Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 16 - Sections 16.1-16.8 - Exercises - Problems by Topic - Page 804: 33c

Answer

$pH = 3.46$

Work Step by Step

1. Draw the ICE table (It is in the end of this answer) 2. Calculate 'x' using the $K_a$ expression. $ 3.5\times 10^{- 4} = \frac{[F^-][H_3O^+]}{[HF]}$ $ 3.5\times 10^{- 4} = \frac{( 0.15 + x )* x}{ 0.15 - x}$ Considering 'x' has a very small value. $ 3.5\times 10^{- 4} = \frac{ 0.15 * x}{ 0.15}$ $ 3.5\times 10^{- 4} = 1x$ $x = 3.5\times 10^{- 4}$ Percent dissociation: $\frac{ 3.5\times 10^{- 4}}{ 0.15} \times 100\% = 0.2333\%$ x = $[H_3O^+]$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 3.5 \times 10^{- 4})$ $pH = 3.46$
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