Problem 47 - General Chemistry

Question

Complete the right side of each of the following molecular equations. Then write the net ionic equations. Assume all salts formed are soluble. Acid salts are possible. a. \(2 \mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) b. \(3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{Al}(\mathrm{OH})_{3}(s) \longrightarrow\) c. \(2 \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) d. \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{NaOH}(a q) \longrightarrow\)

Step-by-Step Solution

Answer

Reaction a: 2OH⁻ + 2H⁺ → 2H₂O; Reaction b: 6H⁺ + 2Al(OH)₃ → 2Al³⁺ + 6H₂O; Reaction c: 2H⁺ + 2OH⁻ → 2H₂O; Reaction d: H₂SO₃ + OH⁻ → HSO₃⁻ + H₂O.

Step 1: Balance Reaction a

The first reaction involves potassium hydroxide reacting with phosphoric acid: \[2 \mathrm{KOH} + \mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{K}_2\mathrm{HPO}_4 + 2\mathrm{H}_2\mathrm{O}\]. Both potassium phosphate salt and water are the products. The coefficients ensure that the number of atoms for each element is equal on both sides of the equation.

Step 2: Write Net Ionic Equation for Reaction a

To write the net ionic equation, first, we dissociate all the aqueous reactants and products: \[2\mathrm{K}^+(aq) + 2\mathrm{OH}^-(aq) + 2\mathrm{H}^+(aq) + \mathrm{HPO}_4^{2-}(aq) \rightarrow 2\mathrm{H}_2\mathrm{O}(l) + 2\mathrm{K}^+(aq) + \mathrm{HPO}_4^{2-}(aq)\]By canceling out the spectator ions, the net ionic equation is: \[2 \mathrm{OH}^- + 2 \mathrm{H}^+ \rightarrow 2 \mathrm{H}_2\mathrm{O}\].

Step 3: Balance Reaction b

The second reaction involves sulfuric acid reacting with aluminum hydroxide:\[3\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{Al(OH)}_3 \rightarrow \mathrm{Al}_2\mathrm{(SO}_4\mathrm{)}_3 + 6\mathrm{H}_2\mathrm{O}\]. Aluminum sulfate and water are the products. Ensure atom balance for each element on both sides.

Step 4: Write Net Ionic Equation for Reaction b

For the net ionic equation, dissociate only the aqueous components: \[6\mathrm{H}^+(aq) + 3\mathrm{SO}_4^{2-}(aq) + 2\mathrm{Al(OH)}_3(s) \rightarrow 2\mathrm{Al}^{3+}(aq) + 3\mathrm{SO}_4^{2-}(aq) + 6\mathrm{H}_2\mathrm{O}(l)\]Cancel the spectator sulfate ions and you get the net ionic equation:\[6 \mathrm{H}^+ + 2\mathrm{Al(OH)}_3 \rightarrow 2\mathrm{Al}^{3+} + 6\mathrm{H}_2\mathrm{O}\].

Step 5: Balance Reaction c

The third reaction involves acetic acid and calcium hydroxide:\[2\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2 + \mathrm{Ca(OH)}_2 \rightarrow \mathrm{Ca(C}_2\mathrm{H}_3\mathrm{O}_2)_2 + 2\mathrm{H}_2\mathrm{O}\].This leads to calcium acetate and water as the products, ensuring the atom balance.

Step 6: Write Net Ionic Equation for Reaction c

For the net ionic equation, we can write: \[2\mathrm{H}^+(aq) + 2\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-(aq) + \mathrm{Ca}^{2+}(aq) + 2\mathrm{OH}^-(aq) \rightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-(aq) + 2\mathrm{H}_2\mathrm{O}(l)\]Canceling the ions that appear on both sides, we get:\[2\mathrm{H}^+ + 2\mathrm{OH}^- \rightarrow 2\mathrm{H}_2\mathrm{O}\].

Step 7: Balance Reaction d

In the fourth reaction, sulfurous acid reacts with sodium hydroxide: \[\mathrm{H}_2\mathrm{SO}_3 + \mathrm{NaOH} \rightarrow \mathrm{NaHSO}_3 + \mathrm{H}_2\mathrm{O}\]. This forms sodium bisulfite and water. Check to ensure each element has the same number of atoms on both sides.

Step 8: Write Net Ionic Equation for Reaction d

For the net ionic equation, we write:\[\mathrm{H}_2\mathrm{SO}_3(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{HSO}_3^-(aq) + \mathrm{H}_2\mathrm{O}(l)\]Sodium ions do not appear as they are spectator ions, and solutions form water with sulfate turning into bisulfite.

Key Concepts

Net Ionic Equations

A net ionic equation serves to simplify chemical reactions by excluding spectator ions. Spectator ions do not participate in the actual chemical change and remain unchanged on both sides of a chemical reaction. This makes net ionic equations helpful tools for focusing on the core chemical process.
To create a net ionic equation, follow these steps:

• Start by writing the balanced molecular equation and identify all substances that are soluble in water. These will ionize in aqueous solutions.
• Dissociate all strong electrolytes into their respective ions. Weak acids, bases, and solids do not dissociate.
• Cancel out identical spectator ions that appear on both sides of the equation.
• The resulting equation is your net ionic equation, showing only the particles that participate in the forming of new products.

This technique is crucial to better understand the actual chemical changes occurring in aqueous reactions and is extensively used in chemistry to predict the products of reactions.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the conservation of mass. This principle dictates that the number of atoms of each element must be the same on both sides of a chemical equation. Here is a basic guideline for balancing equations:

Begin by listing all the elements involved in the reaction on both sides of the equation.

• Adjust coefficients in front of the chemical formulas to match the number of atoms of each element on the reactant side to the product side.
• Start with the most complex molecule, adjusting coefficients for polyatomic ions that appear on both sides as a unit.
• Move next to balance the remaining elements, adjusting the individual atoms’ coefficients as necessary. Oxygen and hydrogen are usually balanced last, especially if they appear in multiple compounds.
• Double-check to ensure that all elements are balanced and that you have the smallest set of whole numbers as coefficients.

Balancing equations not only ensures mastery over stoichiometry but also provides the foundational framework for further chemical calculations and predictions.

Aqueous Reactions

Aqueous reactions are reactions that occur in water, involving two or more substances dissolved in an aqueous solution. These reactions are notable because water is a common solvent, facilitating the dissociation of ionic compounds into their respective ions.
When dealing with aqueous reactions, it is important to note the following:

• Most salts, acids, and bases dissociate in water, forming ions that can react with one another to form new compounds, often resulting in precipitates, gases, or water itself.
• The nature of the solvent, water in this case, supports the mobility and interaction of ions, making it excellent for facilitating chemical reactions.
• Recognizing whether a compound in the reaction is soluble or forms a precipitate is key when predicting products and writing net ionic equations.

Understanding aqueous reactions is essential in fields ranging from analytical chemistry to pharmaceuticals, as it explains the behaviors of substances in a dissolved state.