Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 16 - Applications of Aqueous Equilibria - Section Problems - Page 710: 66


(c) $0.20M HF$ and $0.10M NaOH$

Work Step by Step

- Since we have a strong base and a weak acid, a neutralization reaction will occur: $NaOH(aq) + HF(aq) \lt -- \gt Na^+(aq) + F^-(aq) + H_2O(l)$ $NaOH$ is the limiting reactant (because at equal volumes, it has the lower concentration $(0.10M)$) Therefore: $[NaOH]_{final} = 0.10M (Initial) - 0.10M = 0$ $[HF]_{final} = 0.20M (Initial) - 0.10M = 0.10M$ $[F^-] = 0 (Initial) + 0.10M = 0.10M$ Notice: For reactants, the concentration decreases, because the compound is being consumed. And for the products, the compound is being produced, so, the concentration increases. Since this solution has $HF$ and $F^-$ in equal amounts, it can act as a buffer.
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