pka + pkb is...?

Question

my book says "the sum of pKa and pKb of a conjugate pair of a weak acid and weak base must always be equal to 14 at 25 degrees C"

wud it be more accurate to say that they must equal 14 regardless of the strength of the conjugate pairs?

Answer 1

As I recall, one always uses the relation pKa + pKb = 14 regardless of the strength of the conjugates. It works for strong acid / weak base, weak acid / strong base and weak acid / weak base. Strong acid and weak base do not have dissociation constants for weak solutions, they are considered fully ionized.

Answer 2

Pka Pkb 14

Answer 3

This is a toughie... my guess would be yes. If you assume the Ka or Kb of a strong acid or base is 1, then you're effectively saying it's 1 x 10^0, which gives a pKa or pKb of 0. Solving for the Kb or Ka using Kw, you end up with 1 x 10^14, which gives a pKb or pKa of 14. So in the end the sums would be 14... So that's true, probably, but if you're using this for a titration curve or something I'm not really sure that it's going to be helpful.

This fact is helpful for the fact that when you neutralize a weak acid with a strong base, the pH at the endpoint is basic (>7). The opposite is true when you neutralize a weak base with a strong acid: the pH at the endpoint is acidic (<7). I am pretty sure that's where this stuff will be applicable, but I could be wrong.

Sorry if I wasn't much of a help. Good luck!

Answer 4

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RE:
pka + pkb is...?
my book says "the sum of pKa and pKb of a conjugate pair of a weak acid and weak base must always be equal to 14 at 25 degrees C"

wud it be more accurate to say that they must equal 14 regardless of the strength of the conjugate pairs?

Answer 5

The two are inversely proportional, though.

If pka is large, then pkb must be small because the conjugate base will be much weaker than a stronger respective acid.

The 25 degrees statement is the real experimental part of the statement, because your actual data will deviate from 14 at different temperatures, for a variety of reasons.

Answer 6

If you're working with pure water you can assume that Kw= 1.0 x 10^-14 Kw= Ka x Kb therefore you can divide Kb into Kw and get Ka. Then to get pKa, you take the -log(Ka) Hope this helps!!