General Chemistry (4th Edition)

Published by University Science Books
ISBN 10: 1891389602
ISBN 13: 978-1-89138-960-3

Chapter 20 The Properties of Acids and Bases - Problems - Page 770: 19

Answer

$Ka = 2.6\times 10^{- 8}$

Work Step by Step

1. Calculate the hydronium concentration. $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.79}$ $[H_3O^+] = 1.622 \times 10^{- 5}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is at the end of this answer. -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{[InitialAcid] - x}$ $Ka = \frac{( 1.622\times 10^{- 5})^2}{ 0.01- 1.622\times 10^{- 5}}$ $Ka = \frac{ 2.63\times 10^{- 10}}{ 9.984\times 10^{- 3}}$ $Ka = 2.6\times 10^{- 8}$ ----
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