Variation of conductivity with concentration; Kohlrausch's law
On dilution κ falls but Λm rises. Strong electrolytes follow Λm = Λ°m − A√c; for weak ones, Λ°m comes from Kohlrausch's law of independent migration of ions.
What happens on dilution
Diluting a solution lowers κ, because there are fewer ions in each unit volume. But it raises Λm. Λm = κ × V, where V is the volume that contains 1 mol of electrolyte, and V grows faster than κ falls. As c approaches zero, Λm reaches a limiting value called the limiting molar conductivity, Λ°m. Strong and weak electrolytes reach it in very different ways.
Strong vs weak electrolytes on dilution
How to draw the graph of Λm against √c
- Draw √c on the x-axis and Λm (S cm² mol⁻¹) on the y-axis.
- Strong electrolyte (KCl): a straight line sloping gently downward from left to right. Extend it as a dotted line to the y-axis and label the meeting point Λ°m. The slope is −A.
- Weak electrolyte (CH₃COOH): a curve that is very high close to the y-axis, falls steeply, and then runs low and nearly flat at larger √c.
- Label both curves and both axes.
Kohlrausch's law of independent migration of ions
At infinite dilution, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and its anion. Each ion makes its own fixed contribution, whatever its partner. Λ°m = ν₊λ°₊ + ν₋λ°₋, where ν₊ and ν₋ are the numbers of cations and anions in one formula unit. Some λ° values at 298 K, in S cm² mol⁻¹: H⁺ 349.6, Na⁺ 50.1, K⁺ 73.5, Ca²⁺ 119.0, Mg²⁺ 106.0, OH⁻ 199.1, Cl⁻ 76.3, CH₃COO⁻ 40.9, SO₄²⁻ 160.0. For example, Λ°m(CaCl₂) = 119.0 + 2 × 76.3 = 271.6 S cm² mol⁻¹. Uses: to find Λ°m of a weak electrolyte, and from it the degree of dissociation and the dissociation constant.
Worked example: Λ°m of ethanoic acid
At 298 K, Λ°m of HCl, CH₃COOK and KCl are 425.9, 114.4 and 149.8 S cm² mol⁻¹. Find Λ°m of CH₃COOH.
- Kohlrausch's law: Λ°m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻).
- Add and subtract so that only these two ions remain: Λ°m(CH₃COOH) = Λ°m(HCl) + Λ°m(CH₃COOK) − Λ°m(KCl).
- Check the ions: (H⁺ + Cl⁻) + (K⁺ + CH₃COO⁻) − (K⁺ + Cl⁻) = H⁺ + CH₃COO⁻.
- Λ°m(CH₃COOH) = 425.9 + 114.4 − 149.8 = 390.5 S cm² mol⁻¹.
Λ°m(CH₃COOH) = 390.5 S cm² mol⁻¹
Worked example: degree of dissociation of a weak acid
A 0.002 M solution of a weak monobasic acid HA has κ = 4.0 × 10⁻⁵ S cm⁻¹ at 298 K. Λ°m(HA) = 400 S cm² mol⁻¹. Find Λm, the degree of dissociation α and the dissociation constant Ka.
- Λm = (κ × 1000) ÷ c = (4.0 × 10⁻⁵ × 1000) ÷ 0.002 = (4.0 × 10⁻²) ÷ (2 × 10⁻³) = 20 S cm² mol⁻¹.
- α = Λm ÷ Λ°m = 20 ÷ 400 = 0.05.
- Ka = cα² ÷ (1 − α) = (0.002 × 0.05²) ÷ 0.95 = (0.002 × 0.0025) ÷ 0.95.
- Ka = (5.0 × 10⁻⁶) ÷ 0.95 = 5.3 × 10⁻⁶.
Λm = 20 S cm² mol⁻¹; α = 0.05 (5% dissociated); Ka ≈ 5.3 × 10⁻⁶
Vinegar conducts electricity only weakly compared with salt water of the same strength, because ethanoic acid is a weak electrolyte with a low Λm. The weak acids in a lemon are one reason a lemon battery gives only a feeble current.
2 marks: “State Kohlrausch's law of independent migration of ions, and give one application.” At infinite dilution, the limiting molar conductivity of an electrolyte equals the sum of the limiting molar conductivities of its cation and anion. It is used to find Λ°m of a weak electrolyte such as CH₃COOH, and from it the degree of dissociation.
3 marks: “Why does Λm of CH₃COOH rise steeply on dilution while that of KCl rises only slightly? Show this on a graph.” KCl is fully ionised; dilution only weakens the attraction between ions, giving a gentle straight line against √c. CH₃COOH ionises more on dilution, so the number of ions per mole rises sharply, giving a steep curve near c = 0.
1 marks: “Calculate Λ°m of CaCl₂, given λ°(Ca²⁺) = 119.0 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹.” Λ°m = 119.0 + 2 × 76.3 = 271.6 S cm² mol⁻¹.
Dilute it: kappa drops, lambda lifts. Kohlrausch: every ion pays its own share.