S5 Chemistry – Unit 12: Conductivity of Solutions

12.1

Introduction to Conductivity

Electrolytes and Conductivity Electrolytes are substances that dissolve in water to produce ions, enabling the solution to conduct electricity. The ability of a solution to carry electrical current is called its conductance. Current is carried by the movement of ions — cations move toward the cathode, anions toward the anode.

Strong Electrolytes

Completely dissociate in water into ions. High conductance even at low concentration.

Examples: NaCl, KCl, HCl, H₂SO₄, NaOH, KNO₃, MgCl₂.

NaCl → Na+ + Cl- (100% dissociation)

Weak Electrolytes

Only partially dissociate. Low conductance that increases significantly with dilution.

Examples: CH₃COOH, NH₄OH, HCN, H₂CO₃, HF.

CH3COOH ⇌ CH3COO- + H+ (~1% at 0.1 mol/L)

Non-Electrolytes

Do not produce ions in solution. Cannot conduct electricity. Examples: sucrose, glucose, ethanol, glycerol, urea.

Factors Affecting Conductance of Solutions

Concentration: more ions → greater conductance (up to a point where ion–ion interactions reduce it).
Temperature: higher T → faster ion movement → greater conductance (typically +2% per degree for aqueous solutions).
Nature of electrolyte: charge on ions and degree of dissociation affect conductance.
Ionic charge: higher charge → stronger attraction but also stronger hydration shell → complex effect.
Ionic size/mobility: small, highly mobile ions (H⁺, OH⁻) have exceptionally high conductance due to special transport mechanism (Grotthuss mechanism for H⁺).

12.2

Conductance and Resistance

Basic Electrical Definitions

Resistance (R): R = V/I Units: ohm (Ω) Conductance (G): G = 1/R = I/V Units: siemens (S) = Ω-1 Ohm's Law: V = IR (Voltage = Current x Resistance) For a conductor of length L and cross-sectional area A: R = ρ x L/A (ρ = resistivity, Ω·m) G = κ x A/L (κ = conductivity, S/m)

Measurement of Conductance

DC current cannot be used to measure conductance of electrolyte solutions — it would cause electrolysis (Faradaic current) and change the composition. Instead, an alternating current (AC) bridge (Wheatstone bridge adapted for AC, using a conductance cell) is used.

The conductance cell consists of two platinum electrodes (coated with platinum black to increase surface area) of known dimensions immersed in the solution. The resistance R is measured, then G = 1/R.

Cell Constant (K_cell)

The cell constant is a characteristic of the conductance cell geometry:

K_cell = L/A (units: cm-1 or m-1) Specific conductivity κ = G x K_cell = (1/R) x (L/A)

The cell constant is determined by calibrating the cell with a solution of known conductivity (e.g. 0.100 mol/L KCl, κ = 0.01289 S/cm at 25°C).

12.3

Specific Conductivity (Conductance)

Specific Conductivity (κ) The specific conductivity (or conductance) κ is the conductance of a solution contained in a cube of side 1 cm (or 1 m). It is the reciprocal of resistivity.

κ = G × (L/A) = G × K_cell
Units: S/cm (or S/m = siemens per metre)

Effect of Dilution on Specific Conductivity

For both strong and weak electrolytes, as the solution is diluted (concentration decreases), the specific conductivity decreases because there are fewer ions per unit volume to carry current — even though each individual ion may be moving faster (less interionic attraction).

This seems counterintuitive for weak electrolytes where dilution increases the degree of dissociation α, but the overall decrease in ion concentration still dominates.

Solution	Concentration (mol/L)	κ (S/cm) at 25°C
KCl	1.000	0.11180
KCl	0.100	0.01289
KCl	0.010	0.001412
KCl	0.001	0.0001469
CH₃COOH	0.100	5.2×10⁻⁴
CH₃COOH	0.010	1.63×10⁻⁴
H₂O (pure)	—	5.5×10⁻⁸

12.4

Molar Conductivity (Λ_m)

Molar Conductivity The molar conductivity Λ_m is the conductance of all the ions produced by one mole of electrolyte when the solution is placed between two electrodes 1 cm (or 1 m) apart with sufficient area.

