Chemistry: Atoms First (2nd Edition)

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

Chapter 13 - Exercises - Page 575d: 81

Answer

$K_a (HA) = 5.56 \times 10^{-3}$

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. -$[H_3O^+] = [A^-] = x$ -$[HA] = [HA]_{initial} - x = 0.45M$ 2. Calculate the initial acid concentration: $c(mol/L) = \frac{n(moles)}{v(L)} = \frac{1 mol}{2L} = 0.5 mol/L$ 3. Now, we can find "x": $[HA] = [HA]_{initial} - x$ $0.45M = 0.5M - x$ $0.45 - 0.5 = - x$ $-0.05 = -x$ $x = 0.05M$ So, $[H_3O^+] = [A^-] = x = 0.05M$ 3. Write the Ka expression, and find its value: $K_a = \frac{[H_3O^+][A^-]}{[HA]}$ $K_a = \frac{x * x}{0.45} = \frac{x^2}{0.45}$ $K_a = \frac{(0.05)^2}{0.45} = 5.56 \times 10^{-3}$
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