Answer 1

K_b = 4.27 × 10^–10

c = 0.001M

pH =?

α =?

`k_b = calpha^2`

`4.27 xx 10^(-10)= 0.0014 xx alpha^2`

`4270 xx 10^(-10) = alpha^2`

`65.34 xx 10^(-5) = alpha = 65.34 xx 10^(-4)`

Then [anoin] = `calpha = .001 xx 65.34 xx 10^(-5)`

`= .065 xx 10^(-5)`

`pOH = -log (.065 xx 10^(-5))`

= 6.187

pH = 7.813

Now,

`K_a xx K_b = K_w`

`:. 4.27 xxx 10^(-10) xx K_a = K_w`

`K_a = 10^(-14)/(4.27xx 10^(-10))`

`= 2.34 xx 10^(-5)`

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10^–5.