Psichyk - 19-11-2016 at 12:40
Hello guys. I need some help with this problem. It says: "The melting point of a solution of vinylacetic acid 0.0500 M is -0.096ºC. Determine
its Ka".
I think this can be solved by calculating the molality of the solution from the cryoscopic variation equation, but i can't find the relation between
molality and pH. I hope you can help me 
.
Thank you in advance.
Metacelsus - 19-11-2016 at 16:54
Colligative properties depend on the total number of molecules in solution. This will vary depending on the extent of dissociation of the acid. From
this, you can calculate the Ka.
Here is how I calculate it:
From the freezing point depression, the effective molarity of the solution is 0.0518 M (molarity and molality are nearly equivalent here). By mass
balance, this means that there are 0.0482 M of HA, 0.0018 M of A-, and 0.0018 M of H3O+. Thus, the Ka will
be 6.72 * 10-5.
[Edited on 11-20-2016 by Metacelsus]
CuReUS - 20-11-2016 at 08:00
I am getting Ka as 4.63 * 10-5
[Edited on 20-11-2016 by CuReUS]
Psichyk - 21-11-2016 at 07:26
I see, thank you so much. . I am also getting Ka = 6.72*10^-6.
[Edited on 21-11-2016 by Psichyk]