Answer
$Kb = 9.908\times 10^{- 10}$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [CH_3C_6H_4N{H_3}^+] = x$
-$[CH_3C_6H_4NH_2] = [CH_3C_6H_4NH_2]_{initial} - x$
2. Calculate the [OH^-]
pH + pOH = 14
8.6 + pOH = 14
pOH = 5.4
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 5.4}$
$[OH^-] = 3.981 \times 10^{- 6}$
3. Write the Kb equation, and find its value:
$Kb = \frac{[OH^-][CH_3C_6H_4N{H_3}^+]}{ [CH_3C_6H_4NH_2]}$
$Kb = \frac{x^2}{[Initial CH_3C_6H_4NH_2] - x}$
$Kb = \frac{( 3.981\times 10^{- 6})^2}{ 0.016- 3.981\times 10^{- 6}}$
$Kb = \frac{ 1.585\times 10^{- 11}}{ 0.016}$
$Kb = 9.908\times 10^{- 10}$
