Problem 48 - 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. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) b. \(2 \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) c. \(\mathrm{NaOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) d. \(\mathrm{Sr}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \longrightarrow\)

Step-by-Step Solution

Answer

Complete the equations and write net ionic ones for each reaction.

Step 1: Balance the Molecular Equations

a. \( \text{Ca(OH)}_2(aq) + 2 \text{H}_2\text{SO}_4(aq) \rightarrow \text{CaSO}_4(aq) + 2 \text{H}_2\text{O}(l) \)b. \( 2 \text{H}_3\text{PO}_4(aq) + \text{Ca(OH)}_2(aq) \rightarrow \text{Ca}(\text{H}_2\text{PO}_4)_2(aq) + 2 \text{H}_2\text{O}(l) \)c. \( \text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{NaHSO}_4(aq) + \text{H}_2\text{O}(l) \)c (alternative). \( 2 \text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l) \) d. \( \text{Sr(OH)}_2(aq) + 2 \text{H}_2\text{CO}_3(aq) \rightarrow \text{SrCO}_3(s) + 3\text{H}_2\text{O}(l) + \text{CO}_2(g) \).

Step 2: Write Balanced Ionic Equations

a. \( \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq) + 2\text{H}^+(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{CaSO}_4(s) + 2 \text{H}_2\text{O}(l) \)b. \( 2\text{H}^+(aq) + 2\text{PO}_4^{3-}(aq) + \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Ca}^{2+}(aq) + 2\text{H}_2\text{PO}_4^-(aq) + 2\text{H}_2\text{O}(l) \)c. \( \text{Na}^+(aq) + \text{OH}^-(aq) + \text{H}^+(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{Na}^+(aq) + \text{HSO}_4^-(aq) + \text{H}_2\text{O}(l) \)c(alternative). \( 2 \text{Na}^+(aq) + 2\text{OH}^-(aq) + 2\text{H}^+(aq) + \text{SO}_4^{2-}(aq) \rightarrow 2\text{Na}^+(aq) + \text{SO}_4^{2-}(aq) + 2 \text{H}_2\text{O}(l) \) d. \( \text{Sr}^{2+}(aq) + 2 \text{OH}^-(aq) + 4 \text{H}^+(aq) + 2 \text{CO}_3^{2-}(aq) \rightarrow \text{SrCO}_3(s) + 3 \text{H}_2\text{O}(l) + \text{CO}_2(g) \).

Step 3: Identify Spectator Ions

a. In this reaction, the \( \text{Ca}^{2+}(aq)\) and \(\text{SO}_4^{2-}(aq)\) combine to form \(\text{CaSO}_4(s)\). No spectator ions.b. \(\text{Ca}^{2+}(aq)\) on both sides is a spectator ion.c. \(\text{Na}^{+}(aq)\) remains unchanged in part c.c(alternative). \(2\text{Na}^{+}(aq)\) and \(\text{SO}_4^{2-}(aq)\) are spectators. d. No spectator ions present in the reaction.

Step 4: Write the Net Ionic Equations

a. \( 2\text{OH}^-(aq) + 2 \text{H}^+(aq) \rightarrow 2 \text{H}_2\text{O}(l) \)b. \( 2\text{OH}^-(aq) + 2 \text{H}^+(aq) + 2\text{PO}_4^{3-}(aq) \rightarrow 2 \text{H}_2\text{PO}_4^-(aq) \)c. \( \text{OH}^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l) \)c(alternative). \( 2\text{OH}^-(aq) + 2 \text{H}^+(aq) \rightarrow 2 \text{H}_2\text{O}(l) \) d. \( \text{Sr}^{2+}(aq) + 2\text{CO}_3^{2-}(aq) + 4 \text{H}^+(aq) \rightarrow \text{SrCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g) \).

Key Concepts

Balanced Chemical Equations

In chemistry, a balanced chemical equation is crucial because it shows the reactants and products in a chemical reaction with the correct number of atoms of each element. This balancing reflects the law of conservation of mass, which asserts that atoms are neither created nor destroyed in a chemical reaction. To create a balanced chemical equation, you must ensure that the number of atoms of each element on the reactants side equals the number of atoms of the same element on the products side.
For example, in the reaction of calcium hydroxide with sulfuric acid, \[ \text{Ca(OH)}_2 + 2 \text{H}_2\text{SO}_4 \longrightarrow \text{CaSO}_4 + 2 \text{H}_2\text{O} \] the equation is balanced because the number of calcium, oxygen, hydrogen, and sulfur atoms is the same on both sides.

• Count each type of atom on both sides of the equation.
• Adjust coefficients to balance the number of atoms for each element.
• Recheck all atoms to confirm the balance.

Balancing equations helps us understand the proportions of substances involved and predict the quantities of reactants needed or products formed.

Spectator Ions

Spectator ions are the ions that do not take part in the overall chemical reaction. They exist in the same form on both the reactant and product sides of a reaction equation. Identifying spectator ions is important because they are removed to simplify and rewrite the net ionic equation, which shows only the ions and molecules directly involved in the chemical reaction.
For instance, in the reaction: \[ \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O} \] the sodium ion \(\text{Na}^+\) acts as a spectator ion because it appears unchanged on both sides of the equation.

• Inspect the dissociated ions in the reaction.
• Identify ions unchanged before and after the reaction.
• Exclude these ions when writing the net ionic equation.

Removing spectator ions simplifies the chemical equation and highlights the true chemical change occurring.

Solubility Rules

Solubility rules help predict the solubility of various compounds in water. These rules are guidelines used in chemistry to determine whether a compound is likely to dissolve and form an aqueous solution or form a precipitate. Understandably, these rules are essential when writing ionic and net ionic equations as they reveal which compounds dissociate into ions in aqueous solutions.
Key solubility rules include:

• Most alkali metal salts (e.g., Na⁺, K⁺) and ammonium ( NH₄⁺) salts are soluble.
• Nitrates ( NO₃⁻) and acetates ( CH₃COO⁻) are generally soluble.
• Chlorides ( Cl⁻), bromides ( Br⁻), and iodides ( I⁻) are soluble except for salts with Ag⁺, Pb²⁺, and Hg₂²⁺.
• Most sulfates ( SO₄²⁻) are soluble, with exceptions like BaSO₄, PbSO₄, and CaSO₄.
• Carbonates ( CO₃²⁻), phosphates ( PO₄³⁻), and hydroxides ( OH⁻) are generally insoluble, with exceptions for certain alkali metals and Ca(OH)₂ and Sr(OH)₂.

Using these rules, you can foresee if a compound will dissociate in solution or precipitate out, which significantly assists in predicting the products of a chemical reaction.

Double Displacement Reactions

Double displacement reactions, also known as double replacement or metathesis reactions, occur when parts of two ionic compounds are exchanged and form two new compounds. These reactions generally occur in aqueous solutions where the ions readily dissociate, allowing the ions to interact and form new products.The general form of a double displacement reaction is \[ AB + CD \longrightarrow AD + CB \] where \(AB\) and \(CD\) are reactant compounds, and \(AD\) and \(CB\) are the newly formed products.
A common example is during the reaction of sodium hydroxide and sulfuric acid: \[ \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O} \] Here, \(Na^+\) swaps anions with \(\text{H}^+\) from sulfuric acid, creating new compounds.

• Identify the ions in the reactants.
• Swap the ions to form new combinations.
• Check if new compounds formed follow solubility rules or precipitate.

These reactions are significant because they frequently involve the formation of a precipitate, gas, or weak electrolyte, indicating a chemical change.