dextro88
Hazard to Self
Posts: 60
Registered: 20-10-2018
Location: In the lab
Member Is Offline
Mood: loading...
|
|
Did anyone can find me the pKa of these bases salts
For a little experiment i need to know exactly whats the pKa of :
methylamine acetate
ethylamine acetate
and will be very thanksfull if anyone can show me how i can calculate the pKa of various bases salts.
best regards 
[Edited on 10-6-2020 by dextro88]
|
|
|
Tsjerk
International Hazard
Posts: 3022
Registered: 20-4-2005
Location: Netherlands
Member Is Offline
Mood: Mood
|
|
These compounds are not acids, they are neutral salts, so they don't have a pka.
(M)ethylamine has a pKb and acetic acid has a pka.
|
|
|
dextro88
Hazard to Self
Posts: 60
Registered: 20-10-2018
Location: In the lab
Member Is Offline
Mood: loading...
|
|
thanks for the lesson, i need to know wich one is more acidic, i know most of the bases salts have a pKa by wich they get neutralized and are
liberatet from acidic ions,amonium acetate have pKa of 9.9, but i dont understand why its not the same here.
many thanks, dextro
[Edited on 10-6-2020 by dextro88]
|
|
|
Tsjerk
International Hazard
Posts: 3022
Registered: 20-4-2005
Location: Netherlands
Member Is Offline
Mood: Mood
|
|
Well, methylammonium ion has a pKa of 10.6 and ethylammonium ion 10.8
|
|
|
dextro88
Hazard to Self
Posts: 60
Registered: 20-10-2018
Location: In the lab
Member Is Offline
Mood: loading...
|
|
yes thats already accesible information, but i want to learn the calculation behind how to calculate the pKa of these various salts, for example
anhydrous amonia have a pKa of 37 and in presence of water much lower and acetic acid have pKa of 4.76, how the amonium acetate pKa is calculated 9.9
i just cant understand the math behind that.
[Edited on 10-6-2020 by dextro88]
|
|
|
Tsjerk
International Hazard
Posts: 3022
Registered: 20-4-2005
Location: Netherlands
Member Is Offline
Mood: Mood
|
|
Have a look at the answer given by Bive:
https://chemistry.stackexchange.com/questions/75047/ph-of-am...
|
|
|
dextro88
Hazard to Self
Posts: 60
Registered: 20-10-2018
Location: In the lab
Member Is Offline
Mood: loading...
|
|
thank you for your effort, im still newbee its a litle hard calculations for me, but i will work my way to understand and do these calculations
myself.
Best regards
[Edited on 11-6-2020 by dextro88]
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
