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: 16.60

Answer

This concentration is $0.072M$

Work Step by Step

1. Write the percent ionization equation, and relate the initial acid concentration to 'x': %ionization = $\frac{x}{[Initial Acid]} \times 100$ 2.5= $\frac{x}{[Initial Acid]} \times 100$ 0.025= $\frac{x}{ [Initial Acid]}$ $ [Initial Acid]= \frac{x}{0.025} = \frac{1}{0.025}x = 40x $ 2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x$ For approximation, we consider: $[Acid] = [Acid]_{initial}$ 3. Use the Ka equation to find "x": $Ka = \frac{[H_3O^+][[Conj. Base]]}{ [[Acid]]}$ $Ka = 4.5 \times 10^{- 5}= \frac{x * x}{[Acid]}$ $Ka = 4.5 \times 10^{- 5}= \frac{x^2}{ 40x}$ $4.5 \times 10^{-5} = \frac{x}{40}$ $4.5 \times 10^{-5} * 40 = x$ $x = 1.8 \times 10^{-3}$ 4. Calculate the initial acid concentration: $[Initial Acid] = 40 * x = 40 * 1.8 \times 10^{-3} = 0.072M$
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