Λ_m = κ / c (where c is in mol/cm³ or mol/m³)
If c is in mol/L: Λ_m = (κ × 1000) / c (units: S·cm²/mol)

Effect of Dilution on Molar Conductivity

For strong electrolytes: Λ_m increases slowly and linearly with dilution (decreasing concentration). At infinite dilution (c → 0), interionic attractions become negligible, and each ion moves independently at maximum speed. The value at infinite dilution is called Λ_m° (molar conductivity at infinite dilution).

For weak electrolytes: Λ_m increases rapidly and steeply with dilution, because dilution greatly increases the degree of dissociation α. The value at infinite dilution cannot be obtained by extrapolation (too steep), so it is calculated using Kohlrausch's law.

Molar conductivity vs √c: strong electrolyte (linear, small change) vs weak electrolyte (steep curve)

Example 1

Calculating Molar Conductivity

The specific conductivity of 0.0500 mol/L KCl at 25°C is 6.30×10⁻³ S/cm. Calculate the molar conductivity Λ_m.

Λ_m = (κ × 1000) / c = (6.30×10⁻³ × 1000) / 0.0500

Λ_m = 6.30 / 0.0500 = 126 S·cm²/mol

✓

Compare with Λ_m°(KCl) = 149.9 S·cm²/mol at infinite dilution — value at 0.05 mol/L is about 84% of the infinite-dilution value.

12.5

Kohlrausch's Law

Kohlrausch's Law of Independent Migration of Ions At infinite dilution, every ion makes a definite and independent contribution to the molar conductivity of the electrolyte, regardless of the nature of the other ions present.

Λ_m° = ν₊λ₊° + ν₋λ₋°

where λ₊° and λ₋° are the limiting molar ionic conductivities of the cation and anion respectively, and ν₊, ν₋ are the stoichiometric coefficients.

Debye–Hückel–Onsager Equation (for strong electrolytes)

At low concentrations, the molar conductivity of a strong electrolyte varies with the square root of concentration:

Λm = Λm° - A√c where A is an empirical constant depending on the electrolyte type, solvent, and temperature. This gives the linear plot of Λm vs √c for strong electrolytes, allowing Λm° to be obtained by extrapolation to c = 0.

Ion	λ° (S·cm²/mol) at 25°C	Notes
H⁺	349.8	Exceptionally high (Grotthuss mechanism)
OH⁻	198.6	Very high (proton hole mechanism)
K⁺	73.5	Similar to Cl⁻ — used in KCl bridges
Cl⁻	76.4	—
Na⁺	50.1	Smaller, more hydrated → less mobile
Ca²⁺	119.0	Higher charge, but also higher hydration
SO₄²⁻	160.0	Divalent; high conductance contribution
CH₃COO⁻	40.9	Large organic ion — slow mobility
NH₄⁺	73.4	Similar to K⁺

Example 2

Kohlrausch's Law — Calculating Λ_m°

Calculate Λ_m° for: (a) KCl; (b) NaCl; (c) NaCH₃COO (sodium ethanoate); (d) CH₃COOH (ethanoic acid).
Given: λ°(K⁺)=73.5, λ°(Na⁺)=50.1, λ°(Cl⁻)=76.4, λ°(H⁺)=349.8, λ°(CH₃COO⁻)=40.9 S·cm²/mol

Λ_m°(KCl) = λ°(K⁺) + λ°(Cl⁻) = 73.5 + 76.4 = 149.9 S·cm²/mol

Λ_m°(NaCl) = λ°(Na⁺) + λ°(Cl⁻) = 50.1 + 76.4 = 126.5 S·cm²/mol

Λ_m°(NaCH₃COO) = λ°(Na⁺) + λ°(CH₃COO⁻) = 50.1 + 40.9 = 91.0 S·cm²/mol

Λ_m°(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = 349.8 + 40.9 = 390.7 S·cm²/mol
Note: Λ_m°(CH₃COOH) = Λ_m°(HCl) + Λ_m°(NaCH₃COO) − Λ_m°(NaCl) — the Kohlrausch combination rule.

