## Chemistry 10th Edition

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: ** The image is in the end of this answer. -$[OH^-] = [Conj. Acid] = x$ -$[Benzamide] = [Benzamide]_{initial} - x$ 2. Calculate the $[OH^-]$ $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 2.91}$ $[OH^-] = 1.23 \times 10^{- 3}$ $x = 1.23 \times 10^{- 3}$ 3. Write the Kb equation, and find its value: $Kb = \frac{[OH^-][Conj. Acid]}{ [Benzamide]}$ $Kb = \frac{x^2}{[Initial Benzamide] - x}$ $Kb = \frac{( 1.23\times 10^{- 3})^2}{ 0.068- 1.23\times 10^{- 3}}$ $Kb = \frac{ 1.513\times 10^{- 6}}{ 0.06677}$ $Kb = 2.266\times 10^{- 5}$ 4. Calculate the pKb Value $pKb = -log(Kb)$ $pKb = -log( 2.266 \times 10^{- 5})$ $pKb = 4.645$