Problem 108
Question
Describe the synthesis of each of the following ionic compounds, starting from solutions of sodium and nitrate salts. Then write the net ionic equation for each synthesis. (a) \(\mathrm{Zn}_{3}\left(\mathrm{PO}_{4}\right)_{2};\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2};\) (c) \(\mathrm{NiCO}_{3}.\)
Step-by-Step Solution
Verified Answer
The balanced equations for the syntheses are \(3Zn(NO3)2 + 2Na3PO4 → Zn3(PO4)2 + 6NaNO3\), \(Cu(NO3)2 + 2NaOH → Cu(OH)2 + 2NaNO3\), and \(Ni(NO3)2 + Na2CO3 → NiCO3 + 2NaNO3\). The net ionic equations are \(3Zn^{2+} + 2PO4^{3-} → Zn3(PO4)2\), \(Cu^{2+} + 2OH- → Cu(OH)2\), and \(Ni^{2+} + CO3^{2-} → NiCO3\).
1Step 1: Synthesize Zn3(PO4)2
To form zinc phosphate (\(\mathrm{Zn}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)), you need a nitrate salt of zinc (\(Zn(NO3)2\)) and a sodium phosphate salt (\(Na3PO4\)). Mixing these two, you would use the technique of double displacement, leaving sodium nitrate (\(NaNO3\)), in the solution forming a precipitate of zinc phosphate. The balanced chemical equation for this reaction is: \[3Zn(NO3)2 + 2Na3PO4 → Zn3(PO4)2 + 6NaNO3\]
2Step 2: Write the net ionic equation for Zn3(PO4)2
The net ionic equation cancels out the spectator ions. Sodium and nitrate ions are the spectator ions, so the net ionic equation becomes: \[3Zn^{2+} + 2PO4^{3-} → Zn3(PO4)2\]
3Step 3: Synthesize Cu(OH)2
To form copper(II) hydroxide (\(Cu(OH)2\)), you need a nitrate salt of copper (\(Cu(NO3)2\)) and sodium hydroxide (\(NaOH\)). Mixing these two, the sodium and nitrate ions are left in the solution. The balanced chemical equation for this reaction is: \[Cu(NO3)2 + 2NaOH → Cu(OH)2 + 2NaNO3\]
4Step 4: Write the net ionic equation for Cu(OH)2
Removing the sodium and nitrate ions (spectator ions) from the equation, the net ionic equation is: \[Cu^{2+} + 2OH- → Cu(OH)2\]
5Step 5: Synthesize NiCO3
To form nickel(II) carbonate, you need a nitrate salt of nickel (\(Ni(NO3)2\)) and sodium carbonate (\(Na2CO3\)). Mixing these two, sodium and nitrate ions will be left in the solution. The balanced chemical equation for this reaction is: \[Ni(NO3)2 + Na2CO3 → NiCO3 + 2NaNO3\]
6Step 6: Write the net ionic equation for NiCO3
Eliminating the sodium and nitrate ions (spectator ions) from the equation, the net ionic equation is: \[Ni^{2+} + CO3^{2-} → NiCO3\]
Key Concepts
Net Ionic EquationsDouble Displacement ReactionsSpectator Ions
Net Ionic Equations
Net ionic equations focus on the essence of a chemical reaction by eliminating the ions that do not participate in the formation of the final product. These equations are helpful because they simplify complex equations to show the core chemical change happening in the reaction.
In a chemical reaction, when we dissolve ionic compounds in water, they dissociate into their respective ions. So, when reactions occur in aqueous solutions, it's often useful to distinguish between what changes and what remains unaltered.
In a chemical reaction, when we dissolve ionic compounds in water, they dissociate into their respective ions. So, when reactions occur in aqueous solutions, it's often useful to distinguish between what changes and what remains unaltered.
- To write a net ionic equation, start by writing the complete balanced equation for the reaction.
- Identify the ionic compounds that exist as ions in the solution.
- Remove the spectator ions—ions present on both sides of the equation that do not participate directly in the reaction.
- What remains is the net ionic equation, showing only the species that undergo a change during the reaction.
Double Displacement Reactions
Double displacement reactions, also known as metathesis reactions, involve an exchange of ions between two reacting compounds. This swap usually results in the formation of a precipitate, a gas, or a weak electrolyte. Such reactions are common in aqueous solutions where ions freely move to interact with each other.
During this type of reaction, the general format can be represented as:
\[AB + CD → AD + CB\]Here, A and B are ions from one compound, while C and D come from another. The reaction occurs when A pairs with D and B pairs with C. For example, if we mix zinc nitrate and sodium phosphate, the zinc ions (Zn2+) and phosphate ions (PO43-) come together to form zinc phosphate, an insoluble salt that precipitates out of the solution.These reactions are essential for constructing net ionic equations, as they often result in the creation of a solid precipitate or other non-ion species that signify the actual chemical interaction taking place.
During this type of reaction, the general format can be represented as:
\[AB + CD → AD + CB\]Here, A and B are ions from one compound, while C and D come from another. The reaction occurs when A pairs with D and B pairs with C. For example, if we mix zinc nitrate and sodium phosphate, the zinc ions (Zn2+) and phosphate ions (PO43-) come together to form zinc phosphate, an insoluble salt that precipitates out of the solution.These reactions are essential for constructing net ionic equations, as they often result in the creation of a solid precipitate or other non-ion species that signify the actual chemical interaction taking place.
Spectator Ions
Spectator ions are ions that are present in the reaction mixture but do not participate in the actual chemical transformation. They appear unchanged on both sides of a chemical equation and do not affect the reaction outcome.
Understanding and identifying spectator ions is crucial for writing net ionic equations.
Understanding and identifying spectator ions is crucial for writing net ionic equations.
- First, write out the full balanced equation of the reaction.
- Look for ions that remain in solution unchanged before and after the reaction.
- These unchanged ions are the spectator ions, as they do not participate actively in the formation of the product.
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