12.6

Weak Electrolytes and Degree of Dissociation

Using Conductivity to Find α

For a weak electrolyte at concentration c, the degree of dissociation α at that concentration can be found from:

α = Λm / Λm° where: Λm = molar conductivity at concentration c (measured) Λm° = molar conductivity at infinite dilution (from Kohlrausch) Physical meaning: - At infinite dilution: α = 1 (fully dissociated), Λm = Λm° - At finite concentration: α < 1 (partial dissociation)

Dissociation Constant from Conductivity

For a weak acid HA ⇌ H⁺ + A⁻:

Ka = α²c / (1 - α) (Ostwald dilution law) where α = Λm/Λm° and c is the concentration in mol/L. Substituting: Ka = (Λm/Λm°)² × c / (1 - Λm/Λm°)

Example 3

Finding K_a of Ethanoic Acid from Conductivity

At 25°C, the molar conductivity of 0.100 mol/L CH₃COOH is 5.20 S·cm²/mol. Λ_m°(CH₃COOH) = 390.7 S·cm²/mol. Calculate (a) α and (b) K_a.

α = Λ_m / Λ_m° = 5.20 / 390.7 = 0.01331 (1.33% dissociated at 0.1 mol/L)

K_a = α²c / (1 − α) = (0.01331)² × 0.100 / (1 − 0.01331)

K_a = 1.772×10⁻⁴ × 0.100 / 0.9867 = 1.77×10⁻⁵ / 0.9867 = 1.79×10⁻⁵ mol/L

✓

Literature value: K_a(CH₃COOH) = 1.75×10⁻⁵ mol/L ✓ (agreement is excellent)

12.7

Applications of Conductivity Measurements

Application 1: Conductimetric Titrations

In a conductimetric titration, the conductance of the solution is measured as the titrant is added. The equivalence point is identified by a sharp change in slope of the conductance vs. volume curve — no indicator needed.

Conductimetric titration curves: G vs volume of NaOH added

Interpretation of Conductimetric Titration Curves

Titration	Before EP	After EP	Shape at EP
HCl + NaOH (strong + strong)	G decreases (fast H⁺ replaced by slow Na⁺)	G increases (excess OH⁻, fast)	V-shape minimum at EP
CH₃COOH + NaOH (weak + strong)	G increases slowly (acetate buffer formed)	G increases steeply (excess OH⁻)	Kink/change of slope at EP
NaOH + HCl (base into acid)	G decreases (OH⁻ neutralises H⁺)	G increases (excess HCl?)	V-shape at EP
BaCl₂ + H₂SO₄ (precipitation)	G decreases (Ba²⁺ and SO₄²⁻ removed as BaSO₄)	G increases (excess H⁺)	Minimum at EP

Application 2: Water Purity Testing

The conductivity of water is a measure of its ionic content (dissolved salts). Pure water has extremely low conductivity (~0.055 μS/cm at 25°C). Different water quality standards:

Water Type	Conductivity (μS/cm)	Use
Ultra-pure (Type 1 ASTM)	<0.056	Electronics, HPLC, cell culture
Distilled water	0.5–3	General laboratory
Drinking water (WHO)	<500	Potable
Seawater	~50,000	—
Dead Sea	~270,000	—

Application 3: Degree of Dissociation and K_a/K_b

From conductivity measurements at finite concentration and at infinite dilution (Kohlrausch), we can calculate:

α = Λ_m/Λ_m° (degree of dissociation)
K_a = α²c/(1−α) from Ostwald dilution law
Degree of hydrolysis of salts

Application 4: Solubility of Sparingly Soluble Salts

For sparingly soluble salts (e.g. AgCl, BaSO₄), the solubility is so low that the ions behave ideally and Λ_m ≈ Λ_m°.

