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Aniline, C(6)H(5)NH(2) react with water ...

Aniline, `C_(6)H_(5)NH_(2)` react with water according to the equation
`C_(6)H_(5)NH_(2)(aq)+H_(2)O(I)hArrC_(6)H_(5)NH_(3)^(+)(aq)+OH^(-)(aq)`
In a 0.180 M aqueous aniline solution the `[OH^(-)]=8.80xx10^(-6)M`
The value of the base ionization constant `K_(b)` for `C_(6)H_(5)NH_(2)(aq)` and the percent ionization of `C_(6)H_(5)NH_(2)` in this solution are,

A

`4.3xx10^(-10)`

B

`3.1xx10^(-10)`

C

`4.9xx10^(-3)%`

D

`2.4xx10^(-3)%`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`K_(b)=([C_(6)H_(5)NH_(3)^(+)][OH^(-)])/([C_(6)H_(5)NH_(2)])`
`K_(b)=((8.80xx10^(-6))(8.80xx10^(-6)))/((0.180))=4.3xx10^(-10)`
`((8.80xx10^(-6))/((0.180))xx100%=4.9xx10^(-3)%`
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