CuSO4(aq) + Na2CO3(aq) Write the balanced net ionic equation Q(15) MnSO4(aq) + Na3PO4(aq) Write the balanced net ionic equation
Q(13) CuSO4(aq) + NaOH(aq) Write the balanced net ionic equation Q(14) CuSO4(aq) + Na2CO3(aq) Write the balanced net ionic equation Q(15) MnSO4(aq) + Na3PO4(aq) Write the balanced net ionic equation
The Correct Answer and Explanation is:
Let’s go through the reactions one by one, writing the balanced net ionic equations and providing explanations:
1. CuSO4(aq) + Na2CO3(aq)
When copper sulfate (CuSO4) reacts with sodium carbonate (Na2CO3) in an aqueous solution, a double displacement reaction occurs, forming copper carbonate (CuCO3) and sodium sulfate (Na2SO4).
Molecular Equation:
CuSO4(aq)+Na2CO3(aq)→CuCO3(s)+Na2SO4(aq)\text{CuSO}_4 (aq) + \text{Na}_2\text{CO}_3 (aq) \rightarrow \text{CuCO}_3 (s) + \text{Na}_2\text{SO}_4 (aq)CuSO4(aq)+Na2CO3(aq)→CuCO3(s)+Na2SO4(aq)
To write the net ionic equation, we first dissociate the soluble ionic compounds into their ions:
Complete Ionic Equation:
Cu2+(aq)+SO42−(aq)+2Na+(aq)+CO32−(aq)→CuCO3(s)+2Na+(aq)+SO42−(aq)\text{Cu}^{2+} (aq) + \text{SO}_4^{2-} (aq) + 2 \text{Na}^+ (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{CuCO}_3 (s) + 2 \text{Na}^+ (aq) + \text{SO}_4^{2-} (aq)Cu2+(aq)+SO42−(aq)+2Na+(aq)+CO32−(aq)→CuCO3(s)+2Na+(aq)+SO42−(aq)
In this case, sodium ions (Na+) and sulfate ions (SO4^2-) are spectator ions, as they appear on both sides of the equation. Removing these gives us the net ionic equation:
Net Ionic Equation:
Cu2+(aq)+CO32−(aq)→CuCO3(s)\text{Cu}^{2+} (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{CuCO}_3 (s)Cu2+(aq)+CO32−(aq)→CuCO3(s)
2. MnSO4(aq) + Na3PO4(aq)
Manganese sulfate (MnSO4) reacts with sodium phosphate (Na3PO4), leading to the formation of manganese phosphate (MnPO4) and sodium sulfate (Na2SO4).
Molecular Equation:
MnSO4(aq)+Na3PO4(aq)→MnPO4(s)+Na2SO4(aq)\text{MnSO}_4 (aq) + \text{Na}_3\text{PO}_4 (aq) \rightarrow \text{MnPO}_4 (s) + \text{Na}_2\text{SO}_4 (aq)MnSO4(aq)+Na3PO4(aq)→MnPO4(s)+Na2SO4(aq)
The complete ionic equation dissociates the soluble ions:
Complete Ionic Equation:
Mn2+(aq)+SO42−(aq)+3Na+(aq)+PO43−(aq)→MnPO4(s)+2Na+(aq)+SO42−(aq)\text{Mn}^{2+} (aq) + \text{SO}_4^{2-} (aq) + 3 \text{Na}^+ (aq) + \text{PO}_4^{3-} (aq) \rightarrow \text{MnPO}_4 (s) + 2 \text{Na}^+ (aq) + \text{SO}_4^{2-} (aq)Mn2+(aq)+SO42−(aq)+3Na+(aq)+PO43−(aq)→MnPO4(s)+2Na+(aq)+SO42−(aq)
Again, sodium ions and sulfate ions are spectator ions. Removing these gives the net ionic equation:
Net Ionic Equation:
Mn2+(aq)+PO43−(aq)→MnPO4(s)\text{Mn}^{2+} (aq) + \text{PO}_4^{3-} (aq) \rightarrow \text{MnPO}_4 (s)Mn2+(aq)+PO43−(aq)→MnPO4(s)
3. CuSO4(aq) + NaOH(aq)
Copper sulfate (CuSO4) reacts with sodium hydroxide (NaOH), producing copper(II) hydroxide (Cu(OH)2) and sodium sulfate (Na2SO4).
Molecular Equation:
CuSO4(aq)+2NaOH(aq)→Cu(OH)2(s)+Na2SO4(aq)\text{CuSO}_4 (aq) + 2 \text{NaOH} (aq) \rightarrow \text{Cu(OH)}_2 (s) + \text{Na}_2\text{SO}_4 (aq)CuSO4(aq)+2NaOH(aq)→Cu(OH)2(s)+Na2SO4(aq)
The complete ionic equation dissociates the soluble ions:
Complete Ionic Equation:
Cu2+(aq)+SO42−(aq)+2Na+(aq)+2OH−(aq)→Cu(OH)2(s)+2Na+(aq)+SO42−(aq)\text{Cu}^{2+} (aq) + \text{SO}_4^{2-} (aq) + 2 \text{Na}^+ (aq) + 2 \text{OH}^- (aq) \rightarrow \text{Cu(OH)}_2 (s) + 2 \text{Na}^+ (aq) + \text{SO}_4^{2-} (aq)Cu2+(aq)+SO42−(aq)+2Na+(aq)+2OH−(aq)→Cu(OH)2(s)+2Na+(aq)+SO42−(aq)
Sodium ions and sulfate ions are spectators. Removing them gives the net ionic equation:
Net Ionic Equation:
Cu2+(aq)+2OH−(aq)→Cu(OH)2(s)\text{Cu}^{2+} (aq) + 2 \text{OH}^- (aq) \rightarrow \text{Cu(OH)}_2 (s)Cu2+(aq)+2OH−(aq)→Cu(OH)2(s)
Explanation of Each Reaction:
- In the first reaction, copper sulfate reacts with sodium carbonate, and the insoluble product formed is copper carbonate. This is a typical double displacement reaction where one product precipitates out as a solid.
- In the second reaction, manganese sulfate reacts with sodium phosphate to form manganese phosphate, which is also insoluble and precipitates out. This is another example of a double displacement reaction, where an insoluble phosphate salt forms.
- In the third reaction, copper sulfate reacts with sodium hydroxide to form copper(II) hydroxide, which is a blue precipitate. This is a classic precipitation reaction where a metal hydroxide is formed from a metal ion and hydroxide ion.
In each case, we identify spectator ions (ions that do not participate in the formation of the precipitate) and exclude them from the net ionic equation, focusing on the ions that form the precipitate.
