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: 44

Answer

$Ka = 1.418\times 10^{- 3}$

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. -$[H_3O^+] = [ClCH_2{CO_2}^-] = x$ -$[ClCH_2CO_2H] = [ClCH_2CO_2H]_{initial} - x$ 2. Calculate the $[H_3O^+]$ value: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 1.95}$ $[H_3O^+] = 1.122 \times 10^{- 2}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][ClCH_2{CO_2}^-]}{ [ClCH_2CO_2H]}$ $Ka = \frac{x^2}{[InitialClCH_2CO_2H] - x}$ $Ka = \frac{( 0.01122)^2}{ 0.1- 0.01122}$ $Ka = \frac{ 1.259\times 10^{- 4}}{ 0.08878}$ $Ka = 1.418\times 10^{- 3}$
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