κ(saturated solution) = κ(measured) - κ(water) Λm° (from Kohlrausch) c (solubility) = κ_net / Λm° (in mol/cm³) Ksp = c² (for 1:1 salts like AgCl)

Example 4

Solubility of AgCl from Conductivity

The specific conductivity of a saturated AgCl solution at 25°C is 2.67×10⁻⁶ S/cm, and that of the water used is 0.86×10⁻⁶ S/cm. Λ_m°(AgCl) = 138.3 S·cm²/mol. Find the solubility and K_sp of AgCl.

κ_net = (2.67 − 0.86) × 10⁻⁶ = 1.81×10⁻⁶ S/cm

c = κ_net × 1000 / Λ_m° = (1.81×10⁻⁶ × 1000) / 138.3

c = 1.81×10⁻³ / 138.3 = 1.309×10⁻⁵ mol/L = 1.31×10⁻⁵ mol/L

K_sp(AgCl) = [Ag⁺][Cl⁻] = c² = (1.31×10⁻⁵)² = 1.72×10⁻¹⁰ mol²/L²

✓

Literature K_sp(AgCl) = 1.77×10⁻¹⁰ — excellent agreement ✓

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Exercises

Define: (a) specific conductivity κ; (b) molar conductivity Λ_m; (c) limiting molar conductivity Λ_m°. Give the SI unit for each.

(a) Specific conductivity κ: the conductance of a unit cube (1 cm × 1 cm × 1 cm) of solution; κ = G × K_cell. Units: S/cm (or S/m).
(b) Molar conductivity Λ_m: the conductance due to all ions from one mole of electrolyte at a given concentration; Λ_m = (κ × 1000)/c. Units: S·cm²/mol.
(c) Limiting molar conductivity Λ_m°: the molar conductivity extrapolated to infinite dilution (c → 0), where all interionic interactions vanish. Units: S·cm²/mol.
The specific conductivity of 0.0200 mol/L acetic acid at 25°C is 7.37×10⁻⁴ S/cm. Calculate (a) Λ_m; (b) the degree of dissociation α; (c) K_a. Use Λ_m°(CH₃COOH) = 390.7 S·cm²/mol.

(a) Λ_m = (7.37×10⁻⁴ × 1000)/0.0200 = 0.737/0.0200 = 36.85 S·cm²/mol
(b) α = 36.85/390.7 = 0.0943 (9.43% dissociated at 0.02 mol/L)
(c) K_a = α²c/(1−α) = (0.0943)² × 0.0200/(1−0.0943) = 1.780×10⁻⁴/0.9057 = 1.96×10⁻⁴ mol/L
Using Kohlrausch's law, calculate Λ_m° for H₂SO₄. Given: λ°(H⁺)=349.8, λ°(SO₄²⁻)=160.0 S·cm²/mol.

H₂SO₄ → 2H⁺ + SO₄²⁻
Λ_m°(H₂SO₄) = 2λ°(H⁺) + λ°(SO₄²⁻)
= 2(349.8) + 160.0 = 699.6 + 160.0 = 859.6 S·cm²/mol
A conductance cell has a cell constant of 1.25 cm⁻¹. When filled with 0.0100 mol/L KCl solution, its resistance is 98.2 Ω. Calculate (a) G; (b) κ; (c) Λ_m.

(a) G = 1/R = 1/98.2 = 1.018×10⁻² S
(b) κ = G × K_cell = 1.018×10⁻² × 1.25 = 1.273×10⁻² S/cm
(c) Λ_m = (κ × 1000)/c = (1.273×10⁻² × 1000)/0.0100 = 12.73/0.0100 = 1273 S·cm²/mol
Explain why H⁺ and OH⁻ ions have abnormally high ionic conductivities compared to other ions of similar size.

H⁺ and OH⁻ do not actually migrate through solution like normal ions. Instead, they use the Grotthuss (proton hopping) mechanism:

For H⁺: a proton is transferred from one water molecule to the next along a hydrogen-bond chain (H₃O⁺ → H₂O chain), effectively making the charge move much faster than the ion itself could physically travel. This mechanism is far faster than physical ion diffusion.

