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: 15.43

Answer

pH = 2.599

Work Step by Step

1. Draw the ICE table: (Image on the end) 2. Now, use the Ka value and equation to find "x". $Ka = 6.5 \times 10^{-5} = \frac{(x)*(x)}{0.10-x} $ $6.5 \times 10^{-5} = \frac{(x^2)}{0.10-x} $ $6.5 \times 10^{-6} - 6.5 \times 10^{-5}x = x^2$ $6.5 \times 10^{-6} - 6.5 \times 10^{-5}x - x^2 = 0$ Bhaskara: $\Delta = (-6.5 \times 10^{-5})^2 - 4 * (-1) * (6.5 \times 10^{-6})$ $\Delta = 4.225 \times 10^{-9} + 2.6 \times 10^{-5}$ $\Delta \approx 2.6 \times 10^{-5}$ $x_1 = \frac{-(-6.5 \times 10^{-5}) + \sqrt {2.6 \times 10^{-5}}}{2*(-1)}$ or $x_2 = \frac{-(-6.5 \times 10^{-5}) - \sqrt {2.6 \times 10^{-5}}}{2*(-1)}$ $x_1 = -2.582 \times 10^{-3} (Negative)$ $x_2 = 2.517 \times 10^{-3} $ The concentration can't be negative. $[H^+] = 0 + x = x$ Therefore, the concentration of $[H^+]$ is $2.517 \times 10^{-3}$ 3. Calculate the pH: $pH = -log[H^+] = -log(2.517 \times 10^{-3}) = 2.599$
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