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?

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

the answer is d.) potassium

c

explanation: