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7.52 What is the pH of 0.001M aniline solution? Calculate the degree of ionization of aniline in the solution. Also, calculate the ionization constant of the conjugate acid of aniline.(K_b= $4.27\times 10^{-10}$ )

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We have,
C = 0.001 M
$K_b$ = $4.27\times 10^{-10}$
Degree of ionization of aniline (a) = ?
Ionization constant of the conjugate acid ( $K_a$ ) = ?

We know that
$K_b$ = $C.a^2$
$4.27\times 10^{-10}$ = (0.001) $a^2$
Thus $a={\sqrt{4270\times 10^{-10}}}$
= $65.34\times 10^{-4}$

Then [Base] = C.a = ( $65.34\times 10^{-4}$ )( $0.001$ )
= $0.653\times 10^{-5}$

Now, $P^{OH} = -\log (0.65\times 10^{-5})=6.187$
$P^H = 14-P^{OH}$
= 14 - 6.187
= 7.813

It is known that,

$K_a\times K_b = K_w$
So, $K_a = \frac{10^{-14}}{4.27\times10^{-10}}$
$=2.34\times 10^{-5}$ This is ionization constant.

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