Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 16 - Questions and Problems - Page 772: 16.57

Answer

(a) 10.3% (b) 41.58% (c) 0.986%

Work Step by Step

Ka (formic acid) = $1.7 \times 10^{-4}$ 1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is at the end of this answer. -$[H_3O^+] = [HCOO^-] = x$ -$[HCOOH] = [HCOOH]_{initial} - x$ For approximation, we consider: $[HCOOH] = [HCOOH]_{initial}$ (a) 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$ $Ka = 1.7 \times 10^{- 4}= \frac{x * x}{ 1.6 \times 10^{- 2}}$ $Ka = 1.7 \times 10^{- 4}= \frac{x^2}{ 1.6 \times 10^{- 2}}$ $ 2.72 \times 10^{- 6} = x^2$ $x = 1.64 \times 10^{- 3}$ %Ionization: $\frac{ 1.64 \times 10^{- 3}}{ 1.6 \times 10^{- 2}} \times 100\% = 10.3\%$ (b) 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$ $Ka = 1.7 \times 10^{- 4}= \frac{x * x}{ 5.7 \times 10^{- 4}}$ $Ka = 1.7 \times 10^{- 4}= \frac{x^2}{ 5.7 \times 10^{- 4}}$ $ 9.69 \times 10^{- 8} = x^2$ $x = 3.11 \times 10^{- 4}$ 5% test: $\frac{ 3.11 \times 10^{- 4}}{ 5.7 \times 10^{- 4}} \times 100\% = 54.6\%$ High percent ionization : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 1.7 \times 10^{- 4}= \frac{x^2}{ 5.7 \times 10^{- 4}- x}$ $ 9.69 \times 10^{- 8} - 1.7 \times 10^{- 4}x = x^2$ $ 9.69 \times 10^{- 8} - 1.7 \times 10^{- 4}x - x^2 = 0$ Bhaskara: $\Delta = (- 1.7 \times 10^{- 4})^2 - 4 * (-1) *( 9.69 \times 10^{- 8})$ $\Delta = 2.89 \times 10^{- 8} + 3.87 \times 10^{- 7} = 4.16 \times 10^{- 7}$ $x_1 = \frac{ - (- 1.7 \times 10^{- 4})+ \sqrt { 4.16 \times 10^{- 7}}}{2*(-1)}$ or $x_2 = \frac{ - (- 1.7 \times 10^{- 4})- \sqrt { 4.16 \times 10^{- 7}}}{2*(-1)}$ $x_1 = - 4.07 \times 10^{- 4} (Negative)$ $x_2 = 2.37 \times 10^{- 4}$ Percent Ionization: $\frac{2.37 \times 10^{-4}}{5.7 \times 10^{-4}} \times 100\% = 41.58\%$ (c) 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$ $Ka = 1.7 \times 10^{- 4}= \frac{x * x}{ 1.75 \times 10^{- 0}}$ $Ka = 1.7 \times 10^{- 4}= \frac{x^2}{ 1.75 \times 10^{- 0}}$ $ 2.97 \times 10^{- 4} = x^2$ $x = 1.72 \times 10^{- 2}$ Percent Ionization : $\frac{ 1.72 \times 10^{- 2}}{ 1.75} \times 100\% = 0.986\%$
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