Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 16 - Questions and Problems - Page 772: 16.59

Answer

$Ka = 1.281 \times 10^{-6}$

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. -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x $ 2. Write the percent ionization equation, and find 'x': %ionization = $\frac{x}{[Initial Acid]} \times 100$ 0.92= $\frac{x}{ 0.015} \times 100$ 0.0092= $\frac{x}{ 0.015}$ $1.38\times 10^{- 4}= x$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{[Initial Acid] - x}$ $Ka = \frac{ (1.38\times 10^{- 4})^2}{ 0.015- 1.38\times 10^{- 4}}$ $Ka = \frac{ 1.904\times 10^{- 8}}{ 1.486\times 10^{- 2}}$ $Ka = 1.281\times 10^{- 6}$
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