## Chemistry 10th Edition

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [Conj. Acid] = x$ -$[Base] = [Base]_{initial} - x$ 2. Now, use the percent ionization to find the "x" value. %ionization = $\frac{x}{[Initial Base]} \times 100$ 0.053= $\frac{x}{ 5\times 10^{- 3}} \times 100$ 0.00053= $\frac{x}{ 5\times 10^{- 3}}$ $2.65\times 10^{- 6}= x$ 3. Write the Kb equation, and find its value: $Kb = \frac{[OH^-][Conj. Acid]}{ [Base]}$ $Kb = \frac{x^2}{[Initial Base] - x}$ $Kb = \frac{( 2.65\times 10^{- 6})^2}{ 0.005- 2.65\times 10^{- 6}}$ $Kb = \frac{ 7.022\times 10^{- 12}}{ 4.997\times 10^{- 3}}$ $Kb = 1.405\times 10^{- 9}$ Calculate the pKb Value $pKb = -log(Kb)$ $pKb = -log( 1.45 \times 10^{- 9})$ $pKb = 8.839$