For OH⁻: proton transfer occurs in the reverse direction along H-bond chains. Both ions can effectively “teleport” charge along water chains rather than physically diffuse through the solvent, giving them exceptionally high ionic conductivities (λ°(H⁺) = 349.8 and λ°(OH⁻) = 198.6 S·cm²/mol vs ~50–80 for most other ions).
Describe a conductimetric titration of HCl with NaOH. Sketch the conductance vs volume curve, identifying the shape before and after the equivalence point and explaining the shape in terms of ionic conductivities.

Before EP: The solution contains H⁺ (very high λ°=349.8) and Cl⁻ (λ°=76.4). As NaOH is added, OH⁻ neutralises H⁺ (forming water). The fast H⁺ ions are replaced by slow Na⁺ ions (λ°=50.1). Since λ°(Na⁺) < λ°(H⁺), the total conductance decreases.

At EP: Only Na⁺ and Cl⁻ remain (NaCl solution). Conductance is at a minimum.

After EP: Excess NaOH adds fast OH⁻ ions (λ°=198.6) and more Na⁺. Conductance increases steeply.

Shape: V-shaped curve with a sharp minimum at the equivalence point. The intersection of the two straight-line portions gives the EP volume.

📝

Unit Test

ℹ️

Instructions Total: 50 marks | Time: 55 minutes | Show all working | State units throughout.

Section A — Short Answer

30 marks

Q1 [4 marks]

A conductance cell has cell constant K_cell = 0.845 cm⁻¹. It is filled with 0.0500 mol/L KCl solution and shows resistance R = 216 Ω.
(a) Calculate G; (b) Calculate κ; (c) Calculate Λ_m; (d) If Λ_m°(KCl) = 149.9 S·cm²/mol, what is the ratio Λ_m/Λ_m°?

(a) G = 1/R = 1/216 = 4.63×10⁻³ S
(b) κ = G × K_cell = 4.63×10⁻³ × 0.845 = 3.91×10⁻³ S/cm
(c) Λ_m = (κ × 1000)/c = (3.91×10⁻³ × 1000)/0.0500 = 3.91/0.0500 = 78.2 S·cm²/mol
(d) Λ_m/Λ_m° = 78.2/149.9 = 0.522 (52.2% of infinite dilution value)

Q2 [5 marks]

State Kohlrausch's law of independent migration of ions. Using it, calculate Λ_m° for:
(a) NaOH (b) H₂SO₄ (c) NH₄Cl (d) Ethanoic acid (CH₃COOH) using the combination rule
Data (S·cm²/mol): λ°(H⁺)=349.8; λ°(OH⁻)=198.6; λ°(Na⁺)=50.1; λ°(Cl⁻)=76.4; λ°(SO₄²⁻)=160.0; λ°(NH₄⁺)=73.4; λ°(CH₃COO⁻)=40.9

Kohlrausch's law: Λ_m° = ν₊λ₊° + ν₋λ₋° (sum of individual ionic contributions at infinite dilution).

(a) Λ_m°(NaOH) = λ°(Na⁺) + λ°(OH⁻) = 50.1 + 198.6 = 248.7 S·cm²/mol
(b) Λ_m°(H₂SO₄) = 2λ°(H⁺) + λ°(SO₄²⁻) = 699.6 + 160.0 = 859.6 S·cm²/mol
(c) Λ_m°(NH₄Cl) = λ°(NH₄⁺) + λ°(Cl⁻) = 73.4 + 76.4 = 149.8 S·cm²/mol
(d) Λ_m°(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = 349.8 + 40.9 = 390.7 S·cm²/mol
Alternative: Λ_m°(HCl) + Λ_m°(NaCH₃COO) − Λ_m°(NaCl) = (349.8+76.4) + (50.1+40.9) − (50.1+76.4) = 426.2 + 91.0 − 126.5 = 390.7 ✓

Q3 [6 marks]

The molar conductivity of 0.0500 mol/L formic acid (HCOOH) at 25°C is 46.1 S·cm²/mol. Λ_m°(HCOOH) = 404.5 S·cm²/mol.
(a) Calculate the degree of dissociation α at 0.0500 mol/L.
(b) Using Ostwald dilution law, calculate K_a for formic acid.
(c) At what concentration would α = 0.50 (50% dissociated) for formic acid?

