Chemistry (7th Edition)

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

Chapter 15 - Aqueous Equilibria: Acids and Bases - Worked Example - Page 623: 17

Answer

$Ka = 3.469\times 10^{- 8}$ $pKa = 7.46$ - Yes, the values on the table agree with the calculations.

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. -$[H_3O^+] = [OCl^-] = x$ -$[HOCl] = [HOCl]_{initial} - x$ 2. Calculate the hydronium concentration $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.23}$ $[H_3O^+] = 5.888 \times 10^{- 5}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][OCl^-]}{ [HOCl]}$ $Ka = \frac{x^2}{[InitialHOCl] - x}$ $Ka = \frac{( 5.888\times 10^{- 5})^2}{ 0.1- 5.888\times 10^{- 5}}$ $Ka = \frac{ 3.467\times 10^{- 9}}{ 0.09994}$ $Ka = 3.469\times 10^{- 8}$ 4. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 3.469 \times 10^{- 8})$ $pKa = 7.46$ ------
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