## Chemistry 12th Edition

$Kb \approx 6.96 \times 10^{-7}$
Equations needed: $pH + pOH = 14$ $[OH^-] = 10^{-pOH}$ $Kb = \frac{[OH^-][B^+]}{[BOH]}$** **All these concentrations are the values in the equilibrium. $[OH^-] = [B^+] = x$ $[BOH] = [BOH]_{initial} - x$ 1. Calculate $[OH^-]$. 10.66 + pOH = 14 pOH = 3.34 $[OH^-] = 10^{-3.34} = 4.57 \times 10^{-4}M$ 2. Find $[B^+] and [BOH]$ $[B^+] = [OH^-] = 4.57 \times 10^{-4}$ $[BOH] = 0.30M - 4.57 \times 10^{-4}M \approx 0.30M$ 3. Now, use the Kb formula to find its value. $Kb = \frac{4.57 \times 10^{-4} * 4.57 \times 10^{-4}}{0.30}$ $Kb = 6.96 \times 10^{-7}$