Question 5.3: Writing Equations for Acid–Base Reactions. Write a net ionic......

Analyze

The reactions are neutralization reactions, which means they are of the general form acid + base → salt. Water may also be a product. We can start with the whole formula equation, switch to the ionic equation, and then delete the spectator ions to arrive at the net ionic equation.

Solve

(a)  2 HNO_{3}(aq) + Sr(OH)_{2}(aq) → Sr(NO_{3})_{2}(aq) + 2   H_{2}O(l)

Ionic form:

2 H^{+}(aq) + 2  NO_{3}^{−}(aq) + Sr^{2+}(aq) + 2  OH^{−}(aq) → Sr^{2+}(aq) + 2  NO_{3}^{−}(aq) + 2  H_{2}O(l)

Net ionic equation: Delete the spectator ions (Sr^{2+}  and   NO_{3}^{−}).

2  H^{+}(aq) + 2  OH^{−}(aq) → 2  H_{2}O(l)

or, more simply,

H^{+}(aq) + OH^{−}(aq) → H_{2}O(l)

(b)  Al(OH)_{3}(s) + 3  HCl(aq) → AlCl_{3}(aq) + 3  H_{2}O(l)

Ionic form:

Al(OH)_{3}(s) + 3  H^{+}(aq) + 3  Cl^{−}(aq) → Al^{3+}(aq) + 3  Cl^{−}(aq) + 3  H_{2}O(l)

Net ionic equation: Delete the spectator ion (Cl^{−}).

Al(OH)_{3}(s) + 3  H^{+}(aq) → Al^{3+}(aq) + 3  H_{2}O(l)

Assess

In part (a), the net ionic equation is H^{+}(aq) + OH^{−}(aq) → H_{2}O(l), as is always the case when the neutralization reaction involves a soluble strong acid and a soluble strong base. In part (b), the base was not soluble; thus, the net ionic equation includes a solid.