Question

An acid has an acid dissociation constant of 2.8 10–9. What is the base dissociation constant of its conjugate base?

Answer 1

The answer is 3.6 x 10^-6

Answer 2

Answer: The base dissociation constant for a conjugate base is $3.5\times 10^{-6}$

Explanation:

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant = $2.8\times 10^{-9}$

$K_b$ Base dissociation constant = ?

Putting values in above equation, we get:

$10^{-14}=2.8\times 10^{-9}\times K_b\\\\K_b=0.35\times 10^{-5}=3.5\times 10^{-6}$

Hence, the base dissociation constant for a conjugate base is $3.5\times 10^{-6}$