Textbook Question
Write balanced complete ionic and net ionic equations for each reaction.
c. NaOH(aq) + HC2H3O2(aq) → H2O(l ) + NaC2H3O2(aq)
d. Na3PO4(aq) + NiCl2(aq) → Ni3(PO4)2(s) + NaCl(aq)
Ch.5 - Introduction to Solutions and Aqueous Solutions
Chapter 5, Problem 48a
Write balanced complete ionic and net ionic equations for each reaction. a. K2SO4(aq) + CaI2(aq) → CaSO4(s) + KI(aq)
Verified step by step guidance1
Step 1: Write the balanced molecular equation for the reaction. The given reaction is already balanced: \[ \text{K}_2\text{SO}_4(\text{aq}) + \text{CaI}_2(\text{aq}) \rightarrow \text{CaSO}_4(\text{s}) + 2\text{KI}(\text{aq}) \]
Step 2: Write the complete ionic equation. Break all the aqueous compounds into their respective ions: \[ 2\text{K}^+(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) + \text{Ca}^{2+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow \text{CaSO}_4(\text{s}) + 2\text{K}^+(\text{aq}) + 2\text{I}^-(\text{aq}) \]
Step 3: Identify the spectator ions. Spectator ions are ions that appear on both sides of the complete ionic equation and do not participate in the reaction. In this case, the spectator ions are \( \text{K}^+ \) and \( \text{I}^- \).
Step 4: Write the net ionic equation by removing the spectator ions from the complete ionic equation. The net ionic equation is: \[ \text{SO}_4^{2-}(\text{aq}) + \text{Ca}^{2+}(\text{aq}) \rightarrow \text{CaSO}_4(\text{s}) \]
Step 5: Review the net ionic equation to ensure it is balanced in terms of both mass and charge. Confirm that the number of atoms and the total charge are the same on both sides of the equation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Complete Ionic Equation
A complete ionic equation represents all the ions present in a solution during a chemical reaction. It shows the reactants and products in their ionic forms, highlighting the dissociation of soluble ionic compounds. This equation is essential for understanding how ions interact in solution and sets the stage for identifying the net ionic equation.
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Complete Ionic Equations
Net Ionic Equation
The net ionic equation simplifies the complete ionic equation by removing spectator ions—ions that do not participate in the actual chemical reaction. It focuses solely on the ions and molecules that undergo a change, providing a clearer picture of the chemical process. This equation is crucial for understanding the essence of the reaction and the formation of precipitates or other products.
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Precipitation Reaction
A precipitation reaction occurs when two aqueous solutions react to form an insoluble solid, known as a precipitate. In the context of the given reaction, calcium sulfate (CaSO4) precipitates out of the solution. Recognizing the conditions under which a precipitate forms is vital for writing accurate ionic equations and understanding the overall reaction dynamics.
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Related Practice
Textbook Question
Write balanced complete ionic and net ionic equations for each reaction. d. Na3PO4(aq) + NiCl2(aq) → Ni3(PO4)2(s) + NaCl(aq)
Textbook Question
Write balanced complete ionic and net ionic equations for each reaction.
a. K2SO4(aq) + CaI2(aq) → CaSO4(s) + KI(aq)
b. NH4Cl(aq) + NaOH(aq) → H2O(l) + NH3(g) + NaCl(aq)
c. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Textbook Question
Write balanced complete ionic and net ionic equations for each reaction. d. HC2H3O2(aq) + K2CO3(aq) → H2O(l ) + CO2(g) + KC2H3O2(aq)
Textbook Question
Mercury(I) ions (Hg22+) can be removed from solution by precipitation with Cl- Suppose that a solution contains aqueous Hg2(NO3)2. Write complete ionic and net ionic equations for the reaction of aqueous Hg2(NO3)2 with aqueous sodium chloride to form solid Hg2Cl2 and aqueous sodium nitrate.
Textbook Question
Lead(II) ions can be removed from solution by precipitation with sulfate ions. Suppose that a solution contains lead(II) nitrate. Write complete ionic and net ionic equations for the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II) sulfate and aqueous potassium nitrate.