Q.Suggest a way to determine the $\Lambda^0_m$ value of water.

The key idea is that molar conductivity at infinite dilution ($\Lambda_m^0$) is additive for ions (Kohlrausch’s law), but water is a weak electrolyte — it does not fully dissociate, so direct extrapolation fails.

Reasoning:

1. Water dissociates as $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$. Its $\Lambda_m^0$ cannot be measured directly because the dissociation is incomplete and conductivity is very low.

2. Use Kohlrausch’s law: $\Lambda_m^0(\text{H}_2\text{O}) = \lambda^0(\text{H}^+) + \lambda^0(\text{OH}^-)$.

3. Obtain $\lambda^0(\text{H}^+)$ and $\lambda^0(\text{OH}^-)$ from the $\Lambda_m^0$ values of strong electrolytes containing these ions, e.g.:

   • $\Lambda_m^0(\text{HCl}) = \lambda^0(\text{H}^+) + \lambda^0(\text{Cl}^-)$ …

Why we can’t measure it directly

Water is a weak electrolyte. It dissociates only slightly:

If you try to measure its molar conductivity directly, the concentration of ions is tiny, and the conductivity is dominated by impurities. Extrapolating to infinite dilution is impossible because the dissociation itself changes with concentration. So we need an indirect route.

The key idea is Kohlrausch’s law: at infinite dilution, each ion contributes a fixed amount to the molar conductivity, independent of the other ion it came from. That means:

where $\nu$ are the number of ions per formula unit, and $\lambda^0$ are the limiting ionic conductivities.

For water, we want:

So if we can find $\lambda^0(\ce{H+})$ and $\lambda^0(\ce{OH-})$ from known strong electrolytes, we’re done.

Step-by-step determination

1. Choose three strong electrolytes that contain $\ce{H+}$, $\ce{OH-}$, and a common counterion.

A classic set is:

• $\ce{HCl}$ — gives $\lambda^0(\ce{H+}) + \lambda^0(\ce{Cl-})$
• $\ce{NaOH}$ — gives $\lambda^0(\ce{Na+}) + \lambda^0(\ce{OH-})$
• $\ce{NaCl}$ — gives $\lambda^0(\ce{Na+}) + \lambda^0(\ce{Cl-})$

All three are strong electrolytes, so their $\Lambda_m^0$ values can be measured directly by extrapolating conductivity vs. $\sqrt{c}$ to zero concentration (Kohlrausch’s plot).

2. Write the three equations.

Let:

• $A = \Lambda_m^0(\ce{HCl}) = \lambda^0(\ce{H+}) + \lambda^0(\ce{Cl-})$
• $B = \Lambda_m^0(\ce{NaOH}) = \lambda^0(\ce{Na+}) + \lambda^0(\ce{OH-})$
• $C = \Lambda_m^0(\ce{NaCl}) = \lambda^0(\ce{Na+}) + \lambda^0(\ce{Cl-})$

3. Combine them to isolate $\lambda^0(\ce{H+}) + \lambda^0(\ce{OH-})$.

Notice that:

The $\lambda^0(\ce{Na+})$ and $\lambda^0(\ce{Cl-})$ cancel, leaving:

And that sum is exactly $\Lambda_m^0(\ce{H2O})$. …

Method: Kohlrausch’s Law of Independent Migration of Ions

This method uses the principle that at infinite dilution, each ion contributes a fixed amount to the total molar conductivity, independent of the other ion present.

Why this is needed for water

Water is a weak electrolyte — it does not dissociate completely. So we cannot directly measure $\Lambda^0_m$ for water by extrapolating a graph of $\Lambda_m$ vs. $\sqrt{c}$ (as we do for strong electrolytes). Instead, we use Kohlrausch’s law.

Steps to determine $\Lambda^0_m$ of water

Step 1: Identify the ions in water

Water dissociates as:

So, $\Lambda^0_m(\text{water}) = \lambda^0_{H^+} + \lambda^0_{OH^-}$

Step 2: Use known limiting molar conductivities of strong electrolytes

From Kohlrausch’s law, we can write:

Step 3: Combine to isolate the required sum

Add the first two equations and subtract the third:

Therefore: …

Here are the common mistakes students make when tackling this question, along with the conceptual fixes to avoid them.

1. Forgetting that Water is a Weak Electrolyte

The Mistake:

Students try to extrapolate $\Lambda_m$ vs. $\sqrt{C}$ for water directly, as they would for a strong electrolyte like KCl.

Why it’s wrong:

Water is a very weak electrolyte ($K_w = 1.0 \times 10^{-14}$). Its molar conductivity does not follow the linear Debye-Hückel-Onsager extrapolation. Plotting $\Lambda_m$ vs $\sqrt{C}$ for water gives a curve that cannot be reliably extrapolated to zero concentration.

How to avoid:

Always check the nature of the electrolyte first. For weak electrolytes, you cannot find $\Lambda^0_m$ by direct extrapolation. You must use Kohlrausch’s law of independent migration of ions.

2. Using the Wrong Formula for Kohlrausch’s Law

The Mistake:

Writing $\Lambda^0_m(\text{H}_2\text{O}) = \Lambda^0_m(\text{H}^+) + \Lambda^0_m(\text{OH}^-)$ directly, without realising that water is not a salt.

Why it’s wrong:

Kohlrausch’s law applies to electrolytes that fully dissociate at infinite dilution. Water itself does not dissociate completely — but its ions (H⁺ and OH⁻) do have known limiting molar conductivities from other strong electrolytes.

How to avoid:

Use the indirect method:

This works because:

• $\Lambda^0_m(\text{HCl}) = \lambda^0_{\text{H}^+} + \lambda^0_{\text{Cl}^-}$
• $\Lambda^0_m(\text{NaOH}) = \lambda^0_{\text{Na}^+} + \lambda^0_{\text{OH}^-}$
• $\Lambda^0_m(\text{NaCl}) = \lambda^0_{\text{Na}^+} + \lambda^0_{\text{Cl}^-}$

Subtracting cancels the spectator ions ($\text{Na}^+$ and $\text{Cl}^-$), leaving:

3. Confusing $\Lambda_m$ with $\Lambda^0_m$

The Mistake:

Using the measured molar conductivity of water (which is extremely small, ~$5.5 \times 10^{-6} \, \text{S cm}^2 \text{mol}^{-1}$) as if it were $\Lambda^0_m$.

Why it’s wrong:

The measured $\Lambda_m$ of water is not at infinite dilution — it’s the conductivity of pure water at its natural, very low dissociation. The limiting molar conductivity $\Lambda^0_m$ is a hypothetical value for complete dissociation at infinite dilution, which is much larger (~$550 \, \text{S cm}^2 \text{mol}^{-1}$).

How to avoid:

Remember: $\Lambda^0_m$ is not the conductivity of the pure substance — it’s the conductivity if it were fully dissociated at infinite dilution. For water, you must calculate it via Kohlrausch’s law, never measure it directly.

4. Forgetting Units and Magnitude

The Mistake:

Writing the final answer without units, or giving a value that is orders of magnitude off (e.g., writing $55 \, \text{S cm}^2 \text{mol}^{-1}$ instead of $~550$).

Why it’s wrong: …