Answer
$[BOH]_{initial} \approx 0.155M$
Work Step by Step
Equations needed:
$pH + pOH = 14$
$[OH^-] = 10^{-pOH}$
$Kb = \frac{[OH^-][B^+]}{[BOH]}$**
**All these concentrations are the values in equilibrium.
$[OH^-] = [B^+] = x$
$[BOH] = [BOH]_{initial} - x$
Ammonia Kb: $1.8 \times 10^{-5}$
1. Find $[OH^-]$
$11.22 + pOH = 14$
$pOH = 2.78$
$[OH^-] = 10^{-pOH} = 10^{-2.78} = 1.66 \times 10^{-3}M$
2. Find $[B^+]$.
Pure solution:
$[B^+] = [OH^-] = 1.66 \times 10^{-3}M$
3. Use the Kb formula to find $[BOH]$
$1.8 \times 10^{-5} = \frac{1.66 \times 10^{-3} * 1.66 \times 10^{-3}}{[BOH]}$
$1.8 \times 10^{-5} * [BOH] = 2.76 \times 10^{-6}$
$[BOH] = 0.153M$
- Remember that, this concentration is the one at the equilibrium, and the question asks for the original:
$[BOH] = [BOH]_{initial} - x$
$0.153 = [BOH]_{initial} - 1.66 \times 10^{-3}$
$[BOH]_{initial} = 0.1547M \approx 0.155M$
