Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 16 Principles of Chemical Reactivity: The Chemistry of Acids and Bases - Study Questions - Page 629c: 50

Answer

$[H_3O^+] = 1.265 \times 10^{- 5}M $ $[A^-] = 1.265 \times 10^{- 5}M $ $[HA] \approx 0.04M$

Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: -$[H_3O^+] = [A^-] = x$ -$[HA] = [HA]_{initial} - x = 0.04 - x$ For approximation, we consider: $[HA] = 0.04M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][A^-]}{ [HA]}$ $Ka = 4 \times 10^{- 9}= \frac{x * x}{ 0.04}$ $Ka = 4 \times 10^{- 9}= \frac{x^2}{ 0.04}$ $ 1.6 \times 10^{- 10} = x^2$ $x = 1.265 \times 10^{- 5}$ Percent ionization: $\frac{ 1.265 \times 10^{- 5}}{ 0.04} \times 100\% = 0.03162\%$ %ionization < 5% : Right approximation. Therefore: $[H_3O^+] = [A^-] = x = 1.265 \times 10^{- 5}M $ And, since 'x' has a very small value (compared to the initial concentration): $[HA] \approx 0.04M$
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