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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My apologies, Ruddy. I missed those. Let's get them solved. Titration of 25.0 cm^3 of solution A with 0.100 mol/dm^3 hydrochloric acid. Step 1: Record the burette readings. | Titration | 1st | 2nd | 3rd | | :--------------- | :----- | :----- | :----- | | Final reading | 24.50 | 48.20 | 47.00 | | Initial reading | 0.00 | 24.50 | 48.20 | | Volume used | 24.50 | 23.70 | 18.80 | Step 2: Choose consistent readings. The consistent readings are 23.70 cm^3 and 18.80 cm^3. The average volume of hydrochloric acid used is (23.70 + 18.80) / 2 = 21.25 cm^3. Step 3: Calculate the number of moles of hydrochloric acid used. Formula: moles = concentration × volume (in dm³) Volume of HCl = 21.25 cm^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 = 0.002125 mol Step 4: Write the balanced equation for the reaction between solution A and hydrochloric acid. Assuming solution A is sodium carbonate (Na_2CO_3): Na_2CO_3(aq) + 2HCl(aq) → 2NaCl(aq) + H_2O(l) + CO_2(g) Step 5: 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 = 0.0010625 mol Step 6: Calculate the concentration of solution A in mol/dm^3. Concentration = moles / volume (in dm³) Volume of solution A = 25.0 cm^3 = 0.0250 dm^3 Concentration of Na_2CO_3 = 0.0010625 mol / 0.0250 dm^3 = 0.0425 mol/dm^3 Step 7: Calculate the concentration of solution A in g/dm³ (assuming solution A is sodium carbonate). Molar mass of Na_2CO_3 = (2 × 22.99) + 12.01 + (3 × 16.00) = 45.98 + 12.01 + 48.00 = 105.99 g/mol Concentration in g/dm³ = concentration in mol/dm³ × molar mass Concentration of Na_2CO_3 = 0.0425 mol/dm^3 × 105.99 g/mol = 4.50 g/dm^3 (to 3 significant figures) Step 8: Calculate the percentage purity of sample A. Percentage purity = (Concentration of 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\% = 90.0\% The concentration of solution A is 0.0425 mol/dm^3 or 4.50 g/dm^3. The percentage purity of sample A is 90.0\%. 3 done, 2 left today. You're making progress. Send the next one 📸