(a) α = Λ_m/Λ_m° = 46.1/404.5 = 0.1140 (11.4% dissociated)

(b) K_a = α²c/(1−α) = (0.1140)² × 0.0500/(1−0.1140)
= 0.01300 × 0.0500/0.8860
= 6.498×10⁻⁴/0.8860 = 7.33×10⁻⁴ mol/L
(Literature K_a(HCOOH) = 1.77×10⁻⁴ — note: formic acid K_a ≈ 1.8×10⁻⁴, so our calculation gives a reasonable order of magnitude.)

(c) At α = 0.50: K_a = (0.50)²c/(1−0.50) = 0.25c/0.50 = 0.50c
c = K_a/0.50 = 7.33×10⁻⁴/0.50 = 1.47×10⁻³ mol/L (very dilute solution needed for 50% dissociation)

Q4 [5 marks]

The specific conductivity of a saturated solution of BaSO₄ is 3.648×10⁻⁶ S/cm at 25°C. The specific conductivity of the water used is 0.516×10⁻⁶ S/cm.
λ°(Ba²⁺) = 127.3 and λ°(SO₄²⁻) = 160.0 S·cm²/mol.
Calculate: (a) net κ; (b) Λ_m°(BaSO₄); (c) solubility of BaSO₄ in mol/L; (d) K_sp.

(a) κ_net = (3.648 − 0.516) × 10⁻⁶ = 3.132×10⁻⁶ S/cm
(b) Λ_m°(BaSO₄) = λ°(Ba²⁺) + λ°(SO₄²⁻) = 127.3 + 160.0 = 287.3 S·cm²/mol
(c) c = κ_net × 1000 / Λ_m° = (3.132×10⁻⁶ × 1000)/287.3 = 3.132×10⁻³/287.3 = 1.090×10⁻⁵ mol/L
(d) BaSO₄ → Ba²⁺ + SO₄²⁻; K_sp = [Ba²⁺][SO₄²⁻] = c² = (1.090×10⁻⁵)² = 1.19×10⁻¹⁰ mol²/L²

Q5 [5 marks]

Describe, with a labelled graph, the conductimetric titration of a solution of HCl with NaOH. Explain the shape of the curve in terms of ionic conductivities at each stage. Then describe how the conductimetric titration of CH₃COOH with NaOH would differ.

HCl + NaOH:
Before EP: Solution contains H⁺ (λ°=349.8) and Cl⁻ (λ°=76.4). As NaOH is added, H⁺ is neutralised: H⁺ + OH⁻ → H₂O. Fast H⁺ is replaced by slow Na⁺ (λ°=50.1). Total conductance decreases steadily (steep downward slope).
At EP: Only NaCl in solution — minimum conductance.
After EP: Excess NaOH adds fast OH⁻ (λ°=198.6) and Na⁺. Conductance increases steeply.
Shape: V-shaped with minimum at EP.

CH₃COOH + NaOH differences:
Before EP: Weak acid gives few ions initially — low initial conductance. As NaOH is added, CH₃COONa forms (Na⁺ + CH₃COO⁻). Conductance increases slowly (buffer region — not steep).
At EP: CH₃COONa solution — conductance reaches a bend/kink (not a sharp minimum like strong acid).
After EP: Excess NaOH: fast OH⁻ → steep increase.
Key difference: before EP, HCl curve decreases; CH₃COOH curve increases. Both show sharp changes after EP.

Q6 [5 marks]

Explain why the specific conductivity decreases on dilution but the molar conductivity increases for both strong and weak electrolytes. Why is the increase much more dramatic for weak electrolytes?

κ decreases on dilution: κ = G/volume = ions/volume × (ionic mobility). Even though dilution reduces interionic interactions (slightly increasing mobility), the dominant effect is the reduction in ion concentration per unit volume — fewer charge carriers in 1 cm³ of solution. So κ decreases for both strong and weak electrolytes.

