Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The pH and pOH Scales - Page 744: 40

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