Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575c: 66

Answer

Percent dissociation: $ = 20.79\%$

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. -$[H_3O^+] = [Cl{O_2}^-] = x$ -$[HClO_2] = [HClO_2]_{initial} - x = 0.22 - x$ For approximation, we consider: $[HClO_2] = 0.22M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][Cl{O_2}^-]}{ [HClO_2]}$ $Ka = 1.2 \times 10^{- 2}= \frac{x * x}{ 0.22}$ $Ka = 1.2 \times 10^{- 2}= \frac{x^2}{ 0.22}$ $ 2.64 \times 10^{- 3} = x^2$ $x = 5.138 \times 10^{- 2}$ Percent dissociation: $\frac{ 5.138 \times 10^{- 2}}{ 0.22} \times 100\% = 23.35\%$ %ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 0.012= \frac{x^2}{ 0.22- x}$ $ 2.64 \times 10^{- 3} - 0.012x = x^2$ $ 2.64 \times 10^{- 3} - 1.2 \times 10^{- 2}x - x^2 = 0$ $\Delta = (- 1.2 \times 10^{- 2})^2 - 4 * (-1) *( 2.64 \times 10^{- 3})$ $\Delta = 1.44 \times 10^{- 4} + 1.056 \times 10^{- 2} = 1.07 \times 10^{- 2}$ $x_1 = \frac{ - (- 1.2 \times 10^{- 2})+ \sqrt { 1.07 \times 10^{- 2}}}{2*(-1)}$ or $x_2 = \frac{ - (- 1.2 \times 10^{- 2})- \sqrt { 1.07 \times 10^{- 2}}}{2*(-1)}$ $x_1 = - 5.773 \times 10^{- 2} (Negative)$ $x_2 = 4.573 \times 10^{- 2}$ Percent dissociation: $\frac{ 4.573 \times 10^{- 2}}{ 0.22} \times 100\% = 20.79\%$ - The concentration can't be negative, so $[H_3O^+]$ = $x_2$
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