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K(a) for a weak monobasic acid is 1.0xx1...

`K_(a)` for a weak monobasic acid is `1.0xx10^(-6)`. The `pK_(b)` of its conjugate base base is

A

`8.0`

B

`1.0xx10^(-8)`

C

`1.0xx10^(-4)`

D

`6.0`

Text Solution

Verified by Experts

The correct Answer is:
A
For any conjugate acid-base pair, it is always true that
`K_(a)K_(b)=K_(w)`
`:. K_(b)=(K_(w))/(K_(a))`
`=(1.0xx10^(-14))/(1.0xx10^(-6))`
`=1.0xx10^(-8)`
Now, `pK_(b)=-1og K_(b)= -1og(1.0xx10^(-8))`
`= 8`
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