Neutralizing an Acid Spill in the Lab
During a practical, a student spills 500 mL of 0.10 mol/L hydrochloric acid (HCl) on a bench. The instructor has a bottle of sodium hydroxide solution labeled 0.20 mol/L and a container of solid sodium bicarbonate (NaHCO₃, molar mass 84 g/mol). The safety protocol requires neutralizing the acid to pH near 7 before cleanup, and forbids leaving excess strong base on the bench. A student suggests simply pouring in the entire 0.20 mol/L NaOH bottle (which holds 300 mL) and then wiping up.
1. Calculate the moles of HCl spilled and evaluate the student's proposal quantitatively.
Read the worked answer
Moles HCl: n = C × V = 0.10 mol/L × 0.500 L = 0.050 mol.
Moles NaOH available: 0.20 mol/L × 0.300 L = 0.060 mol.
The reaction HCl + NaOH → NaCl + H₂O is 1:1. Adding all 0.060 mol NaOH to 0.050 mol HCl leaves 0.010 mol NaOH unreacted - an excess of strong base. This violates the protocol (no excess strong base) and leaves a caustic residue.
Verdict: The proposal over-neutralizes. Only 0.050 mol NaOH (250 mL of the 0.20 mol/L solution) is needed for exact neutralization.
2. Compare using solid NaHCO₃ as an alternative and state which approach better satisfies the safety protocol.
Read the worked answer
Reaction: NaHCO₃ + HCl → NaCl + H₂O + CO₂ (1:1 with HCl).
Moles NaHCO₃ needed = 0.050 mol.
Mass = 0.050 mol × 84 g/mol = 4.2 g.
Advantages: NaHCO₃ is a weak base; a slight excess does not create a strongly caustic residue, and the CO₂ evolution gives a visible end-point cue. Cautions: effervescence can splash acid, so add slowly.
Verdict: Adding ~4.2 g NaHCO₃ gradually better satisfies the protocol: it neutralizes to near pH 7 and tolerates slight excess without leaving strong base. The NaOH route is acceptable only if measured to exactly 250 mL, which is harder to control.