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Relationship between Ka and Kb of the exact same molecule?
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Emily_6585
#1 Posted : Tuesday, July 13, 2021 10:07:18 PM
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I know that the Ka (of an acid) times the Kb (of its conjugate base) is equal to Kw.

ie. (Ka of H2PO4-) x (Kb of HPO4--) = 1 x 10^-14 @ room temp


But is there a way to calculate the Ka of a molecule acting as an acid and from the Kb of that exact same molecule if it was to act as a base?

ie. The Ka for HPO4-- + H20 -> Po4(3-) + H3O+ from the Kb of HPO4-- + H2O -> H2PO4- + OH-


Additionally, would it be true to say that as a molecule's Ka increases its Kb decreases (because if something is a strong acid it's a weak base)? Or can you only say that if something's a strong acid then its conjugate base is weak base?
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INSTR_Calla_139
#2 Posted : Thursday, July 15, 2021 4:42:50 PM
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Hi Emily,

To answer your first question, yes you can just solve for the variable that you don't know in the equation you wrote above. If you solve for the Kb you can then plug that into the Ka x Kb =Kw equation.

And for the second question, yes it is true to say that if the molecule's Ka increases the Kb decreases - or if it is a strong acid it will be a weak base. If you look at the equation Ka x Kb = Kw , this shows that they are inversely proportional so if one goes up, the other will go down. I would be careful with the wording of what you said about a strong acid being a weak base because it is in fact not a base at all - it is an acid.

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Emily_6585
#3 Posted : Friday, July 16, 2021 5:24:31 PM
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Thanks for your answer, can I just confirm what you're saying?


Are you saying I can solve for the Ka of HPO4-- by doing 14/[Kb of HPO4--]?

I thought to use that equation I had to use the Kb of the conjugate base, not the Kb of the same molecule?


Perhaps I should have said stronger acid instead of strong. My question is referring to molecules that can act both as an acid and a base (ie. HPO4-- can act as an acid when it donates a proton to become PO4(3-) but could also act as a base by accepting a proton to become H2PO4-).
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INSTR_Molly_129
#4 Posted : Saturday, July 17, 2021 9:03:29 PM
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Hi Emily,

Yes you are correct; we require the conjugate base form to calculate Kb, so for the same molecule, the Ka/Kb could be different, as shown here in the chart (focus on HSO4-):

https://chem.libretexts....etween_Ka_Kb_pKa_and_pKb

In a diprotic system, where:
Ka1 and Kb1 represents values of the first (de)protonation
Ka2 and Kb2 represents the values of the second (de)protonation,

Kw = Ka1 x Kb2.

So, if H2SO4 --1st--> HSO4- --second-->SO4 2-,
Ka of H2SO4 x Kb of HSO4- = Kw.

In your example:

H3PO4 --> H2PO4- --> HPO4 2- --> PO4 3-

If you have Kb of HPO4 2- (I think that's what you meant?), you can find the Ka of H2PO4-, using the Kw relationship. If you want the Ka of HPO4-, you will need the Kb of PO4 3-.

Each Ka x Kb relationship is based on one deprotonation. So if you want to find the Ka of a certain molecule, you will have to have the Kb of its conj. base form.

For your second question, Kb > Ka for bases, and Ka > Kb for acids.

Hope that helps!

Molly
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