Calculate Grams of Sodium Bicarbonate Easily (Step-by-Step)

Problem:

How many grams of sodium bicarbonate are required to neutralize 10.0 ml of 0.902 M vinegar? (1) 8.4 g (2) 1.5 g (3) 0.75 g (4) 1.07 g

Calculate Grams of Sodium Bicarbonate

To determine the mass of sodium bicarbonate (NaHCO₃) required to neutralize vinegar (acetic acid, CH₃COOH), we use principles of stoichiometry.

Step 1: Balanced Chemical Equation

This is a neutralization reaction:

NaHCO₃ (s) + CH₃COOH (aq) → CH₃COONa (aq) + CO₂ (g) + H₂O (l)

The stoichiometric ratio is 1 : 1.

Step 2: Calculate Moles of Acetic Acid

Given:

• Volume (V) = 10.0 mL = 0.0100 L
• Molarity (M) = 0.902 mol/L

n = M × V = 0.902 × 0.0100 = 0.00902 mol

Step 3: Moles of Sodium Bicarbonate

Since ratio is 1:1:

n(NaHCO₃) = 0.00902 mol

Step 4: Calculate Mass

Molar Mass of NaHCO₃:

• Na = 22.99 g/mol
• H = 1.01 g/mol
• C = 12.01 g/mol
• O₃ = 48.00 g/mol

Total = 84.01 g/mol

Mass = moles × molar mass
= 0.00902 × 84.01 = 0.7578 g

Final Answer

Rounded to three significant figures:
0.758 g ≈ 0.75 g

Correct Option: (3) 0.75 g