Conductance of Electrolytic Solutions

From resistance to resistivity. Any conductor — including a column of electrolyte solution held between two electrodes — obeys $R = \rho\,l/A$, where $R$ is resistance, $l$ the length of the solution column, $A$ its cross-sectional area, and $\rho$ (resistivity, or specific resistance) is a material property with unit $\Omega\,\text{m}$. In practice a conductivity cell has a fixed geometry, so $l/A$ is a constant of the cell called the cell constant, $G^* = l/A$, usually quoted in $\text{cm}^{-1}$; you measure $R$ with a bridge and recover $\rho$ (and hence conductivity) without separately measuring $l$ and $A$ each time.

Conductivity. $\kappa$ (kappa) is simply the reciprocal of resistivity,

where: $\kappa$ is conductivity with SI unit $\text{S m}^{-1}$ (commonly reported in $\text{S cm}^{-1}$), $G^*$ is the cell constant, and $R$ is the measured resistance. Conductivity measures how well a unit cube of the solution, at whatever concentration it happens to be, conducts current — it depends on how many ions are packed into that cube.

Molar conductivity — normalising for the amount of electrolyte. Comparing two solutions by $\kappa$ alone is unfair if their concentrations differ, since a more concentrated solution simply has more ions per unit volume regardless of how "good" each ion is at conducting. Molar conductivity fixes this by dividing conductivity by concentration:

where: $C$ is molar concentration in $\text{mol L}^{-1}$, $\kappa$ is in $\text{S cm}^{-1}$, and $\Lambda_m$ comes out in $\text{S cm}^2\,\text{mol}^{-1}$ (the factor 1000 converts the litre in $C$ to the cm³ implicit in $\kappa$'s units). Physically, $\Lambda_m$ is the conducting power contributed by one mole of electrolyte, however that mole happens to be diluted.

Why $\kappa$ and $\Lambda_m$ move in opposite directions on dilution. Diluting a solution always lowers $\kappa$, because there are simply fewer ions in each unit volume to carry current — this is true for every electrolyte. But $\Lambda_m$ rises on dilution for a different reason in each case. In a strong electrolyte (fully dissociated already, e.g. KCl, NaCl), dilution weakens the electrostatic drag ions exert on each other, so each ion moves a little more freely and $\Lambda_m$ climbs slowly and linearly with $\sqrt{C}$ — this near-linear behaviour, discovered empirically by Kohlrausch, lets you extrapolate a $\Lambda_m$ vs $\sqrt{C}$ plot straight to $C\to0$ to read off $\Lambda_m^\circ$, the limiting (infinite-dilution) molar conductivity. In a weak electrolyte (e.g. CH₃COOH), the dominant effect is that dilution shifts the dissociation equilibrium further toward ions, so the fraction dissociated rises — and near infinite dilution it shoots toward completion, so $\Lambda_m$ spikes sharply rather than trending linearly, and the graph cannot be extrapolated to find $\Lambda_m^\circ$.

Kohlrausch's law of independent migration of ions. At infinite dilution, ions are so far apart that each one migrates independently of its partner ion, contributing a fixed amount to $\Lambda_m^\circ$ regardless of what it is paired with:

where: $\lambda_+^\circ$ and $\lambda_-^\circ$ are the limiting molar ionic conductivities of the cation and anion, and $\nu_+$, $\nu_-$ are the numbers of cations and anions produced per formula unit (e.g. $\nu_+=1,\nu_-=1$ for KCl, but $\nu_+=1,\nu_-=2$ for CaCl₂). Because these ionic contributions add independently, $\Lambda_m^\circ$ of an electrolyte that is hard to measure directly (a weak one) can be built from $\Lambda_m^\circ$ values of strong electrolytes that are easy to measure and share its ions. Once $\Lambda_m^\circ$ is known, the degree of dissociation $\alpha$ of a weak electrolyte at any concentration $C$ follows from comparing its actual molar conductivity to the limiting one:

and $\alpha$ then feeds Ostwald's dilution law to give the dissociation constant $K_a$.

Problem. The resistance of a $0.01$ M KCl solution in a conductivity cell is $150\ \Omega$, with cell constant $1.25\ \text{cm}^{-1}$. Find $\kappa$ and $\Lambda_m$ for this solution. Then, using $\Lambda_m^\circ(\text{CH}_3\text{COOH}) = 390.7\ \text{S cm}^2\,\text{mol}^{-1}$ (from the derivation above) and a measured $\Lambda_m^C = 49.2\ \text{S cm}^2\,\text{mol}^{-1}$ for $0.001$ M acetic acid, find its degree of dissociation.

Step 1. Conductivity of the KCl solution: $\kappa = G^*/R = 1.25/150 = 8.33\times10^{-3}\ \text{S cm}^{-1}$.

Step 2. Molar conductivity of the KCl solution, with $C = 0.01\ \text{mol L}^{-1}$: $\Lambda_m = \dfrac{\kappa\times1000}{C} = \dfrac{8.33\times10^{-3}\times1000}{0.01} = 833\ \text{S cm}^2\,\text{mol}^{-1}$.

Step 3. Degree of dissociation of acetic acid, comparing its actual molar conductivity at $0.001$ M to the limiting value found above: $\alpha = \dfrac{\Lambda_m^C}{\Lambda_m^\circ} = \dfrac{49.2}{390.7} = 0.126$.

Answer: $\kappa = 8.33\times10^{-3}\ \text{S cm}^{-1}$, $\Lambda_m(\text{KCl}) = 833\ \text{S cm}^2\,\text{mol}^{-1}$, and acetic acid is only about 12.6 percent dissociated ($\alpha = 0.126$) at 0.001 M — consistent with it being a weak electrolyte.

Competitive stretch

If the same acetic acid solution were diluted further to $0.0001$ M and $\Lambda_m^C$ rose to about $146\ \text{S cm}^2\,\text{mol}^{-1}$, then $\alpha = 146/390.7 \approx 0.374$ — roughly three times more dissociated at ten times greater dilution, illustrating why $\Lambda_m$ for a weak electrolyte climbs so much faster than for a strong one as $C\to0$.

• Mixing up the dilution trends. $\kappa$ always decreases on dilution (fewer ions per volume); $\Lambda_m$ always increases (each mole conducts more effectively) — never state them the same way.
• Extrapolating a weak electrolyte's graph to find $\Lambda_m^\circ$. Only strong electrolytes give a straight $\Lambda_m$ vs $\sqrt{C}$ line reaching $C=0$; weak electrolytes need Kohlrausch's law instead.
• Dropping the factor of 1000 or mixing units in $\Lambda_m = 1000\kappa/C$ — check $\kappa$ is in $\text{S cm}^{-1}$ and $C$ in $\text{mol L}^{-1}$ before substituting.
• Combining the wrong strong electrolytes in Kohlrausch's law. Verify algebraically that the shared ions cancel and the target ions each appear exactly once, as in the acetic acid derivation.

Ask the tutor: want to practice building Λm° for a different weak electrolyte from strong-electrolyte data, or work through how Ostwald's dilution law turns alpha into Ka step by step?