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: 18.65

Answer

$K_a = 4.8 \times 10^{-9} $

Work Step by Step

1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HA ]& [ A^- ]& [ H_3O^+ ]\\ Initial& 0.035 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.035 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 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. Using the pH, find the $H_3O^+$ concentration: $$[H_3O^+] = 10^{-pH} = 10^{-4.88} = 1.3 \times 10^{-5} \space M$$ $[H_3O^+] = x = 1.3 \times 10^{-5} $ 4. Substitute the value of x and calculate the $K_a$: $$K_a = \frac{( 1.3 \times 10^{-5} )^2}{ 0.035 - 1.3 \times 10^{-5} }$$ $K_a = 4.8 \times 10^{-9} $
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