How Solutions Carry Current

Metals conduct electricity by the flow of electrons. Electrolyte solutions conduct by the flow of ions. This section is about measuring and understanding that ionic conduction — a topic that delivers reliable numerical marks.

Start from Ohm's law. A conductor of resistance RR obeys V=IRV = IR. Resistance depends on the conductor's length ll and cross-sectional area AA:

R=ρlAR = \rho\frac{l}{A}

where ρ\rho (rho) is the resistivity (units: Ω\Omega m). Its reciprocal is the conductivity κ\kappa (kappa):

κ=1ρ\kappa = \frac{1}{\rho}

Conductivity has units S m⁻¹ (siemens per metre), though chemists often use S cm⁻¹. Useful conversion: 1 S cm−1=100 S m−11\text{ S cm}^{-1} = 100\text{ S m}^{-1}.

Conductivity (κ\kappa) is the conductance of a solution contained in a cube of side 1 m (or 1 cm), i.e. the conductance of a solution of unit length and unit cross-sectional area.

Cell Constant and Conductance

Conductance GG is the reciprocal of resistance: G=1/RG = 1/R, units siemens (S).

Combining R=ρl/A=l/(κA)R = \rho l/A = l/(\kappa A):

κ=1R⋅lA=G⋅G∗\kappa = \frac{1}{R}\cdot\frac{l}{A} = G \cdot G^*

The geometric factor G∗=lAG^* = \dfrac{l}{A} is the cell constant (units: m⁻¹ or cm⁻¹). It depends only on the electrode geometry, so we measure it once using a solution of known conductivity (standard KCl), then use it for all subsequent measurements:

κ=cell constantresistance=G∗R\kappa = \frac{\text{cell constant}}{\text{resistance}} = \frac{G^*}{R}

Conductivity cell measured with a Wheatstone bridge

Molar Conductivity

Conductivity κ\kappa alone doesn't fairly compare electrolytes, because it depends on how many ions are present. To compare on a per-mole basis we use molar conductivity Λm\Lambda_m (lambda):

Λm=κc\Lambda_m = \frac{\kappa}{c}

where cc is the concentration. When κ\kappa is in S cm⁻¹ and cc in mol L⁻¹ (mol dm⁻³), the working formula is:

Λm=κ×1000c\boxed{\Lambda_m = \frac{\kappa \times 1000}{c}}

with Λm\Lambda_m in S cm² mol⁻¹. (The 1000 converts litres to cm³: 1 L = 1000 cm³.)

Key Point: Molar conductivity is the conductance contributed by all the ions produced by 1 mole of electrolyte when the solution is considered appropriately diluted; it allows fair comparison of electrolytes.

[JEE Tip] Mind the units. If κ\kappa is given in S m⁻¹ and cc in mol m⁻³, then Λm=κ/c\Lambda_m = \kappa/c directly in S m² mol⁻¹. The factor of 1000 only appears in the S cm⁻¹ / mol L⁻¹ version. Getting this conversion wrong is the single most common error in conductance numericals.

Solved Examples

Example 1: Conductivity from resistance and cell constant

A conductivity cell with cell constant 1.25 cm⁻¹ has a resistance of 100 Ω when filled with a solution. Find the conductivity.

Solution: κ=G∗R=1.25100=0.0125\kappa = \dfrac{G^*}{R} = \dfrac{1.25}{100} = 0.0125 S cm⁻¹.

Answer: 0.0125 S cm⁻¹.

Example 2: Molar conductivity

The conductivity of a 0.20 M KCl solution at 298 K is 0.0248 S cm⁻¹. Calculate its molar conductivity.

Solution: Λm=κ×1000c=0.0248×10000.20=124\Lambda_m = \dfrac{\kappa\times1000}{c} = \dfrac{0.0248\times1000}{0.20} = 124 S cm² mol⁻¹.

Answer: 124 S cm² mol⁻¹.

Example 3: Cell constant from a known solution

A cell filled with 0.1 M KCl (conductivity 1.29 S m⁻¹) has resistance 100 Ω. Find the cell constant.

Solution: κ=G∗/R⇒G∗=κ×R=1.29 S m−1×100 Ω=129\kappa = G^*/R \Rightarrow G^* = \kappa\times R = 1.29\text{ S m}^{-1}\times100\,\Omega = 129 m⁻¹.

Answer: 129 m⁻¹ (= 1.29 cm⁻¹).

Example 4: Conductance to conductivity

A solution in a cell of cell constant 0.5 cm⁻¹ has conductance 2×10−32\times10^{-3} S. Find its conductivity.

Solution: κ=G×G∗=2×10−3×0.5=1×10−3\kappa = G\times G^* = 2\times10^{-3}\times0.5 = 1\times10^{-3} S cm⁻¹.

Answer: 1×10−31\times10^{-3} S cm⁻¹.

Example 5: Molar conductivity with unit care

The conductivity of 0.001 M acetic acid is 4.95×10−54.95\times10^{-5} S cm⁻¹. Find its molar conductivity.

Solution: Λm=κ×1000c=4.95×10−5×10000.001=49.5\Lambda_m = \dfrac{\kappa\times1000}{c} = \dfrac{4.95\times10^{-5}\times1000}{0.001} = 49.5 S cm² mol⁻¹.

Answer: 49.5 S cm² mol⁻¹.

Example 6: Resistance from conductivity

A 0.1 M solution has conductivity 0.012 S cm⁻¹ in a cell of cell constant 1.5 cm⁻¹. Find the resistance.

Solution: κ=G∗/R⇒R=G∗/κ=1.5/0.012=125 Ω\kappa = G^*/R \Rightarrow R = G^*/\kappa = 1.5/0.012 = 125\,\Omega.

Answer: 125 Ω.

Example 7: Converting conductivity units

Convert a conductivity of 0.025 S cm⁻¹ into S m⁻¹.

Solution: 1 S cm−1=100 S m−11\text{ S cm}^{-1} = 100\text{ S m}^{-1}, so 0.025 S cm−1=0.025×100=2.50.025\text{ S cm}^{-1} = 0.025\times100 = 2.5 S m⁻¹.

Answer: 2.5 S m⁻¹.

Example 8: Molar conductivity in SI units

A 0.5 mol m⁻³ solution has conductivity 1.0×10−21.0\times10^{-2} S m⁻¹. Find Λm\Lambda_m in S m² mol⁻¹.

Solution: Λm=κc=1.0×10−20.5=2.0×10−2\Lambda_m = \dfrac{\kappa}{c} = \dfrac{1.0\times10^{-2}}{0.5} = 2.0\times10^{-2} S m² mol⁻¹.

Answer: 2.0×10−22.0\times10^{-2} S m² mol⁻¹.

Example 9: Conductance of a solution

A solution has resistance 50 Ω. What is its conductance?

Solution: G=1/R=1/50=0.02G = 1/R = 1/50 = 0.02 S.

Answer: 0.02 S.

Example 10: Finding concentration from molar conductivity

A KCl solution has conductivity 0.014 S cm⁻¹ and molar conductivity 140 S cm² mol⁻¹. Find its concentration.

Solution: Λm=κ×1000c⇒c=κ×1000Λm=0.014×1000140=0.1\Lambda_m = \dfrac{\kappa\times1000}{c} \Rightarrow c = \dfrac{\kappa\times1000}{\Lambda_m} = \dfrac{0.014\times1000}{140} = 0.1 mol L⁻¹.

Answer: 0.1 M.