Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - 17-2 Controlling pH: Buffer Solutions - Review & Check for Section 17-2 - Page 643: 2

Answer

The necessary volume of acetate is equal to 18.2 mL; therefore, the correct answer is (d).

Work Step by Step

1. Calculate the pKa value for acetic acid: $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.74$ 2. Use the Henderson-Hasselbalch equation to find the concentration of the acetate ion: ** Since we are analyzing a single solution, the concentration ratio is equal to the amount (moles) ratio, because they are in the same volume, so: ** 100 mL = 0.100 L Amount of acetic acid: $0.10M \times 0.100L = 0.010 $ moles. So: $pH = pK_a + log\frac{[Base]}{[Acid]}$ $4.00 = 4.74 + log\frac{(Acetate)}{0.010}$ $-0.74 = log(Acetate) - log(0.010)$ $-0.74 = log(Acetate) + 2.00$ $log(Acetate) = -2.74$ $(Acetate) = 10^{-2.74} = 1.82 \times 10^{-3}$ moles. 3. Conver that number to volume: $1.82 \times 10^{-3} moles \times \frac{1 L}{0.10 mol} = 1.82 \times 10^{-2} L = 18.2 mL $
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