Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 15 - Aqueous Equilibria: Acids and Bases - Section Problems - Page 650: 85

Answer

$Ka = 1.434\times 10^{- 4}$ $pKa = 3.843$

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. -$[H_3O^+] = [C_3H_5O_3^-] = x$ -$[C_3H_6O_3] = [C_3H_6O_3]_{initial} - x$ 2. Calculate the $[H_3O^+]$ $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.43}$ $[H_3O^+] = 3.715 \times 10^{- 3}$ Therefore : $x = 3.715\times 10^{- 3}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][C_3H_5O_3^-]}{ [C_3H_6O_3]}$ $Ka = \frac{x^2}{[InitialC_3H_6O_3] - x}$ $Ka = \frac{( 3.715\times 10^{- 3})^2}{ 0.1- 3.715\times 10^{- 3}}$ $Ka = \frac{ 1.38\times 10^{- 5}}{ 0.09628}$ $Ka = 1.434\times 10^{- 4}$ 4. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.434 \times 10^{- 4})$ $pKa = 3.843$
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