Q.The conductivity of $0.001028\ mol\ L^{-1}$ acetic acid is $4.95 \times 10^{-5}\ S\ cm^{-1}$. Calculate its dissociation constant if $\Lambda^0_m$ for acetic acid is $390.5\ S\ cm^2\ mol^{-1}$.

Concept: Molar Conductivity — the conductivity per mole of electrolyte, used to find the degree of dissociation ($\alpha$) and then the dissociation constant ($K_a$) for a weak acid.

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

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

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

The key idea here is that weak electrolytes like acetic acid do not fully dissociate in solution. Their molar conductivity is lower than the limiting value because only a fraction of molecules have actually split into ions. That fraction — the degree of dissociation $\alpha$ — can be extracted by comparing the measured molar conductivity with the maximum possible value. Once we have $\alpha$, the dissociation constant follows directly from the equilibrium expression.

Let’s walk through it step by step.

1. Convert concentration to the right units

The concentration is given as $0.001028\ \text{mol L}^{-1}$. Since $1\ \text{L} = 1000\ \text{cm}^3$, this is the same as:

We’ll keep it in mol L$^{-1}$ because the dissociation constant is conventionally expressed in those units.

2. Calculate the molar conductivity $\Lambda_m$

Molar conductivity is defined as:

where $\kappa$ is the conductivity (in S cm$^{-1}$) and $C$ is the concentration in mol cm$^{-3}$. Watch the units carefully — conductivity is given in S cm$^{-1}$, so we need concentration in mol cm$^{-3}$.

Convert concentration:

Now:

3. Find the degree of dissociation $\alpha$

For a weak electrolyte, the degree of dissociation is the ratio of the actual molar conductivity to the limiting molar conductivity:

Given $\Lambda_m^0 = 390.5\ \text{S cm}^2\ \text{mol}^{-1}$:

So about 12.3% of the acetic acid molecules are dissociated at this concentration.

4. Write the dissociation equilibrium

Acetic acid dissociates as:

Let initial concentration be $C$. At equilibrium: …

Method: Conductometric Determination of Dissociation Constant (Kohlrausch’s Law & Ostwald’s Dilution Law)

Why this method works

For a weak electrolyte like acetic acid, molar conductivity increases with dilution because more ions are produced. The degree of dissociation ($\alpha$) is the ratio of molar conductivity at a given concentration to the limiting molar conductivity. Using $\alpha$, we apply Ostwald’s dilution law to find the dissociation constant.

Step-by-step solution

Step 1: Calculate molar conductivity ($\Lambda_m$) at given concentration

Formula:

Where:

• $\kappa = 4.95 \times 10^{-5}\ \text{S cm}^{-1}$
• $C = 0.001028\ \text{mol L}^{-1}$

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

Given $\Lambda_m^0 = 390.5\ \text{S cm}^2\ \text{mol}^{-1}$:

Step 3: Apply Ostwald’s dilution law

For a weak acid HA $\rightleftharpoons$ H⁺ + A⁻:

Substitute values: …

Common Mistakes in Molar Conductivity & Dissociation Constant Problems

Students often lose marks on this exact type of problem. Here are the most frequent errors and how to avoid each:

1. Unit Confusion (The #1 Mistake)

The Mistake:

Mixing up units of conductivity ($\kappa$) and molar conductivity ($\Lambda_m$).

• Conductivity $\kappa$ is given in $S\ cm^{-1}$
• Molar conductivity $\Lambda_m$ is in $S\ cm^2\ mol^{-1}$
• Students forget to convert concentration from $mol\ L^{-1}$ to $mol\ cm^{-3}$

How to Avoid:

Always write units beside every value before calculating. For $\Lambda_m$:

The factor 1000 converts $L$ to $cm^3$:

• $1\ L = 1000\ cm^3$
• So $C = 0.001028\ mol\ L^{-1} = 0.001028 \times 10^{-3}\ mol\ cm^{-3}$

Correct calculation:

2. Forgetting the 1000 Factor

The Mistake:

Using $\Lambda_m = \kappa / C$ directly without multiplying by 1000.

How to Avoid:

Memorise the formula with the conversion factor:

Where $C$ is in $mol\ L^{-1}$. This is non-negotiable for exam problems.

3. Using Wrong Formula for Degree of Dissociation ($\alpha$)

The Mistake:

Using $\alpha = \Lambda_m / \Lambda^0_m$ for strong electrolytes only — but acetic acid is weak.

How to Avoid:

For weak electrolytes like acetic acid:

This is correct here. But remember: this formula works only when $\Lambda_m$ and $\Lambda^0_m$ are in the same units (both $S\ cm^2\ mol^{-1}$).

Calculation:

4. Incorrect Dissociation Constant Formula

The Mistake:

Using $K_a = \frac{C\alpha^2}{1-\alpha}$ but forgetting the $(1-\alpha)$ denominator, or using $\alpha^2 C$ directly.

How to Avoid:

For a weak acid $HA \rightleftharpoons H^+ + A^-$: …