Λ_m increases on dilution: Λ_m = κ/c. Although κ decreases, c decreases faster in the denominator → Λ_m increases. For strong electrolytes: the increase is small and linear in √c because interionic interactions (electrostatic retardation) decrease — each ion moves slightly faster. The ions are already fully dissociated, so the change is modest.

Larger increase for weak electrolytes: For weak electrolytes, dilution dramatically increases the degree of dissociation α (Le Chatelier's principle — dilution shifts equilibrium toward products). More molecules ionise → many more ions per mole of electrolyte → Λ_m rises steeply, especially at very low concentrations where α approaches 1.

Section B — Extended Response

20 marks

Q7 [10 marks]

(a) Describe how you would determine the molar conductivity of KCl at various concentrations and extrapolate to find Λ_m°. Include: apparatus, procedure, calculations, and graph. [5 marks]

(b) Explain why the same procedure cannot be used to find Λ_m° of CH₃COOH directly. Describe how Λ_m°(CH₃COOH) is instead calculated using Kohlrausch's law. [5 marks]

(a) Determination of Λ_m°(KCl):
Apparatus: Conductance cell (two platinum black electrodes), AC Wheatstone bridge, thermostat bath at 25°C, volumetric flasks, pipettes.
Procedure: Calibrate cell constant using standard KCl (0.100 mol/L, κ known = 0.01289 S/cm). Prepare KCl solutions at several concentrations (e.g. 0.100, 0.050, 0.020, 0.010, 0.005 mol/L). Measure resistance R at each concentration using AC bridge. Calculate G=1/R, κ=G×K_cell, Λ_m=(κ×1000)/c.
Graph: Plot Λ_m (y-axis) vs √c (x-axis). For KCl (strong electrolyte), get a straight line. Extrapolate to √c = 0 (y-intercept) → Λ_m° = 149.9 S·cm²/mol.

(b) Why not for CH₃COOH: Acetic acid is a weak electrolyte. Its Λ_m vs √c plot is not linear — it curves steeply as c → 0 because α increases rapidly (approaches 1 at very low c). The curve approaches the y-axis nearly vertically, making accurate extrapolation to c=0 impossible by graphical methods.

Kohlrausch calculation: Λ_m°(CH₃COOH) cannot be measured directly, but can be calculated from strong electrolytes whose Λ_m° values are obtainable by extrapolation:
Λ_m°(CH₃COOH) = Λ_m°(HCl) + Λ_m°(CH₃COONa) − Λ_m°(NaCl)
= λ°(H⁺) + λ°(Cl⁻) + λ°(Na⁺) + λ°(CH₃COO⁻) − λ°(Na⁺) − λ°(Cl⁻)
= λ°(H⁺) + λ°(CH₃COO⁻) = 349.8 + 40.9 = 390.7 S·cm²/mol
This uses Kohlrausch's law: ion contributions are additive and independent → the appropriate combination of strong electrolytes gives the desired result.

Q8 [10 marks]

(a) Define and compare strong, weak, and non-electrolytes. Give two examples of each. Explain how their behaviour differs in terms of conductance vs concentration graphs. [4 marks]

(b) The molar conductivity data for formic acid (HCOOH) at 25°C are given below. Calculate α and K_a at each concentration. Comment on the trend in α and constancy of K_a. Λ_m°(HCOOH) = 404.5 S·cm²/mol.
c = 0.100 mol/L: Λ_m = 46.1; c = 0.050 mol/L: Λ_m = 63.6; c = 0.020 mol/L: Λ_m = 97.7 S·cm²/mol. [6 marks]

(a)
Strong electrolytes: completely dissociate in water; high conductance; Λ_m increases slightly with dilution (linear in √c). Examples: NaCl, HCl, H₂SO₄, KOH.
Weak electrolytes: partially dissociate; low conductance at normal concentrations; Λ_m increases steeply with dilution (large change in α). Examples: CH₃COOH, NH₃, HCN, HF.
Non-electrolytes: do not ionise; zero conductance (beyond that of pure water). Examples: glucose, sucrose, ethanol, urea.
Graph: strong electrolytes have moderate κ decreasing smoothly; weak electrolytes start with much lower κ at same c. Non-electrolytes: κ ~ 0.

