Q.The molar conductivity of $0.025\ mol\ L^{-1}$ methanoic acid is $46.1\ S\ cm^2\ mol^{-1}$. Calculate its degree of dissociation and dissociation constant. Given $\lambda^0(H^+) = 349.6\ S\ cm^2\ mol^{-1}$ and $\lambda^0(HCOO^-) = 54.6\ S\ cm^2\ mol^{-1}$.

Concept: Molar Conductivity and its use in finding degree of dissociation ($\alpha$) and dissociation constant ($K_a$) for a weak acid.

Step 1 – Find limiting molar conductivity of methanoic acid

Using Kohlrausch’s law:

Step 2 – Degree of dissociation

Step 3 – Dissociation constant

For a weak acid HA $\rightleftharpoons$ H⁺ + A⁻, with initial concentration $c = 0.025\ mol\ L^{-1}$: …

This is a classic problem that connects two big ideas in electrochemistry: molar conductivity as a measure of how well ions carry current, and weak acid equilibrium. The trick is that for a weak acid like methanoic acid (HCOOH), molar conductivity increases as the solution gets more dilute because more molecules dissociate. At infinite dilution, every molecule is dissociated — that’s the limiting molar conductivity $\Lambda_m^0$. By comparing the actual molar conductivity at a given concentration to this limiting value, we get the degree of dissociation $\alpha$. Then the equilibrium constant follows directly.

Let’s walk through it.

1. Find the limiting molar conductivity of methanoic acid

Kohlrausch’s law says that at infinite dilution, the molar conductivity of an electrolyte is the sum of the molar conductivities of its individual ions. For methanoic acid:

Plug in the given values:

This is the conductivity the acid would have if it were fully dissociated.

2. Relate molar conductivity to degree of dissociation

For a weak electrolyte, the degree of dissociation $\alpha$ is the fraction of molecules that have dissociated into ions. The measured molar conductivity $\Lambda_m$ at a given concentration is proportional to $\alpha$, because only the dissociated part contributes to conduction:

This is a direct consequence: if all molecules dissociated, $\Lambda_m$ would equal $\Lambda_m^0$; if none dissociated, $\Lambda_m$ would be zero.

So:

Calculate:

3. Write the dissociation equilibrium

Methanoic acid dissociates as: …

Method: Conductivity-Based Dissociation (Kohlrausch’s Law & Ostwald’s Dilution Law)

Step 1: Find limiting molar conductivity ($\Lambda_m^0$)

Using Kohlrausch’s law of independent migration of ions:

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

The degree of dissociation is the ratio of molar conductivity at given concentration to limiting molar conductivity:

Degree of dissociation = $\boxed{0.114}$ (or $11.4\%$)

Step 3: Calculate dissociation constant ($K_a$)

For a weak acid, using Ostwald’s dilution law:

Where $C = 0.025\ \text{mol L}^{-1}$

Dissociation constant = $\boxed{3.67 \times 10^{-4}\ \text{mol L}^{-1}}$

Key Concept Summary

| Quantity | Formula | Value | …

Here are the most common mistakes students make when solving this exact problem, along with how to avoid each one.

1. Confusing Molar Conductivity with Conductivity

The Mistake:

Students treat the given molar conductivity ($\Lambda_m = 46.1\ \text{S cm}^2 \text{mol}^{-1}$) as if it were the specific conductivity ($\kappa$). This leads to using the wrong formula or misinterpreting units.

How to Avoid:

• Molar conductivity ($\Lambda_m$) is already conductivity per mole — it has units of $\text{S cm}^2 \text{mol}^{-1}$.
• Specific conductivity ($\kappa$) has units $\text{S cm}^{-1}$.
• For this problem, you do not need $\kappa$ at all. You directly use $\Lambda_m$ in the formula for degree of dissociation.

Key formula:

$\alpha = \frac{\Lambda_m}{\Lambda_m^0}$

2. Forgetting to Calculate $\Lambda_m^0$ Correctly

The Mistake:

Using only one of the given limiting molar conductivities (e.g., just $\lambda^0(H^+)$) instead of summing both ions.

How to Avoid:

• The limiting molar conductivity of methanoic acid is the sum of its constituent ions at infinite dilution:

• Plug in the given values:

Always check: For a weak acid like methanoic acid, $\Lambda_m^0$ is the sum of the cation and anion limiting conductivities.

3. Using the Wrong Formula for Degree of Dissociation

The Mistake:

Using $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$ but then treating $\Lambda_m$ as if it were for a different concentration, or using $\Lambda_m$ in place of $\Lambda_m^0$ in the denominator.

How to Avoid:

• The formula is always:

• Here, $\Lambda_m = 46.1$ and $\Lambda_m^0 = 404.2$, so:

• This means 11.4% dissociation — a reasonable value for a weak acid at this concentration.

4. Misapplying the Dissociation Constant Formula

The Mistake:

Using $K_a = \frac{C \alpha^2}{1 - \alpha}$ but forgetting that $C$ must be in mol/L (molarity) and that $\alpha$ is a decimal, not a percentage.

How to Avoid:

• $C = 0.025\ \text{mol L}^{-1}$ (given).
• $\alpha = 0.114$ (from above).
• Plug into:

• Calculate step-by-step:

Final answer: …