Question

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

2.8x 10^-23
3.6x10^-6
2.8x10^5
3.6 x10^22

Answer 1

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

Answer 2

Answer: The correct answer is
`3.6* 10^(-6)`

Step-by-step explanation:

We are given:

Acid dissociation constant
`(k_a)=2.8* 10^(-9)`

Water dissociation constant
`(k_w)=1* 10^(-14)`

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

`k_w=k_a* k_b`

where,

`k_b` = base dissociation constant

Putting values in above equation, we get:

`10^(-14)=2.8* 10^-9}* k_b\\\\k_b=3.6* 10^(-6)`

Hence, the correct answer is
`3.6* 10^(-6)`