Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 18 - Problems - Page 820: 18.70

Answer

(a) $$[H_3O^+] = 6.0 \times 10^{-3}$$ $$[OH^-] = 1.7 \times 10^{-12} \space M$$ $$pH = 2.22 $$ $$pOH = 11.78$$ (b) $K_a = 1.9 \times 10^{-4} $

Work Step by Step

(a) 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HA ]& [ A^- ]& [ H_3O^+ ]\\ Initial& 0.20 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.20 -x&x& x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each pressure is equal to its balance coefficient. $$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ A^- ][ H^+ ]}{[ HA ]}$$ $$K_a = \frac{(x)(x)}{[ HA ]_{initial} - x}$$ 3. Use the percent ionization to find x: $$Percent \space ionization = \frac{x}{[ HA ]_{initial}} \times 100\% $$ $$x = \frac{ 3.0 \% \times 0.20 }{100\%}$$ $x = 6.0 \times 10^{-3} $ $$[H_3O^+] = x = 6.0 \times 10^{-3}$$ $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 6.0 \times 10^{-3} } = 1.7 \times 10^{-12} \space M$$ $$pH = -log[H_3O^+] = -log( 6.0 \times 10^{-3} ) = 2.22 $$ $$pOH = 14 - 2.22 = 11.78$$ (b) 4. Substitute the value of x and calculate the $K_a$: $$K_a = \frac{( 6.0 \times 10^{-3} )^2}{ 0.20 - 6.0 \times 10^{-3} }$$ $K_a = 1.9 \times 10^{-4} $
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