## Chemistry 12th Edition

Published by McGraw-Hill Education

# Chapter 15 - Acids and Bases - Questions & Problems - Page 712: 15.58

#### Answer

The percent of $NH_3$ that is $NH_4^+$ in this solution is 1.5%.

#### Work Step by Step

Equations needed: $Kb = \frac{[OH^-][B^+]}{[BOH]}$** **All these concentrations are the values in the equilibrium. $[OH^-] = [B^+] = x$ $[BOH] = [BOH]_{initial} - x$ Kb ammonia = $1.8 \times 10^{-5}$ -Remember, for ammonia: $BOH = NH_3$ (The base) and $B^+ = NH_4^+$ (The conjugate acid) 1. Find $[NH_4^+]$. It is the conjugate acid: $[NH_4^+] = [B^+]$ - Since 0.08M is a large number, compared to the Kb value, we can consider: $[BOH] \approx [BOH]_{initial}$ Therefore: $1.8 \times 10^{-5} = \frac{x * x}{0.08}$ $1.44 \times 10^{-6} = x^2$ $x = 1.2 \times 10^{-3}M = [OH^-] = [B^+]$ 2. Now we know that: $[NH_4^+] = 1.2 \times 10^{-3}M$ and $[NH_3]_{initial} = 0.08M$ So the percent between these 2 is: $\frac{1.2 \times 10^{-3}}{0.08} \times 100\% = 1.5\%$

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