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If K(h) (hydrolysis constant) for anilin...

If `K_(h)` (hydrolysis constant) for anilinium ion is `2.4xx10^(-5) M`, then `K_(b)` for aniline will be

A

`4.1xx10^(10)`

B

`4.1xx10^(-10)`

C

`2.4xx10^(9)`

D

`2.4xx10^(-19)`

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(h)` for anilinium ion `=2.4xx10^(-5) M`
`K_(h)=(K_(w))/(K_(b))`
`K_(b)=(10^(-14))/(2.4xx10^(-5))=4.1xx10^(-10)`
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