WEB Write a balanced net ionic equation for the reaction of each of the following aqueous solutions with \(\mathrm{OH}^{-}\) ions. (a) ammonium nitrate (b) sodium dihydrogen phosphate \(\left(\mathrm{NaH}_{2} \mathrm{PO}_{4}\right)\) (c) \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\)

We are given the aqueous solution ammonium nitrate. The formula for ammonium nitrate is \(\mathrm{NH}_{4} \mathrm{NO}_{3}\). The hydroxide ions are denoted by \(\mathrm{OH}^{-}\). The reaction can be written as: $$\mathrm{NH}_{4} \mathrm{NO}_{3} + \mathrm{OH}^{-} \rightarrow$$ Now, we need to predict the products. Ammonium will react with hydroxide ions to produce water and ammonia. The products will be ammonia (\(\mathrm{NH}_{3}\)), nitrate ion (\(\mathrm{NO}_{3}^{-}\)), and water (\(\mathrm{H}_{2}\mathrm{O}\)). Write the balanced chemical equation: $$\mathrm{NH}_{4} \mathrm{NO}_{3} + \mathrm{OH}^{-} \rightarrow \mathrm{NH}_{3} + \mathrm{NO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O}$$ Finally, we write the balanced net ionic equation by removing the spectator ions: $$\mathrm{NH}_{4}^{+} + \mathrm{OH}^{-} \rightarrow \mathrm{NH}_{3} + \mathrm{H}_{2}\mathrm{O}$$

The given aqueous solution is sodium dihydrogen phosphate, which has the formula \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\). The hydroxide ions are denoted by \(\mathrm{OH}^{-}\). Write the reaction as: $$\mathrm{NaH}_{2} \mathrm{PO}_{4} + \mathrm{OH}^{-} \rightarrow$$ Now, we need to predict the products. When the sodium dihydrogen phosphate reacts with hydroxide ions, it forms water and sodium monohydrogen phosphate. Write the balanced chemical equation: $$\mathrm{NaH}_{2} \mathrm{PO}_{4} + \mathrm{OH}^{-} \rightarrow \mathrm{NaHPO}_{4} + \mathrm{H}_{2}\mathrm{O}$$ Our net ionic equation does not include sodium ions as it's a spectator ion, and we get: $$\mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O}$$

We have the complex ion \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) reacting with hydroxide ions, \(\mathrm{OH}^{-}\). $$\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+} + \mathrm{OH}^{-} \rightarrow$$ When the complex ion reacts with hydroxide ions, it will form an insoluble aluminum hydroxide and release water. Write the balanced chemical equation: $$\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+} + 3\mathrm{OH}^{-} \rightarrow \mathrm{Al}\left(\mathrm{OH}\right)_{3} + 6\mathrm{H}_{2}\mathrm{O}$$ Since there are no spectator ions in this reaction, the net ionic equation is the same as the balanced chemical equation: $$\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+} + 3\mathrm{OH}^{-} \rightarrow \mathrm{Al}\left(\mathrm{OH}\right)_{3} + 6\mathrm{H}_{2}\mathrm{O}$$