## Chemistry and Chemical Reactivity (9th Edition)

$Kb = 6.642\times 10^{- 9}$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [H_3NOH^+] = x$ -$[H_2NOH] = [H_2NOH]_{initial} - x$ 2. Calculate the $[OH^-]$ value: pH + pOH = 14 9.11 + pOH = 14 pOH = 4.89 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 4.89}$ $[OH^-] = 1.288 \times 10^{- 5}$ 3. Write the Kb equation, and find its value: $Kb = \frac{[OH^-][H_3NOH^+]}{ [H_2NOH]}$ $Kb = \frac{x^2}{[Initial H_2NOH] - x}$ $Kb = \frac{( 1.288\times 10^{- 5})^2}{ 0.025- 1.288\times 10^{- 5}}$ $Kb = \frac{ 1.66\times 10^{- 10}}{ 0.02499}$ $Kb = 6.642\times 10^{- 9}$