Question

Kb for an unknown base is 3.7 x 10-9. What is the acid dissociation constant for the conjugate acid?

Answer 1

$Ka$ is the acid dissociation constant.

$Ka*Kb=Kw$

where kw is the water dissociation constant.

and $Kw$ =1.0 * 10^^-14

$Ka*(3.7*10^{-9})=1.0*10^{-14}$

$Ka=\dfrac{1.0*10^{-14}}{3.7*10^{-9}}$

$=2.7*10^{-6}$