Q.Conductivity of $0.00241\ M$ acetic acid is $7.896 \times 10^{-5}\ S\ cm^{-1}$. Calculate its molar conductivity. If $\Lambda^0_m$ for acetic acid is $390.5\ S\ cm^2\ mol^{-1}$, what is its dissociation constant?

Concept: Molar Conductivity — the conductivity of all ions produced from one mole of electrolyte in solution.

Step 1: Molar conductivity

where $\kappa = 7.896 \times 10^{-5}\ \text{S cm}^{-1}$, $C = 0.00241\ \text{mol L}^{-1}$.

Step 2: Degree of dissociation

Step 3: Dissociation constant (for weak acid HA $\rightleftharpoons$ H⁺ + A⁻) …

The problem gives you the conductivity ($\kappa$) of a dilute acetic acid solution and asks for two things: the molar conductivity ($\Lambda_m$), and then the dissociation constant ($K_a$) using the given limiting molar conductivity ($\Lambda^0_m$). This is a classic weak electrolyte problem — the key is that acetic acid does not fully dissociate, so its molar conductivity is much smaller than $\Lambda^0_m$, and the ratio tells you the fraction of molecules that have actually split into ions.

Let’s go step by step.

1. Molar conductivity from conductivity and concentration Molar conductivity is defined as the conductivity of the solution divided by the molar concentration:

But watch the units carefully. Conductivity $\kappa$ is given in $S\ cm^{-1}$, and concentration $c$ is in $mol\ L^{-1}$. For molar conductivity in $S\ cm^2\ mol^{-1}$, you need to convert concentration from $mol/L$ to $mol/cm^3$:

Now plug in:

So the molar conductivity of this solution is about $32.8\ S\ cm^2\ mol^{-1}$.

2. Degree of dissociation from molar conductivities For a weak electrolyte, the degree of dissociation $\alpha$ is given by the ratio of its molar conductivity at a given concentration to its limiting molar conductivity (at infinite dilution):

This works because at infinite dilution, every molecule is dissociated, so $\Lambda_m^0$ represents the conductivity if all molecules were ions. Here:

$\alpha = \frac{32.76}{390.5} = 0.0839$ …

Method: Conductometric Determination of Degree of Dissociation and Dissociation Constant

This method uses the definition of molar conductivity and the degree-of-dissociation relation for a weak electrolyte.

Step 1: Calculate molar conductivity ($\Lambda_m$)

Given $\kappa = 7.896 \times 10^{-5}\ S\ cm^{-1}$ and $C = 0.00241\ mol\ L^{-1}$:

Step 2: Calculate degree of dissociation ($\alpha$)

For a weak electrolyte, the degree of dissociation is:

Using the given $\Lambda_m^0 = 390.5\ S\ cm^2\ mol^{-1}$ (no Kohlrausch summation is needed here since the limiting molar conductivity is provided directly):

So about 8.4% of acetic acid molecules are dissociated.

Step 3: Calculate dissociation constant ($K_a$) …

Common Mistakes & How to Avoid Them — Molar Conductivity & Dissociation Constant of Acetic Acid

Mistake 1: Forgetting the 1000 Factor in $\Lambda_m = \kappa \times 1000 / C$

Conductivity $\kappa$ is effectively per cm³, but concentration is given per litre ($1000\ cm^3$). Skipping the 1000 factor gives a value 1000× too small.

Fix: Always use $\Lambda_m = \dfrac{1000\,\kappa\,(S\,cm^{-1})}{C\,(mol\,L^{-1})}$.

Mistake 2: Inverting the Degree-of-Dissociation Ratio

Some students divide $\Lambda_m^0$ by $\Lambda_m$ instead of the other way round, giving $\alpha > 1$, which is physically impossible (a fraction dissociated can never exceed 1).

Fix: $\alpha = \Lambda_m / \Lambda_m^0$ is always $\le 1$; if your ratio exceeds 1, you've inverted it.

Mistake 3: Dropping the $(1-\alpha)$ Term in Ostwald's Dilution Law

Writing $K_a = c\alpha^2$ alone is only a valid approximation when $\alpha \ll 1$ (typically $< 0.05$). Here $\alpha \approx 0.084$, so omitting $(1-\alpha)$ changes the answer by about 8% ($1.70\times10^{-5}$ instead of the correct $1.85\times10^{-5}$).

Fix: Always use the full expression $K_a = \dfrac{c\alpha^2}{1-\alpha}$ unless the question explicitly says $\alpha$ is small enough to neglect.

Mistake 4: Mixing Up $S\ m^{-1}$ and $S\ cm^{-1}$

If $\kappa$ were given in $S\ m^{-1}$ (as in some other problems), forgetting to convert ($1\ S\ m^{-1} = 10^{-2}\ S\ cm^{-1}$) throws the answer off by a factor of 100. In this problem $\kappa$ is already in $S\ cm^{-1}$, so no conversion is needed — but always check the given unit first.

Mistake 5: Rounding $\alpha$ Too Early …