Answer
$Kb \approx 6.96 \times 10^{-7}$
Work Step by Step
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.
$Kb = \frac{4.57 \times 10^{-4} * 4.57 \times 10^{-4}}{0.30}$
$Kb = 6.96 \times 10^{-7}$
