Resistance, conductance and conductivity
Metals conduct with electrons. A solution of an electrolyte (a substance that splits into ions in water) conducts with ions. The ions themselves move.
A solution obeys Ohm's law, so it has a resistance R. Like a wire, R = ρ × l/A. Here l is the gap between the plates, A is their area, and ρ (rho) is the resistivity.
- Conductance G = 1/R. Unit: siemens (S) = Ω⁻¹.
- Conductivity κ (kappa) = 1/ρ. It is the conductance of 1 cm³ (or 1 m³) of solution between plates 1 cm (or 1 m) apart. Unit: S cm⁻¹ or S m⁻¹. 1 S m⁻¹ = 0.01 S cm⁻¹.
Conductivity depends on the nature of the electrolyte, the size and charge of the ions, the solvent, the temperature and the concentration.
Measuring conductivity: the cell constant
A conductivity cell has two platinum plates coated with platinum black. We measure R with an AC bridge (Wheatstone bridge with a detector). We use AC, not DC, so that the solution does not get electrolysed while we measure.
Measuring l and A exactly is hard. So we find the cell constant G* = l/A using a standard KCl solution whose κ is known:
G* = κ × R and then, for any other solution in the same cell, κ = G*/R.
Molar conductivity
Two solutions of different strength cannot be compared by κ alone. So we use molar conductivity Λm: the conductivity of a volume of solution that has 1 mole of the electrolyte, between plates 1 cm apart.
Λm = κ / c. If κ is in S cm⁻¹ and c in mol L⁻¹, use Λm = κ × 1000 / c and the unit is S cm² mol⁻¹. (1 S cm² mol⁻¹ = 10⁻⁴ S m² mol⁻¹.)
Change of conductivity with concentration
When we dilute a solution:
- κ decreases, because there are fewer ions in each cm³.
- Λm increases, because the volume holding one mole gets bigger and the ions have more room.
Strong electrolytes (KCl, NaCl, HCl)
They are fully split into ions at all concentrations. On dilution, ions are farther apart and pull on each other less, so they move faster. Λm rises slowly. A graph of Λm against √c is almost a straight line:
Λm = Λ°m − A√c
Extend the line to c = 0 to read Λ°m, the limiting molar conductivity. The slope A depends on the type of salt (1-1, 2-1 etc.).
Weak electrolytes (CH₃COOH, NH₄OH)
Only a small part splits into ions. Dilution makes much more of it split, so Λm rises very sharply at low c. The graph is a steep curve, and we cannot extend it to read Λ°m. Kohlrausch's law solves this.
Kohlrausch's law of independent migration of ions
At infinite dilution, every ion moves on its own and adds a fixed share to the molar conductivity.
Λ°m = ν₊ λ°₊ + ν₋ λ°₋, where ν₊ and ν₋ are the numbers of cations and anions in one formula unit.
Example: Λ°m(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻).
Uses of Kohlrausch's law
- Λ°m of a weak electrolyte from strong ones: Λ°m(CH₃COOH) = Λ°m(HCl) + Λ°m(CH₃COONa) − Λ°m(NaCl).
- Degree of dissociation: α = Λm / Λ°m.
- Dissociation constant of a weak acid: Ka = cα² / (1 − α).
Try it: salt makes water conduct
Fill a glass with distilled (or RO) water and test it with a battery, two wires and an LED: it barely glows. Add a pinch of salt and stir: it glows brighter. Add more salt: brighter still (κ goes up). In the 3D, predict first: when you slide c down for CH₃COOH, will Λm rise a little or a lot? Then check the dot.
Key formulas and definitions
- R = ρ l/A ; G = 1/R ; κ = 1/ρ
- Cell constant G* = l/A = κ × R
- κ = G*/R
- Λm = κ × 1000 / c (κ in S cm⁻¹, c in mol L⁻¹)
- Λm = Λ°m − A√c (strong electrolytes)
- Λ°m = ν₊λ°₊ + ν₋λ°₋ (Kohlrausch)
- α = Λm/Λ°m ; Ka = cα²/(1 − α)
Worked examples
1. A cell has plates 1 cm apart, each of area 2 cm². Its resistance with a solution is 50 Ω. Find G* and κ.
Step 1: G* = l/A = 1/2 = 0.5 cm⁻¹. Step 2: κ = G*/R = 0.5/50. Step 3: κ = 0.01 S cm⁻¹.
2. 0.1 M KCl (κ = 0.0129 S cm⁻¹) shows R = 100 Ω in a cell. Find the cell constant.
Step 1: G* = κ × R. Step 2: G* = 0.0129 × 100. Step 3: G* = 1.29 cm⁻¹.
3. κ of 0.1 M KCl is 0.0129 S cm⁻¹. Find its molar conductivity.
Step 1: Λm = κ × 1000 / c. Step 2: Λm = 0.0129 × 1000 / 0.1 = 12.9 / 0.1. Step 3: Λm = 129 S cm² mol⁻¹.
4. A 0.01 M solution in a cell of G* = 1.5 cm⁻¹ has R = 1500 Ω. Find κ and Λm.
Step 1: κ = G*/R = 1.5/1500 = 0.001 S cm⁻¹. Step 2: Λm = 0.001 × 1000 / 0.01. Step 3: Λm = 100 S cm² mol⁻¹.
5. λ°(Na⁺) = 50.1 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹. Find Λ°m(NaCl).
Step 1: Kohlrausch: Λ°m = λ°₊ + λ°₋. Step 2: Λ°m = 50.1 + 76.3. Step 3: Λ°m(NaCl) = 126.4 S cm² mol⁻¹.
6. Λ°m of HCl, CH₃COONa and NaCl are 426.1, 91.0 and 126.4 S cm² mol⁻¹. Find Λ°m(CH₃COOH).
Step 1: CH₃COOH = HCl + CH₃COONa − NaCl (the Na⁺ and Cl⁻ cancel). Step 2: Λ°m = 426.1 + 91.0 − 126.4. Step 3: Λ°m(CH₃COOH) = 390.7 S cm² mol⁻¹.
7. 0.001 M acetic acid has Λm = 39.05 S cm² mol⁻¹; Λ°m = 390.5 S cm² mol⁻¹. Find α and Ka.
Step 1: α = Λm/Λ°m = 39.05/390.5 = 0.1. Step 2: Ka = cα²/(1 − α) = 0.001 × 0.01 / 0.9. Step 3: Ka = 1.11 × 10⁻⁵.
8. Find Λ°m(CaCl₂) given λ°(Ca²⁺) = 119.0 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹.
Step 1: One CaCl₂ gives 1 Ca²⁺ and 2 Cl⁻. Step 2: Λ°m = 119.0 + 2 × 76.3 = 119.0 + 152.6. Step 3: Λ°m = 271.6 S cm² mol⁻¹.
Common mistakes
- Thinking conductivity κ rises on dilution. κ falls (fewer ions per cm³); it is Λm that rises.
- Forgetting the 1000 in Λm = κ × 1000/c when c is in mol L⁻¹ and κ in S cm⁻¹.
- Trying to read Λ°m of a weak electrolyte by extending its graph. The curve is too steep; use Kohlrausch's law.
- Forgetting the ion count in Kohlrausch's law: CaCl₂ needs 2 × λ°(Cl⁻).