Neutralising an Acid Spill: When Is It Done?
A technician spills 50 mL of dilute sulfuric acid, approximately 0.10 mol/L, on a bench. She sprinkles sodium hydrogencarbonate onto the puddle. The mixture fizzes vigorously and the bench warms noticeably. After three minutes a fresh pH strip touched to the liquid reads pH 5. She keeps adding solid until fizzing stops; a new strip then reads pH 8. A conductivity meter shows the reading falling during the first additions but never reaching zero. A colleague says the acid is now fully neutralised because the pH is above 7, and that the remaining conductivity is simply from sodium sulfate. The technician is not convinced and asks whether the endpoint is really complete neutralisation, what the fizzing and warming show, and why the conductivity behaves as it does.
1. Write the balanced equation for the reaction and explain the fizzing and warming in terms of ions and energy.
Read the worked answer
Balanced equation:
H₂SO₄(aq) + 2NaHCO₃(s) → Na₂SO₄(aq) + 2CO₂(g) + 2H₂O(l)
Net ionic equation:
H⁺(aq) + HCO₃⁻(aq) → CO₂(g) + H₂O(l)
The fizzing is carbon dioxide escaping from solution. The warming is because the neutralisation is exothermic: H⁺ and HCO₃⁻ react to form water and CO₂, releasing energy to the surroundings. Sulfuric acid is diprotic; with excess bicarbonate, both acidic protons are consumed.
2. Calculate the amount of sodium hydrogencarbonate needed to neutralise the acid completely. Does pH 8 indicate excess base, incomplete reaction, or both?
Read the worked answer
Amount of acid:
- n(H₂SO₄) = 0.0500 L × 0.100 mol/L = 5.0 × 10⁻³ mol.
- Complete neutralisation needs 2 × 5.0 × 10⁻³ = 1.0 × 10⁻² mol NaHCO₃.
- Mass = 0.010 mol × 84 g/mol = 0.84 g.
If fizzing has stopped and pH is 8, the acid has been consumed. pH above 7 indicates excess hydrogencarbonate, which is weakly basic:
HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻
So the acid is neutralised in the sense that no significant acid remains, but the solution is not neutral. The endpoint is basic because of excess bicarbonate, not because the reaction stopped early.
3. Why does conductivity fall but not reach zero? Is the colleague's explanation adequate?
Read the worked answer
Initially the solution contains H⁺ and SO₄²⁻ from H₂SO₄. H⁺ has very high ionic mobility, so conductivity is high. Adding NaHCO₃ replaces H⁺ with Na⁺ and forms CO₂ and water. Na⁺ is much less mobile than H⁺, so conductivity falls.
It cannot reach zero because Na₂SO₄ is a strong electrolyte:
Na₂SO₄ → 2Na⁺ + SO₄²⁻
Excess NaHCO₃ also adds ions. Per formula unit, H₂SO₄ and Na₂SO₄ both give three ions, but the mobility difference explains the drop. The colleague is partly right that sodium sulfate keeps conductivity nonzero, but excess bicarbonate contributes too, and pH above 7 does not mean zero ions.