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: 15.44

Answer

$[H^+] = 5.71 \times 10^{-4} M $ $[CH_3COO^-] = 5.71 \times 10^{-4}M$ $[CH_3COOH] = 0.018129M$

Work Step by Step

1. Calculate the initial acid concentration. - Find the number of moles. Acetic acid: Molar Mass = 60g/mol. $mm = \frac{mass}{n(moles)}$ $60 = \frac{0.0560}{n(moles)}$ $n(moles) = \frac{0.056}{60} = 9.333 \times 10^{-4}$ - Find the concentration: $Concentration(mol/L) = \frac{n(moles)}{V(L)} = \frac{9.333 \times 10^{-4}}{0.050} = 0.0187 M$ 2. Draw the ICE table: (Image on the end) 3. Now, use the Ka value and equation to find "x". $Ka = 1.8 \times 10^{-5} = \frac{(x)*(x)}{0.0187-x} $ $Ka = 1.8 \times 10^{-5} = \frac{(x^2)}{0.0187-x} $ $3.366 \times 10^{-7} - 1.8 \times 10^{-5}x = x^2$ $3.366 \times 10^{-7} - 1.8 \times 10^{-5}x - x^2 = 0$ Bhaskara: $\Delta = (-1.8 \times 10^{-5})^2 - 4 * (-1) * (3.366 \times 10^{-7})$ $\Delta = 3.24 \times 10^{-10} + 1.346 \times 10^{-6}$ $\Delta \approx 1.346 \times 10^{-6}$ $x_1 = \frac{-(-1.8 \times 10^{-5}) + \sqrt {1.346 \times 10^{-6}}}{2*(-1)}$ or $x_2 = \frac{-(-1.8 \times 10^{-5}) - \sqrt {1.346 \times 10^{-6}}}{2*(-1)}$ $x_1 = -5.89 \times 10^{-4} (Negative)$ $x_2 = 5.71 \times 10^{-4} $ The concentration can't be negative. $[H^+] = [A^-] = 0 + x = x = 5.71 \times 10^{-4}$ $[HA] = 0.0187 - x = 0.0187 - 5.71 \times 10^{-4} = 0.018129$ --------
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