## Chemistry and Chemical Reactivity (9th Edition)

$Kb = 4.187\times 10^{- 4}$
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: $Kb = \frac{[OH^-][CH_3N{H_3}^+]}{ [CH_3NH_2]}$ $Kb = \frac{x^2}{[Initial CH_3NH_2] - x}$ $Kb = \frac{( 5.012\times 10^{- 3})^2}{ 0.065- 5.012\times 10^{- 3}}$ $Kb = \frac{ 2.512\times 10^{- 5}}{ 0.05999}$ $Kb = 4.187\times 10^{- 4}$