## Chemistry 12th Edition

Equations needed: $pH + pOH = 14$ $[OH^-] = 10^{-pOH}$ $Kb = \frac{[OH^-][B^+]}{[BOH]}$** **All these concentrations are the values in the equilibrium. $[OH^-] = [B^+] = x$ $[BOH] = [BOH]_{initial} - x$ Kb (Ammonia) = $1.8 \times 10^{-5}$ Kb (Pyridine) = $1.7 \times 10^{-9}$ (a) Since the concentration has a large value, compared to the ka, we can assume: $[BOH] \approx [BOH]_{initial}$ Therefore: $1.8 \times 10^{-5} = \frac{x * x}{0.10}$ $1.8 \times 10^{-6} = x^2$ $x = 1.34 \times 10^{-3}M = [OH^-]$~ - Now, find the pH: $pOH = -log[OH^-]$ $pOH = -log( 1.34 \times 10^{- 3})$ $pOH = 2.87$ $pH + pOH = 14$ $pH + 2.87 = 14$ $pH = 11.13$ (b) Since the concentration has a large value, compared to the ka, we can assume: $[BOH] \approx [BOH]_{initial}$ Therefore: $1.7 \times 10^{-9} = \frac{x * x}{0.05}$ $8.5 \times 10^{-11} = x^2$ $x = 9.22 \times 10^{-6} = [OH^-]$ - Now, find the pH: $pOH = -log[OH^-]$ $pOH = -log( 9.22 \times 10^{- 6})$ $pOH = 5.03$ $pH + pOH = 14$ $pH + 5.03 = 14$ $pH = 8.97$