Question

Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate): (a) \(\operatorname{HBr}(a q)+\mathrm{NH}_{3}(a q) \longrightarrow\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow\)

Short answer

The balanced chemical equations are \n\n(a) \(HBr(aq) + NH_{3}(aq) \longrightarrow NH_{4}Br(aq)\) \n(b) \(Ba(OH)_{2}(aq) + 2H_{3}PO_{4}(aq) \longrightarrow Ba(H_{2}PO_{4})_{2}(aq) + 2H_{2}O(l)\) \n(c) \(2HClO_{4}(aq) + Mg(OH)_{2}(s) \longrightarrow Mg(ClO_{4})_{2}(aq) + 2H_{2}O(l)\) \n\nThe final net ionic equations are \n\n(a) \(H^{+}(aq) + NH_{3}(aq) \longrightarrow NH_{4}^{+}(aq)\) \n(b) \(2OH^{-}(aq) + 2H_{3}PO_{4}(aq) \longrightarrow 2H_{2}PO_{4}^{-}(aq) + 2H_{2}O(l)\) \n(c) \(2H^{+}(aq) + Mg^{2+}(s) + 2OH^{-}(s) \longrightarrow Mg^{2+}(aq) + 2H_{2}O(l)\)

Step-by-step solution

Balancing the chemical equations

Firstly, the given equations need to be balanced. Balanced equations ensure that the number of atoms for each element are equal on both the reactant and product side of the equation. \n\n(a) \(HBr(aq) + NH_{3}(aq) \longrightarrow NH_{4}Br(aq)\) \n(b) \(Ba(OH)_{2}(aq) + 2H_{3}PO_{4}(aq) \longrightarrow Ba(H_{2}PO_{4})_{2}(aq) + 2H_{2}O(l)\) \n(c) \(2HClO_{4}(aq) + Mg(OH)_{2}(s) \longrightarrow Mg(ClO_{4})_{2}(aq) + 2H_{2}O(l)\)

Writing the ionic equations

The next step is to transform the balanced equations into ionic equations. Every component in the equations is split into its constituent ions. \n\n(a) \(H^{+}(aq) + Br^{-}(aq) + NH_{3}(aq) \longrightarrow NH_{4}^{+}(aq) + Br^{-}(aq)\) \n(b) \(Ba^{2+}(aq) + 2OH^{-}(aq) + 2H_{3}PO_{4}(aq) \longrightarrow Ba^{2+}(aq) + 2H_{2}PO_{4}^{-}(aq) + 2H_{2}O(l)\) \n(c) \(2H^{+}(aq) + 2ClO_{4}^{-}(aq) + Mg^{2+}(s) + 2OH^{-}(s) \longrightarrow Mg^{2+}(aq) + 2ClO_{4}^{-}(aq) + 2H_{2}O(l)\)

Writing the net ionic equations

The final step is to write the net ionic equations by removing the spectator ions, which are ions present on both sides of the ionic equation. \n\n(a) \(H^{+}(aq) + NH_{3}(aq) \longrightarrow NH_{4}^{+}(aq)\) \n(b) \(2OH^{-}(aq) + 2H_{3}PO_{4}(aq) \longrightarrow 2H_{2}PO_{4}^{-}(aq) + 2H_{2}O(l)\) \n(c) \(2H^{+}(aq) + Mg^{2+}(s) + 2OH^{-}(s) \longrightarrow Mg^{2+}(aq) + 2H_{2}O(l)\)

Key concepts

Ionic Equations

Ionic equations play a pivotal role in understanding chemical reactions in solution. These equations provide a clearer picture of the species that actively participate in the reaction by representing compounds as their constituent ions. For example, when hydrobromic acid (\textbf{HBr}) reacts with ammonia (\textbf{NH}\(_3\)), the ionic equation shows the dissociation of \textbf{HBr} into hydrogen (\textbf{H}\(^+\)) and bromide (\textbf{Br}\(^-\)) ions, and ammonia remains as a molecule because it is a weak electrolyte. So, the ionic equation for this reaction would be \textbf{H}\(^+\)(aq) + \textbf{Br}\(^-\)(aq) + \textbf{NH}\(_{3}\)(aq) \rightarrow \textbf{NH}\(_{4}^+\)(aq) + \textbf{Br}\(^-\)(aq).

Understanding these equations is crucial for students to grasp how reactions occur at the ionic level. It helps in visualizing how compounds break apart into ions in an aqueous solution and how these ions recombine to form new substances. To write an ionic equation, start with a balanced molecular equation and then separate the strong electrolytes into their ions. Remember that covalent compounds and weak electrolytes like \textbf{NH}\(_3\) are not dissociated in their ionic form.

Net Ionic Equations

Net ionic equations further refine our understanding of chemical reactions by removing spectator ions, which are ions that do not undergo a chemical change. These equations are especially useful for simplifying complex reactions and focusing on the essence of the chemical change. Taking the earlier example, once the ionic equation is written, you can determine that \textbf{Br}\(^-\) is a spectator ion because it appears on both sides of the equation and does not participate in the reaction. After removing spectator ions, the net ionic equation for the reaction between \textbf{HBr} and \textbf{NH}\(_3\) simplifies to \textbf{H}\(^+\)(aq) + \textbf{NH}\(_{3}\)(aq) \rightarrow \textbf{NH}\(_{4}^+\)(aq).

For students to write net ionic equations effectively, it is essential to identify the strong electrolytes which fully dissociate in solution, and weak electrolytes or insoluble compounds that remain intact. After canceling out the spectator ions, what remains is the net ionic equation, bringing forward the actual chemical change taking place. This allows for a much clearer and specific delineation of the reaction's outcome.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is fundamental to the study of chemistry, quantitatively describing the relationship between the amounts of reactants and products in a chemical reaction. It is key to accurately predict the quantities of substances consumed and produced in these reactions. Stoichiometry is rooted in the law of conservation of mass which tells us that atoms are neither created nor destroyed in a chemical reaction. Consequently, you must have the same number of each type of atom on both sides of a balanced equation.

Let's analyze the reaction between \textbf{Mg}(\textbf{OH})\(_2\) and \textbf{HClO}\(_4\). To balance the reaction, it's necessary to observe that two moles of \textbf{HClO}\(_4\) react with one mole of \textbf{Mg}(\textbf{OH})\(_2\) to produce one mole of \textbf{Mg}(\textbf{ClO}\(_4\))\(_2\) and two moles of water (\textbf{H}\(_2\)\textbf{O}). Understanding stoichiometry allows students not only to balance chemical equations but also to perform calculations involving reactant and product quantities, limiting reagents, and theoretical yields, fundamental tasks for any chemist.