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If a base has Kb = 4.09 x 10-8, the value of Ka for the conjugate...

If a base has Kb = 4.09 x 10-8, the value of Ka for the conjugate acid is___________________

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Answer #1

The required answer is 2.44 * 10-7 .

EXPLANATION -

Let the base is given by B.

B + H2O = BH+ + OH

Kb for the base will be given by

[BH+][ОН- К) =

[It is assumed that the concentration of water does not change much, so it is excluded.]

BH+ is the conjugate acid.

Now the conjugate acid has its own equilibrium.

BH+ =B + H+

Ka for the conjugate acid will be given by

K_a=\frac{[B][H^+]}{[BH^+]}

  [Holl_] -Holl+Ha] – al

  Kw K

8-01 X 601 - 11-01

= 2.44 * 10-7

So the required answer is 2.44 * 10-7 .

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