Problem 58 - Chemistry The Science in Context

Question

Complete and balance each of the following neutralization reactions, name the products, and write the overall ionic and net ionic equations. a. \(\mathrm{HBr}(a q)+\mathrm{KOH}(a q) \rightarrow\) b. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) c. \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \rightarrow\) d. \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{Sr}(\mathrm{OH})_{2}(a q) \rightarrow\)

Step-by-Step Solution

Answer

a. HBr(aq) + KOH(aq) b. H3PO4(aq) + Ba(OH)2(aq) c. Al(OH)3(s) + HCl(aq) d. CH3COOH(aq) + Sr(OH)2(aq) Answer: a. H^(+)(aq) + OH^(-)(aq) -> H2O(l) b. 6H^(+)(aq) + 6OH^(-)(aq) -> 6H2O(l) c. 3H^(+)(aq) + 3OH^(-)(aq) -> 3H2O(l) d. 2H^(+)(aq) + 2OH^(-)(aq) -> 2H2O(l)

Step 1: Balancing the chemical equation

The balanced equation will be: \(\mathrm{HBr}(aq) + \mathrm{KOH}(aq) \rightarrow \mathrm{KBr}(aq) + \mathrm{H_2O}(l)\)

Step 2: Naming the products

The products are: Potassium bromide (KBr) and Water (H2O).

Step 3: Writing the ionic and net ionic equations

Ionic equation: \(\mathrm{H}^+(aq) + \mathrm{Br}^-(aq) + \mathrm{K}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{K}^+(aq) + \mathrm{Br}^-(aq) + \mathrm{H_2O}(l)\) Net ionic equation: \(\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{H_2O}(l)\) b. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\)

Step 4: Balancing the chemical equation

The balanced equation will be: \(2 \mathrm{H}_{3} \mathrm{PO}_{4}(aq) + 3 \mathrm{Ba}(\mathrm{OH})_{2}(aq) \rightarrow \mathrm{Ba}_3(\mathrm{PO_4})_2(s) + 6 \mathrm{H_2O}(l)\)

Step 5: Naming the products

The products are: Barium phosphate (Ba3(PO4)2) and Water (H2O).

Step 6: Writing the ionic and net ionic equations

Ionic equation: \(6 \mathrm{H}^+(aq) + 6 \mathrm{OH}^-(aq) + 2 \mathrm{PO}_4^{3-}(aq) + 3 \mathrm{Ba}^{2+}(aq) \rightarrow \mathrm{Ba}_3(\mathrm{PO_4})_2(s) + 6 \mathrm{H_2O}(l)\) Net ionic equation: \(6 \mathrm{H}^+(aq) + 6 \mathrm{OH}^-(aq) \rightarrow 6\mathrm{H_2O}(l)\) c. \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \rightarrow\)

Step 7: Balancing the chemical equation

The balanced equation will be: \(\mathrm{Al}(\mathrm{OH})_{3}(s) + 3\mathrm{HCl}(aq) \rightarrow \mathrm{AlCl_3}(aq) + 3\mathrm{H_2O}(l)\)

Step 8: Naming the products

The products are: Aluminum chloride (AlCl3) and Water (H2O).

Step 9: Writing the ionic and net ionic equations

Ionic equation: \(\mathrm{Al}^{3+}(aq) + 3 \mathrm{OH}^-(aq) + 3 \mathrm{H}^+(aq) + 3\mathrm{Cl}^-(aq) \rightarrow \mathrm{Al}^{3+}(aq) + 3\mathrm{Cl}^-(aq) + 3 \mathrm{H_2O}(l)\) Net ionic equation: \(3 \mathrm{H}^+(aq) + 3 \mathrm{OH}^-(aq) \rightarrow 3 \mathrm{H_2O}(l)\) d. \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{Sr}(\mathrm{OH})_{2}(a q) \rightarrow\)

Step 10: Balancing the chemical equation

The balanced equation will be: \(2 \mathrm{CH}_{3} \mathrm{COOH}(aq) + \mathrm{Sr}(\mathrm{OH})_{2}(aq) \rightarrow \mathrm{Sr}(\mathrm{CH}_3 \mathrm{COO})_{2}(aq) + 2 \mathrm{H_2O}(l)\)

Step 11: Naming the products

The products are: Strontium acetate (Sr(CH3COO)2) and Water (H2O).

Step 12: Writing the ionic and net ionic equations

Ionic equation: \(2 \mathrm{CH}_{3} \mathrm{COO}^-(aq) + 2 \mathrm{H}^+(aq) + \mathrm{Sr}^{2+}(aq) + 2 \mathrm{OH}^-(aq) \rightarrow \mathrm{Sr}^{2+}(aq) + 2 \mathrm{CH}_{3} \mathrm{COO}^-(aq) + 2 \mathrm{H_2O}(l)\) Net ionic equation: \(2 \mathrm{H}^+(aq) + 2 \mathrm{OH}^-(aq) \rightarrow 2 \mathrm{H_2O}(l)\)

Key Concepts

Balancing Chemical Equations

To understand neutralization reactions, one must start with the basics of balancing chemical equations. It's a fundamental practice in chemistry that ensures the law of conservation of mass is upheld—that is, the same number of atoms of each element is present on both sides of the equation.

When balancing equations, first write down the unbalanced equation. Then count the number of atoms of each element on both the reactant and product sides. Adjust coefficients—never subscripts, as this changes the substance—to balance the atoms. For example, in the reaction between hydrobromic acid (HBr) and potassium hydroxide (KOH), balancing leads to one molecule of HBr reacting with one molecule of KOH to form one molecule of potassium bromide (KBr) and one molecule of water (H2O).

Ionic Equations

In neutralization reactions involving strong acids and bases, it's useful to represent what's actually happening in solution. This is where ionic equations come in, which show the ions in an aqueous solution.

Consider the reaction between HBr and KOH. In an ionic equation, each strong electrolyte (substances that dissociate completely into ions in solution) is written as its constituent ions. So HBr and KOH dissociate into H⁺, Br^-, K⁺, and OH^-. The ionic equation is a realistic representation of the species that are in the solution before and after the reaction.

Net Ionic Equations

The net ionic equation simplifies the ionic equation by removing spectator ions—those that do not participate in the actual reaction. They appear unchanged on both sides of the ionic equation.

For the same reaction between HBr and KOH, K⁺ and Br^- are spectator ions. Removing them leaves us with the essence of the reaction, where H⁺ ions react with OH^- ions to form water, H₂O. Thus, the net ionic equation is H⁺(aq) + OH^-(aq) → H₂O(l), providing a clear picture of the chemical change.

Acid-Base Reactions

Neutralization reactions are a type of acid-base reaction in which an acid reacts with a base to produce a salt and usually water. In essence, these reactions are about the transfer of protons (H⁺) from the acid to the base.

This proton transfer neutralizes the acid and base, often resulting in a pH that is closer to neutral (7), depending on the strength of the reactants. In our examples, we have strong acids and strong bases, so the products will tend to be neutral. However, if a weak acid reacts with a strong base or vice versa, the resulting solution will have a pH that is not neutral.