Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575d: 75

Answer

$Ka = 1.392\times 10^{- 4}$

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 dissociation equation, and find 'x': %dissociation = $\frac{x}{[Initial Acid]} \times 100$ 3= $\frac{x}{ 0.15} \times 100$ 0.03= $\frac{x}{ 0.15}$ $ 4.5\times 10^{- 3}= x$ Therefore: $[Conj. Base] and [H_3O^+] = 4.5\times 10^{- 3}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{[InitialAcid] - x}$ $Ka = \frac{( 4.5\times 10^{- 3})^2}{ 0.15- 4.5\times 10^{- 3}}$ $Ka = \frac{ 2.025\times 10^{- 5}}{ 0.1455}$ $Ka = 1.392\times 10^{- 4}$
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