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### pKa and pKb

Posted: Sun Jan 19, 2020 10:50 pm
I am a little unsure of how these are calculated and how they share a relationship. Can anyone give me a brief explanation?

### Re: pKa and pKb

Posted: Sun Jan 19, 2020 10:55 pm
PKa is calculated by the -log(kA) and pKb is calculated by the -log(kB). If you add pKa and pKb together, it will equal 14.

### Re: pKa and pKb

Posted: Sun Jan 19, 2020 11:01 pm
These are just logarithmic versions of Ka and Kb, their relationship is that they add up to 14.

### Re: pKa and pKb

Posted: Mon Jan 20, 2020 8:06 am
always go back to pH= -log[H30+] and pOH= -log[OH-]. From this you can get the other relationships

### Re: pKa and pKb

Posted: Mon Jan 20, 2020 9:26 am
pKa is the negative base-10 logarithm of the acid dissociation constant (Ka) of a solution and pKb is the negative base-10 logarithm of the base dissociation constant (Kb). pKa and pKb added together equals 14.

### Re: pKa and pKb

Posted: Mon Jan 20, 2020 9:55 am
Think of Pka and Pkb as being calculated pretty much the same exact way as pH, except in this case you could calculate -log(ka) and ka is the acid dissociation constant used for weak acids.

### Re: pKa and pKb

Posted: Mon Jan 20, 2020 11:48 am
pka = -log(ka)
pkb=-log(kb)
pka+pkb=pkw
pkw = 14.00 at 25 degrees celsius
ka= 10^-pka
kb=10^-pkb
ka x kb = kw

### Re: pKa and pKb

Posted: Mon Jan 20, 2020 10:39 pm
pKa and pKb are acidity and basicity constants which are given as their negative log:
pKa=-log(ka)
pKb=-log(kb)

### Re: pKa and pKb

Posted: Tue Jan 21, 2020 1:18 am
p just means its the log of something. so pKa is log(Ka) and pKb is log(Kb).

### Re: pKa and pKb

Posted: Wed Jan 22, 2020 12:28 pm
pka and pkb are added together to get 14; so, if you know one, you can subtract it from 14 to get the other

### Re: pKa and pKb

Posted: Thu Jan 23, 2020 7:55 pm
@stephaniekim2k , That is a really great way of explaining it, thank you! It helped me a lot!

### Re: pKa and pKb

Posted: Fri Jan 24, 2020 12:21 pm
when you are given Ka and Kb, you take -log of that to find pka and pkb. Pka + Pkb = 14

### Re: pKa and pKb

Posted: Fri Jan 24, 2020 9:47 pm
pKa + pKa = Kw (autoprotolysis constant) = 14

This equation can be flipped around respectively to find a value you need.

### Re: pKa and pKb

Posted: Fri Jan 24, 2020 10:15 pm
pKa's are used for acids and pKb's are used for bases. The K value for a reaction between an acid and liquid water is deemed a Ka value. A Kb value is the K value for a reaction between a base and liquid water. pKa's are calculated by taking the negative log of the Ka. pKb's are calculated by taking the negative log of the Kb.

### Re: pKa and pKb

Posted: Fri Jan 24, 2020 11:54 pm
Ka and Kb are calculated the same way as K but for weak acids and bases, with the concentration of products on the top and the concentration of reactants on the bottom. PKa and Pkb are the -log(Ka) and -log(Kb)

### Re: pKa and pKb

Posted: Sat Jan 25, 2020 12:32 am
the sum of pka and pkb is equivalent to the pkw which is 14.
the product of ka and kb is kw, which is 10-14.

ka is the equilibrium constant for an acid.
kb is the equilibrium constant for a base.

it's good to know that the lower the pKa, the higher the Ka, and the stronger the acid; and vice versa. the same also applies to the pKb

### Re: pKa and pKb

Posted: Sun Jan 26, 2020 11:44 pm
pKa is basically the hydronium concentration while pKb is the hydroxide concentration of a solution. pKa and pKb when added together always equals 14. You can calculate pKa or pKb from having Ka(equilibrium constant for acids) or Kb (equilibrium constant for bases) and taking the negative log.

### Re: pKa and pKb

Posted: Tue Jan 28, 2020 10:22 am
pKa = -logKa
pKb = -logKb

pka + pkb = 14