i = 1 + (n-1)α. Why NaCl gives i ≈ 2.

Van't Hoff Factor — Effect of Dissociation on Colligative Properties

Question

A 0.1 m aqueous solution of NaCl shows a boiling point elevation of 0.10 °C. The boiling point elevation constant for water is $K_b = 0.52$ K·kg/mol. Calculate the Van’t Hoff factor $i$ and the degree of dissociation $\alpha$ for NaCl in this solution.

Solution — Step by Step

The boiling point elevation formula with the Van’t Hoff factor is:

\Delta T_b = i \cdot K_b \cdot m

We know $\Delta T_b = 0.10$ °C, $K_b = 0.52$ K·kg/mol, and $m = 0.1$ mol/kg. Plugging in to solve for $i$ :

i = \frac{\Delta T_b}{K_b \cdot m} = \frac{0.10}{0.52 \times 0.1}

i = \frac{0.10}{0.052} \approx 1.92

This tells us each formula unit of NaCl effectively produces about 1.92 particles in solution. For complete dissociation we’d expect $i = 2$ exactly — so this solution is nearly (but not fully) dissociated.

NaCl dissociates as: $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$

Here $n = 2$ (two ions per formula unit). The general formula linking $i$ and the degree of dissociation $\alpha$ is:

i = 1 + (n - 1)\alpha

Why does this formula work? Start with 1 mole. After dissociation, $(1 - \alpha)$ moles of undissociated NaCl remain, and $n\alpha$ moles of ions form. Total moles $= 1 - \alpha + n\alpha = 1 + (n-1)\alpha$ . Since $i$ is just the ratio of total particles to original particles, we get this expression directly.

Substituting $i = 1.92$ and $n = 2$ :

1.92 = 1 + (2 - 1)\alpha = 1 + \alpha

\alpha = 0.92

NaCl is 92% dissociated in this 0.1 m solution.

Why This Works

The Van’t Hoff factor $i$ captures the ratio of actual particles in solution to the number of formula units dissolved. For non-electrolytes like glucose, $i = 1$ because no dissociation happens. For electrolytes, $i > 1$ and it scales every colligative property — boiling point elevation, freezing point depression, osmotic pressure — by the same factor.

The reason NaCl doesn’t give exactly $i = 2$ even though it’s a strong electrolyte is inter-ionic attraction. At finite concentrations, $\text{Na}^+$ and $\text{Cl}^-$ ions don’t behave completely independently — they cluster slightly, reducing the effective particle count. At infinite dilution, strong electrolytes would give integer values of $i$ .

This concept has high weightage in both JEE Main and CBSE Class 12 — questions routinely give you $\Delta T_f$ or osmotic pressure and ask you to back-calculate $\alpha$ or identify whether the solute is associating (like acetic acid in benzene, where $i < 1$ ) or dissociating.

Alternative Method — Using Freezing Point Depression

If the question gave freezing point depression instead, the approach is identical:

\Delta T_f = i \cdot K_f \cdot m

For water, $K_f = 1.86$ K·kg/mol. Say you observe $\Delta T_f = 0.344$ °C for the same 0.1 m NaCl solution:

i = \frac{0.344}{1.86 \times 0.1} = \frac{0.344}{0.186} \approx 1.85

Same logic, different constant. The $\alpha$ calculation then follows identically. JEE Main 2023 had a variant using osmotic pressure ( $\pi = iMRT$ ) — the $i$ extraction is the same step regardless of which colligative property you’re given.

Common Mistake

Students confuse $n$ (number of ions per formula unit) with $i$ (the Van’t Hoff factor). For NaCl, $n = 2$ always — that’s fixed by the formula. But $i$ depends on the actual concentration and degree of dissociation; it equals $n$ only when $\alpha = 1$ . Writing $i = 2$ directly without calculating it is wrong and will cost you marks in numerical problems.

A related trap: for $\text{Na}_2\text{SO}_4$ , the dissociation gives $\text{Na}^+$ , $\text{Na}^+$ , and $\text{SO}_4^{2-}$ , so $n = 3$ , not 2. Always count ions, not just the number of types of ions.

For association problems (like acetic acid dimerising in benzene), the same formula applies but $n < 1$ . If acetic acid fully dimerises, $n = 0.5$ , so $i = 1 + (0.5 - 1)\alpha = 1 - 0.5\alpha$ . Watch the sign — $i$ will come out less than 1, confirming association.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Freezing Point Depression

Common Mistake

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