## Chemistry 12th Edition

Published by McGraw-Hill Education

# Chapter 15 - Acids and Bases - Questions & Problems - Page 712: 15.57

#### 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: $[BOH] = [BOH]_{initial} - x$ $0.153 = [BOH]_{initial} - 1.66 \times 10^{-3}$ $[BOH]_{initial} = 0.1547M \approx 0.155M$

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