Answer
$Kb = 6.642\times 10^{- 9}$
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.
-$[OH^-] = [H_3NOH^+] = x$
-$[H_2NOH] = [H_2NOH]_{initial} - x$
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:
$Kb = \frac{[OH^-][H_3NOH^+]}{ [H_2NOH]}$
$Kb = \frac{x^2}{[Initial H_2NOH] - x}$
$Kb = \frac{( 1.288\times 10^{- 5})^2}{ 0.025- 1.288\times 10^{- 5}}$
$Kb = \frac{ 1.66\times 10^{- 10}}{ 0.02499}$
$Kb = 6.642\times 10^{- 9}$
