Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 4 - Reactions in Aqueous Solution - Exercises: 4.40a

Answer

Balanced equation: $2HCH_3COO(aq) + Ba(OH)_2(aq) -> 2H_2O(l) + {Ba(CH_3COO)_2}(aq)$ Net ionic equation: $2HCH_3COO(aq) + 2OH^-(aq) -> 2H_2O(l) + 2{CH_3COO}^-(aq)$

Work Step by Step

1. Write the equation: $HCH_3COO(aq) + Ba(OH)_2(aq) -> H_2O(l) + {Ba(CH_3COO)_2}(aq)$ 2. Balance the equation: $2HCH_3COO(aq) + Ba(OH)_2(aq) -> 2H_2O(l) + {Ba(CH_3COO)_2}(aq)$ 3. Write the net ionic equation: $2HCH_3COO(aq) + Ba^{2+}(aq) + 2OH^-(aq) -> 2H_2O(l) + Ba^{2+}(aq) + 2{CH_3COO}^-(aq)$ 4. Remove the spectator ions: $2HCH_3COO(aq) + 2OH^-(aq) -> 2H_2O(l) + 2{CH_3COO}^-(aq)$
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