General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

Published by Pearson
ISBN 10: 0321967461
ISBN 13: 978-0-32196-746-6

Chapter 11 - Acids and Bases - 11.7 Reactions of Acids and Bases - Questions and Problems - Page 425: 11.60


a. $$H_3PO_4(aq) + 3NaOH(aq) \longrightarrow H_2O(l) + Na_3PO_4(aq)$$ b. $$HI(aq) + LiOH(aq) \longrightarrow H_2O(l) + LiI$$ c. $$2HNO_3(aq) + Ca(OH)_2 \longrightarrow H_2O(l) + Ca(NO_3)_2$$

Work Step by Step

1. All neutralization reactions follow this pattern: $$Acid + Base \longrightarrow H_2O(l) + salt$$ 2. Balance the $H^+$ in the acid with the $OH^-$ in the base. a. $H_3PO_4$ provides 3 $H^+$, so we have to put a coefficient of 3 in front of the $NaOH$, since it only provides 1 $OH^-$. $$H_3PO_4(aq) + 3NaOH(aq) \longrightarrow H_2O(l) + salt$$ b. $HI$ provides 1 $H^+$, and $LiOH$ provides 1 $OH^-$. The ions are already balanced. $$HI(aq) + LiOH(aq) \longrightarrow H_2O(l) + salt$$ c. $Ca(OH)_2$ provides 2 $OH^-$, so we have to put a coefficient of 2 in front of the $HNO_3$, since it only provides 1 $H^+$ $$2HNO_3(aq) + Ca(OH)_2 \longrightarrow H_2O(l) + salt$$ 3. Balance the $H_2O$ with the $H^+$ and $OH^-$. a. The reaction uses 3 $H^+$ and 3 $OH^-$. Thus, there should be 3 $H_2O$ in the products. $$H_3PO_4(aq) + 3NaOH(aq) \longrightarrow H_2O(l) + salt$$ b. The reaction uses 1 $H^+$ and 1 $OH^-$. Thus, there should be 1 $H_2O$ in the products. $$HI(aq) + LiOH(aq) \longrightarrow H_2O(l) + salt$$ c. The reaction uses 2 $H^+$ and 2 $OH^-$. Thus, there should be 2 $H_2O$ in the products. $$2HNO_3(aq) + Ca(OH)_2 \longrightarrow H_2O(l) + salt$$ 4. Use the remaining ions to produce a salt. a. Remaining ions: $1P{O_4}^{3-}$ and $3Na^+$: $$H_3PO_4(aq) + 3NaOH(aq) \longrightarrow H_2O(l) + Na_3PO_4(aq)$$ b. Remaining ions: $1I^-$ and $1Li^{+}$: $$HI(aq) + LiOH(aq) \longrightarrow H_2O(l) + LiI$$ c. Remaining ions: $2N{O_3}^{-}$ and $Ca^{2+}$: $$2HNO_3(aq) + Ca(OH)_2 \longrightarrow H_2O(l) + Ca(NO_3)_2$$
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