# chem

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show why ka(kb)=kw for this pair of conjugate acid-base: H2C6H6O6 and HC6H66 ^-1

• chem -

It's too long and too confusing using long formulas. Let's cut that to ANY acid and its conjugate base, HA.
The conjugate base is A^- so if we add that to water we have
..........A^- + HOH ==> HA^- + OH^-

(HA)(OH^-)
----------- = Kb for the conjugate base.
(A^-)

Now let's multiply that by (H^+)/(H^+)
I must use ..... for spacing and it still may not look very good.
(HA)(OH^-)......(H^+)
----------- x ------ = Kb
(A^-)............(H^)
Now look at the equation.
In the numerator, (H^+)(OH^-) = Kw.
In the denominator, we have the reciprocal of Ka. Note that is (A^-)/(H^+)(A^-) which is just 1/Ka. Therefore, Kb = Kw*1/Kb or just Kw/Kb.

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