van't Hoff factor for association and dissociation: deriving i with α, and why NaCl never shows exactly 2

6 answers & replies

• [+][–]Nisha Verma12thSample· 8d ago

  Derivation for dissociation. Start with 1 mol of solute, fraction $\alpha$ dissociates into $n$ particles each:

  • left undissociated: $1 - \alpha$
  • ions formed: $n\alpha$
  • total after dissociation: $1 - \alpha + n\alpha = 1 + (n-1)\alpha$

  Since the colligative property depends on the number of particles, $i = \frac{1 + (n-1)\alpha}{1} = 1 + (n-1)\alpha$.

  Example: BaCl₂ with $\alpha = 0.9$: $n = 3$, $i = 1 + 2(0.9) = 2.8$.

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• [+][–]Mehak AroraDropperSample· 8d ago

  For counting particles, K₄[Fe(CN)₆] gives 4 K⁺ ions, so i = 4. The complex ion is one unit so it doesn't count separately, it just stays together like one block... I think the 4 is from potassium only. That's how I did it in the last two tests and I got those right by luck maybe?

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  • [+][–]Ishita GoyalDropperSample· 8d ago

    Mehak, the complex anion [Fe(CN)₆]⁴⁻ is also a particle in solution, just a single (big) one. So the dissociation is

    $K_4[Fe(CN)_6] \to 4K^+ + [Fe(CN)_6]^{4-}$

    That's 4 + 1 = 5 particles, so $i = 5$ for complete dissociation. The complex stays intact (it doesn't break into Fe and CN), but it still counts as one particle. Similarly [Cu(NH₃)₄]SO₄ gives 2 and K₃[Fe(CN)₆] gives 4.

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• [+][–]Pooja Iyer12thSample· 7d ago

  Association mirrors it. Say $n$ molecules join to make 1 big molecule, with degree of association $\alpha$. Start with 1 mol of monomer:

  • remaining monomer: $1 - \alpha$
  • associated particles formed: $\alpha/n$
  • total: $1 - \alpha + \alpha/n$

  $i = 1 - \alpha\left(1 - \frac1n\right)$

  For acetic acid dimerising in benzene ($n = 2$), full association $\alpha = 1$ gives $i = 0.5$. If $\alpha = 0.8$, $i = 1 - 0.8\times0.5 = 0.6$. So $i<1$ for association, $i>1$ for dissociation.

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  • [+][–]Siddharth RaoDropperSample· 7d agoAuthor

    Both derivations are clear now. The 'particles count' idea is the key, I will stop memorising two separate formulas. Thank you, I owe you all chai.

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• [+][–]Zoya Khan12thSample· 7d ago

  On (3): in a solution that is not infinitely dilute, oppositely charged ions pull on each other and some of them stay close for a bit (ion pairing). Each one doesn't behave as a fully independent particle, so the effective number of free particles is slightly less than 2, and $i$ is slightly less than 2. As the solution gets more dilute the ions are farther apart, the attraction matters less, and $i$ comes closer and closer to 2. NaCl is a strong electrolyte, so it's 'fully dissociated into ions', but it's not 'ideal'.

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