Chemistry: The Molecular Science (5th Edition)

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

Chapter 3 - Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 148c: 42c


Acid and base used to form this ion, respectively, : $HI$ and $NaOH$, Complete ionic equation: $H^+(aq) + I^-(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + I^-(aq) + H_2O(l) $ Net ionic equation: $H^+(aq) + OH^-(aq) -- \gt H_2O(l)$

Work Step by Step

1. Identify the ions of the salt: $(NaI)$: $Na^+$ and $I^-$: To the cation, add a hydroxide ion: $NaOH$; this is the base. To the anion, add a hydrogen ion: $HI$; this is the acid. 2. Now, write the balanced overall equation between them, which is: $Acid + Base -- \gt Salt + Water$ We already have the salt so: $HI(aq) + NaOH(aq) -- \gt NaI(aq) + H_2O(l)$ 3. Write the complete ionic equation. - For the completely dissociated/ionized compounds, separate them by their ions: $H^+(aq) + I^-(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + I^-(aq) + H_2O(l) $ * NaOH is a strong base, and $NaI$ is soluble according to the table 3.1, because it is a compound with the $Na^+$ ion. * HI is a strong acid, so it is completely dissociated in water. 4. Remove the repeated ions: $H^+(aq) + OH^-(aq) -- \gt H_2O(l)$ This is the net ionic equation.
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