3. Suppose you had titrated your vinegar sample with barium hydroxide instead of sodium hydroxide:
Ba(OH)2(aq)+2CH3COOH(aq)⟶Ba(CH3COO)2(aq)+2H2O(l)
Consider a 0.586 M aqueous solution of barium hydroxide,
What volume (in mL) of 0.586 M Ba(OH)2 solution are required to neutralize 10 ml of vinegar containing 2.78 g of acetic acid?
Answer:
39.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction
Ba(OH)₂(aq) + 2 CH₃COOH(aq) ⟶ Ba(CH₃COO)₂(aq) + 2 H₂O(l)
Step 2: Calculate the moles corresponding to 2.78 g of CH₃COOH
The molar mass of CH₃COOH is 60.05 g/mol.
2.78 g × 1 mol/60.05 g = 0.0463 mol
Step 3: Calculate the moles of Ba(OH)₂ needed to react with 0.0463 moles of CH₃COOH
The molar ratio of Ba(OH)₂ to CH₃COOH is 1:2. The moles of Ba(OH)₂ needed are 1/2 × 0.0463 mol = 0.0232 mol.
Step 4: Calculate the volume of 0.586 M solution that contains 0.0232 moles of Ba(OH)₂
0.0232 mol × 1 L/0.586 mol = 0.0396 L = 39.6 mL