Q.$\Lambda^0_{m(NH_4OH)}$ is equal to ______________.

The key idea is Kohlrausch’s law of independent migration of ions: the limiting molar conductivity of a weak electrolyte is the sum of the limiting conductivities of its constituent ions. For $NH_4OH$, the ions are $NH_4^+$ and $OH^-$.

We need to combine known strong electrolytes to get these ions.

$NH_4Cl$ gives $NH_4^+ + Cl^-$ and $NaOH$ gives $Na^+ + OH^-$. Adding them gives $NH_4^+ + OH^- + Na^+ + Cl^-$.

To cancel the extra $Na^+$ and $Cl^-$, subtract $NaCl$ (which gives $Na^+ + Cl^-$).

So: …

The key idea here is Kohlrausch’s law: at infinite dilution, each ion contributes a fixed amount to the molar conductivity, independent of the other ion it travels with. So the limiting molar conductivity of any electrolyte is simply the sum of the limiting conductivities of its constituent ions.

For a weak base like $NH_4OH$, we cannot measure $\Lambda_m^0$ directly by extrapolation (because it doesn’t fully dissociate even at low concentrations). But we can build it from the $\Lambda_m^0$ values of strong electrolytes that contain the same ions — $NH_4^+$ and $OH^-$ — by adding and subtracting known values to cancel out the unwanted ions.

Let’s see how.

1. Write what we want in terms of ions.

   $\Lambda_{m(NH_4OH)}^0 = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$

   That’s our target.

2. Find strong electrolytes that give us these ions.

   • $NH_4Cl$ gives $\lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0$
   • $NaOH$ gives $\lambda_{Na^+}^0 + \lambda_{OH^-}^0$
   • $NaCl$ gives $\lambda_{Na^+}^0 + \lambda_{Cl^-}^0$

3. Combine them to isolate the target sum.

   If we add the first two and subtract the third:

The $\lambda_{Na^+}^0$ and $\lambda_{Cl^-}^0$ cancel perfectly, leaving:

4. Translate back to electrolyte notation. So: …

Method: Kohlrausch’s Law of Independent Migration of Ions

Why this method?

Weak electrolytes like $NH_4OH$ do not fully dissociate, so their limiting molar conductivity ($\Lambda^0_m$) cannot be measured directly by extrapolation. Kohlrausch’s law allows us to calculate it by combining $\Lambda^0_m$ values of strong electrolytes that share the same ions.

Steps

1. Write the dissociation of the target weak electrolyte

So,

2. Identify strong electrolytes that contain these ions

   • $NH_4Cl$ gives $\lambda^0_{NH_4^+} + \lambda^0_{Cl^-}$
   • $NaOH$ gives $\lambda^0_{Na^+} + \lambda^0_{OH^-}$
   • $NaCl$ gives $\lambda^0_{Na^+} + \lambda^0_{Cl^-}$

3. Combine to cancel spectator ions

   Add $\Lambda^0_{m(NH_4Cl)}$ and $\Lambda^0_{m(NaOH)}$:

   $(\lambda^0_{NH_4^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-})$ …

Common Mistakes & How to Avoid Them

Mistake 1: Not Understanding Kohlrausch’s Law

Students often try to memorise the answer without knowing why the formula works.

How to avoid:

Kohlrausch’s Law states that at infinite dilution, molar conductivity is the sum of independent ionic contributions:

For a weak base like $NH_4OH$, you cannot measure $\Lambda_m^0$ directly (it doesn’t fully dissociate). So you build it from strong electrolytes whose $\Lambda_m^0$ values are known.

Mistake 2: Forgetting to Cancel Ions Correctly

Students pick options without checking if the unwanted ions cancel out.

How to avoid:

Write each electrolyte as its ions, then cancel common ions.

For correct option (ii):

Now do:

Cancel $\lambda_{Na^+}$ and $\lambda_{Cl^-}$ → you get:

Mistake 3: Confusing Addition/Subtraction Signs

Students misplace the minus sign, especially in options like (iii) or (iv).

How to avoid:

Always write the ionic breakdown before deciding the sign. The target is $NH_4^+ + OH^-$.

• You need $NH_4^+$ from $NH_4Cl$
• You need $OH^-$ from $NaOH$
• You must subtract the common ion pair ($Na^+ + Cl^-$) that appears twice — that’s $NaCl$.

So the correct structure is:

Mistake 4: Picking Option (i) Without Checking

Option (i) looks similar but uses $HCl$ instead of $NaOH$ and $NaCl$. …