Chemistry Lab: Neutralizing Nitric Acid Spill with Sodium Bicarbonate
How to neutralize a spilled nitric acid in the lab with sodium bicarbonate?
A student has accidentally spilled 100.0 mL of 3.0 mol/L nitric acid onto the lab bench. What mass of sodium bicarbonate would the teacher need to sprinkle on this spill to neutralize and clean it up?
Answer:
25 g
To neutralize the 100.0 mL of 3.0 mol/L nitric acid spill, the teacher would need 50.0 mL of sodium bicarbonate with the same molar concentration.
Explanation:
Step 1: Write the balanced equation
HNO₃ + NaHCO₃ ⇒ NaNO₃ + H₂O + CO₂
Step 2: Calculate the reacting moles of HNO₃
100.0 mL of 3.0 mol/L HNO₃ reacted.
0.1000 L × 3.0 mol/L = 0.30 mol
Step 3: Calculate the reacting moles of NaHCO₃
The molar ratio of HNO₃ to NaHCO₃ is 1:1. The reacting moles of NaHCO₃ are 1/1 × 0.30 mol = 0.30 mol.
Step 4: Calculate the mass corresponding to 0.30 moles of NaHCO₃
The molar mass of NaHCO₃ is 84.01 g/mol.
0.30 mol × 84.01 g/mol = 25 g
Final answer:
To neutralize the 100.0 mL of 3.0 mol/L nitric acid spill, the teacher would need 50.0 mL of sodium bicarbonate with the same molar concentration.
Explanation:
To neutralize the spill of 100.0 mL of 3.0 mol/L nitric acid, the teacher would need to sprinkle sodium bicarbonate (NaHCO₃) on it. The balanced chemical equation for the neutralization reaction between nitric acid and sodium bicarbonate is:
2HNO₃(aq) + NaHCO₃(aq) → NaNO₃(aq) + CO₂(g) + H₂O(l)
From this balanced equation, we can see that 2 moles of nitric acid react with 1 mole of sodium bicarbonate. Therefore, to neutralize 1 mL of 3.0 mol/L nitric acid, we would need 0.5 mL of sodium bicarbonate with the same molar concentration.
Using this information, to neutralize the 100.0 mL of 3.0 mol/L nitric acid, the teacher would need:
0.5 mL/mL × 100.0 mL/mL = 50.0 mL of sodium bicarbonate with the same molar concentration.