Why Does the Salt Water Bulb Stay Lit? Conductivity and Bonding
Students build a conductivity tester: a battery, a small bulb, and two graphite electrodes. They test four liquids: distilled water (bulb dim/off), 0.1 M NaCl(aq) (bulb bright), 0.1 M glucose(aq) (bulb off), and glacial acetic acid (bulb dim). A student is confused because glucose and acetic acid both contain O–H groups and both dissolve in water, yet only some solutions conduct. Another student notes that solid NaCl does not conduct but molten NaCl does. The teacher asks them to explain the observations using bonding and intermolecular forces, and to predict what happens if they use 0.1 M CaCl2 instead of NaCl.
1. Explain why 0.1 M NaCl conducts brightly, 0.1 M glucose does not conduct, and 0.1 M acetic acid conducts only dimly, using the concepts of electrolytes and bonding.
Read the worked answer
Conductivity in solution requires mobile charge carriers (ions).
NaCl is an ionic compound; in water it dissociates fully into Na+ and Cl−. A 0.1 M solution gives ~0.2 mol/L of ions, so the bulb is bright - a strong electrolyte.
Glucose is a molecular (covalent) compound. It dissolves by hydrogen bonding to water but does not ionize; the solution contains neutral molecules, no mobile ions, so no current flows - a nonelectrolyte. The presence of O–H groups enables H-bonding (solubility) but not ionization.
Acetic acid is covalent but partially ionizes: CH3COOH ⇌ CH3COO− + H+. At 0.1 M only a small fraction (Ka ≈ 1.8×10⁻⁵) dissociates, so the ion concentration is low (~1.3×10⁻³ M). Few charge carriers → dim bulb. It is a weak electrolyte. The key distinction is degree of ionization, not whether the solute is covalent or ionic per se.
2. Reconcile the observation that solid NaCl does not conduct but molten NaCl does, in terms of structure and mobility of charge carriers.
Read the worked answer
In solid NaCl, ions are locked in a rigid crystal lattice by strong electrostatic (ionic) bonds. The ions are charged but cannot translate, so there is no sustained current - the ions are not mobile.
Melting supplies enough thermal energy to overcome the lattice energy, freeing the ions to move. In the liquid, Na+ and Cl− migrate under an applied field, carrying current. Likewise, dissolving in water separates the ions and hydrates them, giving mobility.
The unifying principle: conduction by ions requires both charge carriers and mobility. Solids with a fixed lattice fail the mobility test; liquids (molten or aqueous) pass. This is why the exam distinguishes 'has ions' from 'has mobile ions.'
3. Predict the relative brightness for 0.1 M CaCl2 versus 0.1 M NaCl, and explain any assumption about ion pairing.
Read the worked answer
CaCl2 dissociates into one Ca²⁺ and two Cl⁻ per formula unit: CaCl2 → Ca²⁺ + 2 Cl⁻. At 0.1 M this gives 0.1 M Ca²⁺ and 0.2 M Cl⁻, total ion concentration 0.3 M versus 0.2 M for NaCl. Because conductivity also depends on ion charge and mobility, the divalent Ca²⁺ contributes more per ion than Na⁺ (roughly, conductivity scales with concentration × charge × mobility). So 0.1 M CaCl2 should glow noticeably brighter than 0.1 M NaCl.
Assumption: we treat both as fully dissociated strong electrolytes. In reality, Ca²⁺ forms ion pairs with Cl⁻ and is less mobile than Na⁺ (larger hydrated radius, higher charge), so the brightness increase is less than a naive 3:2 ion-count ratio predicts. The judgment being tested is that conductivity is not simply proportional to ion count - charge and mobility matter, and ion pairing can mute the effect.