(b)
At c = 0.100 mol/L:
α = 46.1/404.5 = 0.1140
K_a = (0.1140)²×0.100/(1−0.1140) = 1.300×10⁻³×0.100/0.886 = 1.30×10⁻⁴/0.886 = 1.47×10⁻⁴

At c = 0.050 mol/L:
α = 63.6/404.5 = 0.1572
K_a = (0.1572)²×0.050/(1−0.1572) = 2.471×10⁻³×0.050/0.8428 = 1.236×10⁻⁴/0.8428 = 1.47×10⁻⁴

At c = 0.020 mol/L:
α = 97.7/404.5 = 0.2415
K_a = (0.2415)²×0.020/(1−0.2415) = 5.832×10⁻³×0.020/0.7585 = 1.166×10⁻⁴/0.7585 = 1.54×10⁻⁴

Comment: α increases with dilution (0.114 → 0.157 → 0.242) — confirms Le Chatelier's principle: dilution shifts dissociation equilibrium forward. K_a values are approximately constant (~1.47–1.54 ×10⁻⁴) across the concentration range, confirming it is a true thermodynamic constant (small variation is due to non-ideal activity effects and measurement precision).

← Unit 11: Solutions & Titration Unit 13: Electrolysis →

Conductivity of Solutions

Introduction to Conductivity

Strong Electrolytes

Weak Electrolytes

Non-Electrolytes

Factors Affecting Conductance of Solutions

Conductance and Resistance

Basic Electrical Definitions

Measurement of Conductance

Cell Constant (K_cell)

Specific Conductivity (Conductance)

Effect of Dilution on Specific Conductivity

Molar Conductivity (Λ_m)

Effect of Dilution on Molar Conductivity

Calculating Molar Conductivity

Kohlrausch's Law

Debye–Hückel–Onsager Equation (for strong electrolytes)

Kohlrausch's Law — Calculating Λ_m°

Weak Electrolytes and Degree of Dissociation

Using Conductivity to Find α

Dissociation Constant from Conductivity

Finding K_a of Ethanoic Acid from Conductivity

Applications of Conductivity Measurements

Application 1: Conductimetric Titrations

Interpretation of Conductimetric Titration Curves

Application 2: Water Purity Testing

Application 3: Degree of Dissociation and K_a/K_b

Application 4: Solubility of Sparingly Soluble Salts

Solubility of AgCl from Conductivity

Exercises

Interactive Quiz

Unit 12 Quiz — Conductivity of Solutions

Unit Test

Section A — Short Answer

Section B — Extended Response

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Introduction to Conductivity

Strong Electrolytes

Weak Electrolytes

Non-Electrolytes

Factors Affecting Conductance of Solutions

Conductance and Resistance

Basic Electrical Definitions

Measurement of Conductance

Cell Constant (Kcell)

Specific Conductivity (Conductance)

Effect of Dilution on Specific Conductivity

Molar Conductivity (Λm)

Effect of Dilution on Molar Conductivity

Calculating Molar Conductivity

Kohlrausch's Law

Debye–Hückel–Onsager Equation (for strong electrolytes)

Kohlrausch's Law — Calculating Λm°

Weak Electrolytes and Degree of Dissociation

Using Conductivity to Find α

Dissociation Constant from Conductivity

Finding Ka of Ethanoic Acid from Conductivity

Applications of Conductivity Measurements

Application 1: Conductimetric Titrations

Interpretation of Conductimetric Titration Curves

Application 2: Water Purity Testing

Application 3: Degree of Dissociation and Ka/Kb

Application 4: Solubility of Sparingly Soluble Salts

Solubility of AgCl from Conductivity

Exercises

Interactive Quiz

Unit 12 Quiz — Conductivity of Solutions

Unit Test

Section A — Short Answer

Section B — Extended Response

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Cell Constant (K_cell)

Molar Conductivity (Λ_m)

Kohlrausch's Law — Calculating Λ_m°

Finding K_a of Ethanoic Acid from Conductivity

Application 3: Degree of Dissociation and K_a/K_b