[brb725]

I have a quick question for you chemists who are much brighter than I. How can I determine the Ka value of a weak, monoprotic acid (with a concentration of 0.10 m) through finding the percent ionization by freezing point depression? I completely understand how to find the percent ionization, but I am confused on how to translate that data into the ka value for the weak acid. Thanks in advance.

[Borek]

Think about stoichiometry of dissociation. How does dissociation change number of molecules/ions dissolved?

[brb725]

Think about stoichiometry of dissociation. How does dissociation change number of molecules/ions dissolved?

Dissociation will increase the number of molecules dissolved. So, if I was working with a 0.10 molal solution of HCl, for example, I would have an effective molality of 0.20, correct? But for a weak acid it will be much less than that.

But, I am still confused on how to find the Ka value. Will the percent ionization be the Ka value for the weak acid? Or, would the Ka value be equal to the van't Hoff factor?

[xenon]

what do you mean by effective molality

[Borek]

But, I am still confused on how to find the Ka value. Will the percent ionization be the Ka value for the weak acid? Or, would the Ka value be equal to the van't Hoff factor?

None. Look at the definition of percent ionization. Think, how you can use it to calculate concentrations of HA and H⁺/A^-. Once you will get through these calculations, all you need is to put these values into Ka definition.

[brb725]

what do you mean by effective molality

Again, if I was using 0.10 molal of HCL, it would completely dissociate into a proton and a chloride ion, so you'd effectively have 0.20 moles of particles affecting the freezing point for every kilogram of solvent. So, as far as freezing point (or boiling point) is concerned, the solution has an effective molality of 0.20 molal.

[brb725]

None. Look at the definition of percent ionization. Think, how you can use it to calculate concentrations of HA and H⁺/A^-. Once you will get through these calculations, all you need is to put these values into Ka definition.

Ok, I think I'm following you. So, if I have a 0.10 molal solution of a weak acid, and I experimentally find that it has a percent ionization of 5%, I would multiply 0.10 by 0.05, which would give me a concentration of 5.0 x 10^-3 molal for both H⁺ and A^-, correct? And I would subtract 5.0 x 10^-3 from 0.10, giving me a concentration of 0.095 molal of HA.

So, would Ka = ([5.0 x 10^-3] X [5.0 x 10^-3])/ [0.095] = 2.6 x 10^-4
or did I do that completely wrong?

[Borek]

Wasn't that hard

[brb725]

Definitely a lot easier than I thought . Thanks for all the help, it is very appreciated.