Question

An acid has an acid dissociation constant of 2.8x10^-9. What is the base dissociation constant of its conjugate base?

Answer 1

Answer: The correct answer is $3.6\times 10^{-6}$

Explanation:

We are given:

Acid dissociation constant $(k_a)=2.8\times 10^{-9}$

Water dissociation constant $(k_w)=1\times 10^{-14}$

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

$k_w=k_a\times k_b$

where,

$k_b$ = base dissociation constant

Putting values in above equation, we get:

$10^{-14}=2.8\times 10^-9}\times k_b\\\\k_b=3.6\times 10^{-6}$

Hence, the correct answer is $3.6\times 10^{-6}$

Answer 2

PH=-lg[H⁺]
pH=-lg(2.8×10⁻⁹)=8.553

pOH=14-pH
pOH=14-8.553=5.447

[OH⁻]=10^(-pOH)
[OH⁻]=10⁻⁵·⁴⁴⁷=3.6×10⁻⁶