Answer
$Kb = 4.187\times 10^{- 4}$
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.
-$[OH^-] = [CH_3N{H_3}^+] = x$
-$[CH_3NH_2] = [CH_3NH_2]_{initial} - x$
2. Calculate the hydroxide concentration:
pH + pOH = 14
11.7 + pOH = 14
pOH = 2.3
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 2.3}$
$[OH^-] = 5.012 \times 10^{- 3}$
3. Write the Kb equation, and find its value:
$Kb = \frac{[OH^-][CH_3N{H_3}^+]}{ [CH_3NH_2]}$
$Kb = \frac{x^2}{[Initial CH_3NH_2] - x}$
$Kb = \frac{( 5.012\times 10^{- 3})^2}{ 0.065- 5.012\times 10^{- 3}}$
$Kb = \frac{ 2.512\times 10^{- 5}}{ 0.05999}$
$Kb = 4.187\times 10^{- 4}$
