[Atome]

Hello all:

I am having some trouble understanding why my attempt at solving the question is incorrect.

Could someone please explain the solution?

Thank you!

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1. The freezing point of a 0.0830 molal aqueous acetic acid is -0.159°C. Acetic acid, HC₂H₃O₂, is partially dissociated according to the equation:

 HC₂H₃O_{2 (aq)} H⁺ _(aq) + C₂H₃O_{2 (aq)}

Calculate the percentage of  HC₂H₃O₂ molecules that are dissociated, assuming that the equation of freezing-point depression holds for the total concentration of molecules and ions in the solution.

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Attempt:

I assumed that the solvent was 1 kg H₂O, so there would be 0.0830 moles of HC₂H₃O₂.

I used  : to find that there were 0.0856 moles of both the hydron and acetate.

Therefore, I thought that the percentage of acetic acid molecules dissociated = 0.0830 mols/0.0856 mols = 97.0%.

However, the textbook's answer is:

SPOILER

3.1%.

[Borek]

I thought that the percentage of acetic acid molecules dissociated = 0.0830 mols/0.0856 mols = 97.0%.

Why?

HA + A^- + H⁺ = 0.0856

HA + A^- = 0.0830

[Atome]

Hello Borek:

Thanks for your response.

Could you please explain what A denotes?

Also, I am not familiar with equilibrium concepts so I must solve this problem using concepts on colligative properties.

[Borek]

Sorry, this is shorthand notation I use whenever I need to write about an acid. A^- stands for any anion from acid dissociation. This time HA stands for acetic acid and A^- for acetate.

HA <-> H⁺ + A^-

Also, I am not familiar with equilibrium concepts so I must solve this problem using concepts on colligative properties.

Well, you have to use percentage dissociation definition and for sure things like mass preservation (which leads to mass balance) and stoichiometry of dissociation can be used as well. In fact not using these things you can't solve the problem.

[Atome]

Thanks for your reply, Borek.

I got the answer!