Concentration of Na_2CO_3 = 4.50 g/dm^3 (to 3 significant figures) Step 7: Calculate the percentage purity of sample A. The original concentration of sample A was 5.00 g/dm^3. Percentage purity = (Concentration of pure Na_2CO_3 in g/dm³ / Original concentration of sample A in g/dm³) × 100% Percentage purity = (4.50 g/dm^3 / 5.00 g/dm^3) × 100\% Percentage purity = 0.900 × 100\% Percentage purity = 90.0\% Hope this is much clearer! What's next?