Q.$\Lambda^0_m$ for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 $S\ cm^2\ mol^{-1}$ respectively. Calculate $\Lambda^0$ for HAc.

Concept: Molar Conductivity at infinite dilution — Kohlrausch’s law of independent migration of ions.

Step 1: Write Kohlrausch’s law for each salt:

• $\Lambda^0_m(\text{NaCl}) = \lambda^0_{\text{Na}^+} + \lambda^0_{\text{Cl}^-} = 126.4$
• $\Lambda^0_m(\text{HCl}) = \lambda^0_{\text{H}^+} + \lambda^0_{\text{Cl}^-} = 425.9$
• $\Lambda^0_m(\text{NaAc}) = \lambda^0_{\text{Na}^+} + \lambda^0_{\text{Ac}^-} = 91.0$

Step 2: For acetic acid (HAc):

$\Lambda^0_m(\text{HAc}) = \lambda^0_{\text{H}^+} + \lambda^0_{\text{Ac}^-}$

Step 3: Combine the given values: …

The problem asks for the limiting molar conductivity of acetic acid (HAc), a weak electrolyte. You cannot measure it directly by extrapolation because weak acids don’t fully dissociate even at infinite dilution — the conductivity curve bends. So we need a different route.

That route is Kohlrausch’s law of independent migration of ions. At infinite dilution, each ion contributes a fixed amount to the total molar conductivity, regardless of what other ion it’s paired with. So the limiting molar conductivity of any electrolyte is simply the sum of the limiting conductivities of its constituent ions:

This means we can add and subtract $\Lambda_m^0$ values of strong electrolytes to get the $\Lambda_m^0$ of a weak electrolyte — as long as the ionic combinations cancel out correctly.

Here’s how it works for HAc. Write the target:

We don’t know either ion’s value individually. But we do know three strong electrolytes that share these ions:

1. HCl gives $\lambda^0(\text{H}^+) + \lambda^0(\text{Cl}^-) = 425.9$
2. NaAc gives $\lambda^0(\text{Na}^+) + \lambda^0(\text{Ac}^-) = 91.0$
3. NaCl gives $\lambda^0(\text{Na}^+) + \lambda^0(\text{Cl}^-) = 126.4$

Notice: if we add the first two and subtract the third, the sodium and chloride ions cancel out, leaving exactly what we need. …

Method: Kohlrausch’s Law of Independent Migration of Ions

This law states that at infinite dilution, each ion contributes a fixed amount to the molar conductivity of an electrolyte, independent of the other ion it is paired with.

Steps

1. Write the expression for $\Lambda^0_m$ of each given salt

   At infinite dilution:

   • $\Lambda^0_m(\text{NaCl}) = \lambda^0_{\text{Na}^+} + \lambda^0_{\text{Cl}^-} = 126.4$
   • $\Lambda^0_m(\text{HCl}) = \lambda^0_{\text{H}^+} + \lambda^0_{\text{Cl}^-} = 425.9$
   • $\Lambda^0_m(\text{NaAc}) = \lambda^0_{\text{Na}^+} + \lambda^0_{\text{Ac}^-} = 91.0$

2. Identify the target

   We need $\Lambda^0_m(\text{HAc}) = \lambda^0_{\text{H}^+} + \lambda^0_{\text{Ac}^-}$

3. Combine the given equations

   Add $\Lambda^0_m(\text{HCl})$ and $\Lambda^0_m(\text{NaAc})$:

This gives:

Now subtract $\Lambda^0_m(\text{NaCl})$: …

Here are the most common mistakes students make when solving this Kohlrausch’s law problem — and exactly how to avoid each one.

1. ✗ Forgetting the Kohlrausch’s Law Formula

Mistake: Trying to guess or recall the combination incorrectly — e.g., adding all three given values.

Why it’s wrong:

Kohlrausch’s law says:

At infinite dilution, the molar conductivity of an electrolyte is the sum of the molar conductivities of its constituent ions.

For a weak acid like HAc ($\text{H}^+ + \text{Ac}^-$), you cannot measure $\Lambda^0_m$ directly — you build it from known strong electrolytes.

How to avoid:

Always write the ionic breakdown:

• $\Lambda^0_m(\text{HAc}) = \lambda^0(\text{H}^+) + \lambda^0(\text{Ac}^-)$
• You have: $\Lambda^0_m(\text{HCl}) = \lambda^0(\text{H}^+) + \lambda^0(\text{Cl}^-)$ $\Lambda^0_m(\text{NaAc}) = \lambda^0(\text{Na}^+) + \lambda^0(\text{Ac}^-)$ $\Lambda^0_m(\text{NaCl}) = \lambda^0(\text{Na}^+) + \lambda^0(\text{Cl}^-)$

Then combine:

2. ✗ Sign Errors in the Combination

Mistake: Subtracting the wrong term or adding all three.

Why it’s wrong:

You need to cancel the common ions ($\text{Na}^+$ and $\text{Cl}^-$). Adding all three gives a meaningless sum.

How to avoid:

Use the “add and subtract” logic:

• Start with $\lambda^0(\text{H}^+) + \lambda^0(\text{Ac}^-)$
• Add $\lambda^0(\text{Cl}^-)$ and $\lambda^0(\text{Na}^+)$ to match given data, then subtract them out.

Check:

If you get a negative or absurd number, you’ve swapped signs.

3. ✗ Unit Confusion

Mistake: Using values in different units (e.g., $\text{mS}$ vs $\text{S}$, or $\text{cm}^2$ vs $\text{m}^2$).

Why it’s wrong:

All given values are in $S\ cm^2\ mol^{-1}$. Mixing units gives a wrong numeric answer.

How to avoid:

• Always check units before calculation.
• If needed, convert: $1\ \text{S cm}^2\ \text{mol}^{-1} = 10^{-4}\ \text{S m}^2\ \text{mol}^{-1}$.
• In this problem, all three are already consistent — just use them as given.

4. ✗ Misidentifying Strong vs Weak Electrolytes

Mistake: Thinking HAc’s $\Lambda^0_m$ can be measured directly (like for NaCl).

Why it’s wrong: …