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.