Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 15 - Acids and Bases - Questions & Problems - Page 711: 15.53

Answer

$pH = 10.96 $

Work Step by Step

- $[BOH]_{initial} = 0.24M$ Since it is a pure solution: - $[OH^-] = [B^+] = x$ - $[BOH] = [BOH]_{inital} - x$ 1. Now, use the Kb value and equation to find the 'x' value. $Kb = 3.5 \times 10^{- 6}= \frac{x * x}{ 2.4 \times 10^{- 1}- x}$ $Kb = 3.5 \times 10^{- 6}= \frac{x^2}{ 2.4 \times 10^{- 1}- x}$ $ 8.4 \times 10^{- 7} - 3.5 \times 10^{- 6}x = x^2$ $ 8.4 \times 10^{- 7} - 3.5 \times 10^{- 6}x - x^2 = 0$ $\Delta = (- 3.5 \times 10^{- 6})^2 - 4 * (-1) *( 8.4 \times 10^{- 7})$ $\Delta = 1.22 \times 10^{- 11} + 3.36 \times 10^{- 6} = 3.36 \times 10^{- 6}$ $x_1 = \frac{ - (- 3.5 \times 10^{- 6})+ \sqrt { 3.36 \times 10^{- 6}}}{2*(-1)}$ or $x_2 = \frac{ - (- 3.5 \times 10^{- 6})- \sqrt { 3.36 \times 10^{- 6}}}{2*(-1)}$ $x_1 = - 9.18 \times 10^{- 4} (Negative)$ $x_2 = 9.14 \times 10^{- 4}$ The concentration can't be negative, therefore: x = $9.14 \times 10^{- 4}M$ $[OH^-] = x = 9.14 \times 10^{- 4}M$ 2. Calculate the pOH, and then the pH. - $pOH = -log[OH^-] = -log(9.14 \times 10^{- 4}) = 3.04$ - $pH + pOH = 14$ - $pH + 3.04 = 14$ - $pH = 14 - 3.04 = 10.96$
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