Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Topical Questions - Page 693a: 27b


$1L $ of $0.20M$ $NaOH$ and 0.30 moles of acetic acid are capable of making a buffer, because the solution has a weak conjugate acid-base pair in adequate amounts.

Work Step by Step

- There is a weak acid and a strong base, so we have to consider their reaction: $CH_3COOH(aq)+NaOH(aq)\\ -->CH_3COO^−(aq)+Na^+(aq)+H_2O(l)$ - The limiting reactant is $NaOH$ (0.20 mol), so these are the final concentrations: $CH_3COOH$: 0.30 − 0.20=0.10 mol $NaOH $: 0.20 − 0.20=0 mol $CH3COO^−$: 0 + 0.20= 0.20mol - Since this solution has a conjugate acid-base pair ($CH_3COOH$ and $CH_3COO^-$), in adequate proportions, it is considered a buffer.
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