Degree of dissociation from van't Hoff factor — solve for weak electrolyte

Question

A 0.1 M solution of a weak electrolyte AB shows a van’t Hoff factor $i = 1.2$ . If AB dissociates as $\text{AB} \rightarrow \text{A}^+ + \text{B}^-$ , find the degree of dissociation $\alpha$ . Also calculate the expected boiling point elevation if $K_b = 0.52$ K·kg/mol.

(JEE Main 2023, similar pattern)

Solution — Step by Step

For a substance that dissociates into $n$ ions per formula unit:

i = 1 + (n - 1)\alpha

Here AB gives 2 ions ( $\text{A}^+$ and $\text{B}^-$ ), so $n = 2$ .

1.2 = 1 + (2 - 1)\alpha = 1 + \alpha

\alpha = 0.2 = \mathbf{20\%}

So 20% of the AB molecules have dissociated into ions.

The colligative property formula with the van’t Hoff factor:

\Delta T_b = i \times K_b \times m

Assuming molality $\approx$ molarity for dilute solutions: $m \approx 0.1$ mol/kg.

\Delta T_b = 1.2 \times 0.52 \times 0.1 = \mathbf{0.0624 \text{ K}}

Why This Works

The van’t Hoff factor $i$ tells us how many particles actually exist in solution compared to what we’d expect without dissociation. If there’s no dissociation, $i = 1$ . If every molecule dissociates completely into $n$ ions, $i = n$ .

For partial dissociation, we’re somewhere in between. Starting with 1 mole of AB: $\alpha$ fraction dissociates, giving $\alpha$ moles of A⁺ and $\alpha$ moles of B⁻, while $(1 - \alpha)$ moles of AB remain undissociated. Total moles = $(1 - \alpha) + \alpha + \alpha = 1 + \alpha$ . The van’t Hoff factor is simply the ratio of actual particles to initial particles: $i = (1 + \alpha)/1 = 1 + \alpha$ .

This generalises to $i = 1 + (n-1)\alpha$ when the electrolyte gives $n$ ions.

Alternative Method

You can derive $\alpha$ from first principles using an ICE-style table:

	AB	A⁺	B⁻
Initial	1	0	0
Change	$-\alpha$	$+\alpha$	$+\alpha$
Equilibrium	$1-\alpha$	$\alpha$	$\alpha$

Total particles = $1 - \alpha + \alpha + \alpha = 1 + \alpha$

So $i = \frac{1 + \alpha}{1} = 1 + \alpha$ , giving $\alpha = i - 1 = 0.2$ .

For association (like acetic acid dimerising in benzene), the formula changes to $i = 1 + (1/n - 1)\alpha$ where $n$ is the number of molecules that combine. Here $i < 1$ . JEE loves mixing up dissociation and association in the same question — read carefully whether the solute breaks apart or comes together.

Common Mistake

Students often plug $n = 1$ into the formula for AB. The value of $n$ is the number of ions produced per formula unit upon complete dissociation, not the number of formula units. AB gives A⁺ + B⁻ = 2 ions, so $n = 2$ . For something like $\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}$ , $n = 3$ . Getting $n$ wrong throws off the entire calculation.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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