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 - 16-7 Calculations with Equilibrium Constants - Review & Check for Section 16-7 - Page 615: 2

Answer

A

Work Step by Step

$pH=-log[H_3O^+]$ $[H_3O^+]=10^{-3.02}$ $[H_3O^+]=9.55\times10^{-4}\ M=[A^-], [HA]=0.039\ M$ $K_a=[H_3O^+][A^-]/[HA]$ $K_a=(9.55\times10^{-4})^2/0.039$ $K_a=2.34\times10^{-5}$
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