pKa/pKb and Ka/Kb

Moderators: Chem_Mod, Chem_Admin

Diana Andrade_4F
Posts: 50
Joined: Tue Nov 05, 2019 12:18 am

pKa/pKb and Ka/Kb

Postby Diana Andrade_4F » Thu Jan 23, 2020 4:21 pm

What is the difference between pKa/pKb and Ka/Kb?

Amanda Lin 2I
Posts: 101
Joined: Sat Aug 17, 2019 12:15 am

Re: pKa/pKb and Ka/Kb

Postby Amanda Lin 2I » Thu Jan 23, 2020 4:26 pm

Ka is the acid dissociation constant. Kb is the base dissociation constant.

pKa = -log(Ka)
pKb = -log(Kb)

Connor Chappell 2B
Posts: 59
Joined: Wed Feb 20, 2019 12:16 am

Re: pKa/pKb and Ka/Kb

Postby Connor Chappell 2B » Thu Jan 23, 2020 4:28 pm

Ka is the acidity constant, and Kb is the basicity constant.

pKa = -log(Ka)
pKb = -log(Kb)

Jasmine Kim 1L
Posts: 71
Joined: Fri Aug 02, 2019 12:16 am

Re: pKa/pKb and Ka/Kb

Postby Jasmine Kim 1L » Thu Jan 23, 2020 5:11 pm

The p before Ka and Kb just shows that it is the -log value of Ka and Kb, like how pH and pOH means -log[H3O+] and -log[OH-]

Kallista McCarty 1C
Posts: 212
Joined: Wed Sep 18, 2019 12:18 am

Re: pKa/pKb and Ka/Kb

Postby Kallista McCarty 1C » Thu Jan 23, 2020 5:12 pm

Ka = 10^-pKa
Kb = 10^-pKb

Jialun Chen 4F
Posts: 108
Joined: Sat Sep 07, 2019 12:16 am

Re: pKa/pKb and Ka/Kb

Postby Jialun Chen 4F » Thu Jan 23, 2020 5:44 pm

The relationship is analogous to that of pH/pOH and [H3O+]/[OH-].
Essentially,
10^-pKa=Ka (same applies for Kb)
pKa=-log(Ka) (same applies for Kb).


Return to “Non-Equilibrium Conditions & The Reaction Quotient”

Who is online

Users browsing this forum: No registered users and 2 guests