Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 744: 47

Answer

Percent ionization $\approx$ 0.57%.

Work Step by Step

1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [N{H_4}^+] = x$ -$[NH_3] = [NH_3]_{initial} - x = 0.56 - x$ For approximation, we consider: $[NH_3] = 0.56M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.56}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.56}$ $ 1.008 \times 10^{- 5} = x^2$ $x = 3.175 \times 10^{- 3}$ Percent ionization: $\frac{ 3.175 \times 10^{- 3}}{ 0.56} \times 100\% = 0.5669\%$ %ionization < 5% : Right approximation.
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