Question

Write the complete ionic equation and the net ionic equation for the reaction that occurs when the following solutions are mixed, and name the salt formed in each case: (a) \(\mathrm{HBr}(a q)+\mathrm{NaOH}(a q)\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HBr}(a q)\) (d) \(\mathrm{Mg}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q)\)

Short answer

(a) The net ionic equation for HBr(aq) + NaOH(aq) is \(H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)\) and the salt formed is NaBr. (b) The net ionic equation for Ca(OH)2(aq) + HNO3(aq) is \(2OH^-(aq) + 2H^+(aq) \rightarrow 2H_2O (l)\) and the salt formed is Ca(NO3)2. (c) The net ionic equation for Ca(OH)2(aq) + HBr(aq) is \(2OH^-(aq) + 2H^+(aq) \rightarrow 2H_2O (l)\) and the salt formed is CaBr2. (d) The net ionic equation for Mg(OH)2(aq) + HCl(aq) is \(2OH^-(aq) + 2H^+(aq) \rightarrow 2H_2O (l)\) and the salt formed is MgCl2.

Step-by-step solution

In this reaction, the acids will neutralize the bases, forming water (H2O) and a salt. The salt will be formed by the exchange of ions – the positive Na+ ion will pair with the negative Br- ion, resulting in NaBr. The reaction will be written as: HBr(aq) + NaOH(aq) -> NaBr(aq) + H2O(l) #Step 2: Writing the complete ionic equation#

We can now write out the complete ionic equation by separating the aqueous compounds into their constituent ions: \(H^+ (aq) + Br^- (aq) + Na^+ (aq) + OH^- (aq) \rightarrow Na^+(aq) + Br^-(aq) + H_2O (l)\) #Step 3: Writing the net ionic equation#

By crossing out the spectator ions (those that don't change during the reaction), Na+ and Br-, we can write the net ionic equation: \(H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)\) #Case (b): Ca(OH)2(aq) + HNO3(aq)# #Step 1: Writing the products of the reaction#

The reaction between Ca(OH)2 and HNO3 will again produce water and a salt. In this case, the salt will be formed by the positive Ca^2+ ion pairing with the negative NO3- ion, resulting in Ca(NO3)2. The reaction will be written as: Ca(OH)2(aq) + 2HNO3(aq) -> Ca(NO3)2(aq) + 2H2O(l) #Step 2: Writing the complete ionic equation#

The complete ionic equation, separating the ions, will be: \(Ca^{2+}(aq) + 2OH^-(aq) + 2H^+(aq) + 2NO_3^-(aq) \rightarrow Ca^{2+}(aq) + 2NO_3^-(aq) + 2H_2O (l)\) #Step 3: Writing the net ionic equation#

The net ionic equation, after crossing out the spectator ions, Ca^2+ and NO3-, will be: \(2OH^-(aq) + 2H^+(aq) \rightarrow 2H_2O (l)\) #Case (c): Ca(OH)2(aq) + HBr(aq)# #Step 1: Writing the products of the reaction#

This reaction will also produce water and a salt. Here, the salt will be formed by the positive Ca^2+ ion pairing with the negative Br- ion, resulting in CaBr2. The reaction will be written as: Ca(OH)2(aq) + 2HBr(aq) -> CaBr2(aq) + 2H2O(l) #Step 2: Writing the complete ionic equation#

The complete ionic equation, separating the ions, will be: \(Ca^{2+}(aq) + 2OH^-(aq) + 2H^+(aq) + 2Br^-(aq) \rightarrow Ca^{2+}(aq) + 2Br^-(aq) + 2H_2O (l)\) #Step 3: Writing the net ionic equation#

The net ionic equation, after crossing out the spectator ions, Ca^2+ and Br-, will be: \(2OH^-(aq) + 2H^+(aq) \rightarrow 2H_2O (l)\) #Case (d): Mg(OH)2(aq) + HCl(aq)# #Step 1: Writing the products of the reaction#

The reaction between Mg(OH)2 and HCl will produce water and a salt. In this case, the salt will be formed by the positive Mg^2+ ion pairing with the negative Cl- ion, resulting in MgCl2. The reaction will be written as: Mg(OH)2(aq) + 2HCl(aq) -> MgCl2(aq) + 2H2O(l) #Step 2: Writing the complete ionic equation#

The complete ionic equation, separating the ions, will be: \(Mg^{2+}(aq) + 2OH^-(aq) + 2H^+(aq) + 2Cl^-(aq) \rightarrow Mg^{2+}(aq) + 2Cl^-(aq) + 2H_2O (l)\) #Step 3: Writing the net ionic equation#

The net ionic equation, after crossing out the spectator ions, Mg^2+ and Cl-, will be: \(2OH^-(aq) + 2H^+(aq) \rightarrow 2H_2O (l)\)

Key concepts

Complete Ionic Equation

Understanding the complete ionic equation is crucial for anyone studying chemistry, particularly when evaluating reactions in an aqueous solution. This type of equation breaks down soluble ionic compounds into individual ions that exist in the solution. To masterfully solve a complete ionic equation, first write the balanced molecular equation, then split the aqueous compounds into their corresponding ions. For instance, when sodium hydroxide reacts with hydrobromic acid to form water and sodium bromide, the complete ionic equation is:
\(H^+ (aq) + Br^- (aq) + Na^+ (aq) + OH^- (aq) \rightarrow Na^+(aq) + Br^-(aq) + H_2O (l)\).
What's key here is that you only separate the compounds that dissolve into ions in water. This excludes compounds like water and solids, which remain intact in these equations.

Spectator Ions

The term 'spectator ions' might evoke sporting events where fans watch the game without playing in it. Similarly, spectator ions are present in the reaction mixture but don't actually participate in the reaction itself. They begin and end in the same form and simply 'watch' the reaction happen around them. To identify these ions, write out the complete ionic equation and pinpoint which ions remain unchanged on both sides of the equation, such as Na+ and Br- in our previous example. These ions can then be omitted when writing the net ionic equation. Eliminating spectator ions simplifies the equation and focuses on the actual chemical change occurring, which is immensely beneficial for understanding the reaction's dynamics.

Acid-Base Neutralization Reactions

The dance between acids and bases can end in a beautiful display of chemical harmony known as an acid-base neutralization reaction. This delightful tango typically results in the formation of water and a salt, and it's a perfect example of a double displacement reaction. Taking hydrochloric acid mixing with magnesium hydroxide as an illustration, both reagents neutralize each other. The equation for the reaction is:
\(Mg(OH)_2(aq) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)\).
Identifying the products of these reactions is a key step that involves understanding the nature of acids and bases, as well as being able to predict the salt that will be produced.

Solubility in Reactions

Solubility plays a starring role in chemical reactions, particularly when dealing with ionic compounds in aqueous solutions. The solubility rules determine whether a compound will dissociate into ions and participate actively in the reaction or form a precipitate and sit out the action. If a compound is soluble, it will break apart into ions in solution, whereas insoluble compounds will remain intact. Being adept at applying the solubility rules helps in predicting the outcomes of reactions and writing both complete and net ionic equations. For instance, while sodium bromide dissolves completely and is present as Na+ and Br- ions in solution, lead iodide might form a precipitate, leaving the solution as solid PbI2. This understanding directly informs which ions become part of the complete ionic equation versus those who emerge as solid precipitates.