Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 16 - Questions and Problems - Page 773: 16.73

Answer

$pH = 11.981 $

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. -$[OH^-] = [HB^+] = x$ -$[B] = [B]_{initial} - x = 0.61 - x$ For approximation, we consider: $[B] = 0.61M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][HB^+]}{ [B]}$ $Ka = 1.5 \times 10^{- 4}= \frac{x * x}{ 6.1 \times 10^{- 1}}$ $Ka = 1.5 \times 10^{- 4}= \frac{x^2}{ 6.1 \times 10^{- 1}}$ $ 9.15 \times 10^{- 5} = x^2$ $x = 9.566 \times 10^{- 3}$ Percent ionization: $\frac{ 9.566 \times 10^{- 3}}{ 6.1 \times 10^{- 1}} \times 100\% = 1.568\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = x = 9.566 \times 10^{- 3} $ $pOH = -log[OH^-]$ $pOH = -log( 9.566 \times 10^{- 3})$ $pOH = 2.019$ $pH + pOH = 14$ $pH + 2.019 = 14$ $pH = 11.981$
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