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: 15.48

Answer

(a) 3.31% (b) 31.3% (c) 75.4% - When the initial acid concentration decreases, the percent ionization increases.

Work Step by Step

$Ka (HF) = 6.6 \times 10^{-4}$ - Drawing the ICE table, we get: -$[H_3O^+] = [A^-] = x$ -$[HA] = [HA]_{initial} - x$ For approximation, we consider: $[HA] = [HA]_{initial}$ (a) Use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][A^-]}{ [HA]}$ $Ka = 6.6 \times 10^{- 4}= \frac{x * x}{ 6 \times 10^{- 1}}$ $Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 6 \times 10^{- 1}}$ $ 3.95 \times 10^{- 4} = x^2$ $x = 1.98 \times 10^{- 2}$ %Ionization: $\frac{ 1.98 \times 10^{- 2}}{ 6 \times 10^{- 1}} \times 100\% = 3.31\%$ -------- (b) Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][A^-]}{ [HA]}$ $Ka = 6.6 \times 10^{- 4}= \frac{x * x}{ 4.6 \times 10^{- 3}}$ $Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 4.6 \times 10^{- 3}}$ $ 3.03 \times 10^{- 6} = x^2$ $x = 1.74 \times 10^{- 3}$ Ionization: $\frac{ 1.74 \times 10^{- 3}}{ 4.59 \times 10^{- 3}} \times 100\% = 37.8\%$ Ionization > 5%, so we to consider the "x" of the acid concentration: $Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 4.6 \times 10^{- 3}- x}$ $ 3.03 \times 10^{- 6} - 6.6 \times 10^{- 4}x = x^2$ $ 3.03 \times 10^{- 6} - 6.6 \times 10^{- 4}x - x^2 = 0$ Bhaskara: $\Delta = (- 6.6 \times 10^{- 4})^2 - 4 * (-1) *( 3.03 \times 10^{- 6})$ $\Delta = 4.35 \times 10^{- 7} + 1.21 \times 10^{- 5} = 1.25 \times 10^{- 5}$ $x_1 = \frac{ - (- 6.6 \times 10^{- 4})+ \sqrt { 1.25 \times 10^{- 5}}}{2*(-1)}$ or $x_2 = \frac{ - (- 6.6 \times 10^{- 4})- \sqrt { 1.25 \times 10^{- 5}}}{2*(-1)}$ $x_1 = - 2.1 \times 10^{- 3} (Negative)$ $x_2 = 1.44 \times 10^{- 3}$ Ionization: $\frac{ 1.44 \times 10^{- 3}}{ 4.6 \times 10^{-3}} \times 100\% = 31.3\%$ ----- (c) Use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][A^-]}{ [HA]}$ $Ka = 6.6 \times 10^{- 4}= \frac{x * x}{ 2.8 \times 10^{- 4}}$ $Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 2.8 \times 10^{- 4}}$ $ 1.84 \times 10^{- 7} = x^2$ $x = 4.29 \times 10^{- 4}$ 5% test: $\frac{ 4.29 \times 10^{- 4}}{ 2.8 \times 10^{- 4}} \times 100\% = 153\%$ Ionization > 5%, so we to consider the "x" of the acid concentration: $Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 2.8 \times 10^{- 4}- x}$ $ 1.84 \times 10^{- 7} - 6.6 \times 10^{- 4}x = x^2$ $ 1.84 \times 10^{- 7} - 6.6 \times 10^{- 4}x - x^2 = 0$ Bhaskara: $\Delta = (- 6.6 \times 10^{- 4})^2 - 4 * (-1) *( 1.84 \times 10^{- 7})$ $\Delta = 4.35 \times 10^{- 7} + 7.39 \times 10^{- 7} = 1.17 \times 10^{- 6}$ $x_1 = \frac{ - (- 6.6 \times 10^{- 4})+ \sqrt { 1.17 \times 10^{- 6}}}{2*(-1)}$ or $x_2 = \frac{ - (- 6.6 \times 10^{- 4})- \sqrt { 1.17 \times 10^{- 6}}}{2*(-1)}$ $x_1 = - 8.71 \times 10^{- 4} (Negative)$ $x_2 = 2.11 \times 10^{- 4}$ Ionization: $\frac{ 2.11 \times 10^{- 4}}{ 2.8 \times 10^{- 4}} \times 100\% = 75.4\%$
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