Question

Write a balanced net ionic equation for each of the following reactions in water. (a) acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2},\) with strontium hydroxide (b) diethylamine, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{NH}\) with sulfuric acid (c) aqueous hydrofluoric acid with sodium hydroxide

Short answer

Question: Write the net ionic equations for the following reactions in water: (a) acetic acid and strontium hydroxide, (b) diethylamine and sulfuric acid, (c) aqueous hydrofluoric acid and sodium hydroxide. Answer: (a) $$2 \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} (\mathrm{aq}) + \mathrm{Sr(OH)}_{2} (\mathrm{aq}) \rightarrow \mathrm{Sr(C}_{2}\mathrm{H}_{3}\mathrm{O}_{2})_{2} (\mathrm{aq}) + 2 \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$ (b) $$(\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{SO}_{4} (\mathrm{aq}) \rightarrow (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}_{2} \mathrm{SO}_{4} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$ (c) $$\mathrm{F}^{-} (\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \rightarrow \mathrm{F}^{-} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$

Step-by-step solution

(a) Full balanced molecular equation for the reaction between acetic acid and strontium hydroxide

To find the full balanced molecular equation, we first need to recognize the products of the reaction between acetic acid and strontium hydroxide. Acetic acid is a weak acid, and strontium hydroxide is a strong base; hence, the products will be a salt (strontium acetate) and water. The balanced molecular equation is: $$2 \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} (\mathrm{aq}) + \mathrm{Sr(OH)}_{2} (\mathrm{aq}) \rightarrow \mathrm{Sr(C}_{2}\mathrm{H}_{3}\mathrm{O}_{2})_{2} (\mathrm{aq}) + 2 \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$

(a) Finding the net ionic equation

Next, we need to identify any spectator ions. In this case, there are none, as all the ions are involved in the reaction. Therefore, the net ionic equation is the same as the full balanced molecular equation: $$2 \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} (\mathrm{aq}) + \mathrm{Sr(OH)}_{2} (\mathrm{aq}) \rightarrow \mathrm{Sr(C}_{2}\mathrm{H}_{3}\mathrm{O}_{2})_{2} (\mathrm{aq}) + 2 \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$

(b) Full balanced molecular equation for the reaction between diethylamine and sulfuric acid

The reaction between diethylamine (a weak base) and sulfuric acid (a strong acid) will produce a salt (diethylammonium sulfate) and water. The balanced molecular equation for this reaction is: $$(\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{SO}_{4} (\mathrm{aq}) \rightarrow (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}_{2} \mathrm{SO}_{4} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$

(b) Finding the net ionic equation

Once again, there are no spectator ions in this reaction. Therefore, the net ionic equation is the same as the full balanced molecular equation: $$(\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{SO}_{4} (\mathrm{aq}) \rightarrow (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}_{2} \mathrm{SO}_{4} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$

(c) Full balanced molecular equation for the reaction between aqueous hydrofluoric acid and sodium hydroxide

The reaction between hydrofluoric acid (a weak acid) and sodium hydroxide (a strong base) will produce a salt (sodium fluoride) and water. The balanced molecular equation for this reaction is: $$\mathrm{HF} (\mathrm{aq}) + \mathrm{NaOH} (\mathrm{aq}) \rightarrow \mathrm{NaF} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$

(c) Finding the net ionic equation

In this case, we have one spectator ion: Na+. We can remove it from both sides of the reaction, and the net ionic equation becomes: $$\mathrm{F}^{-} (\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \rightarrow \mathrm{F}^{-} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O} (\mathrm{l})$$