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:
1. Net Ionic Equation for CuSO4(aq) + NaOH(aq):
Step 1: Write the balanced 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)
Step 2: Dissociate all aqueous compounds into their ions.
- CuSO4 (aq) → Cu²⁺ (aq) + SO₄²⁻ (aq)
- NaOH (aq) → Na⁺ (aq) + OH⁻ (aq)
- Na₂SO₄ (aq) → 2Na⁺ (aq) + SO₄²⁻ (aq)
Step 3: Cancel spectator ions.
- The sulfate ions (SO₄²⁻) are present on both sides of the equation, so they are spectator ions and can be canceled out.
Step 4: Write the 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)
2. Net Ionic Equation for CuSO4(aq) + Na2CO3(aq):
Step 1: Write the balanced 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)
Step 2: Dissociate all aqueous compounds into their ions.
- CuSO₄ (aq) → Cu²⁺ (aq) + SO₄²⁻ (aq)
- Na₂CO₃ (aq) → 2Na⁺ (aq) + CO₃²⁻ (aq)
- Na₂SO₄ (aq) → 2Na⁺ (aq) + SO₄²⁻ (aq)
Step 3: Cancel spectator ions.
- The sodium ions (Na⁺) and sulfate ions (SO₄²⁻) are spectator ions and can be canceled out.
Step 4: Write the 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)
3. Net Ionic Equation for MnSO4(aq) + Na3PO4(aq):
Step 1: Write the balanced molecular equation.MnSO4(aq)+Na3PO4(aq)→Mn3(PO4)2(s)+Na2SO4(aq)\text{MnSO}_4(aq) + \text{Na}_3\text{PO}_4(aq) \rightarrow \text{Mn}_3\text{(PO}_4)_2(s) + \text{Na}_2\text{SO}_4(aq)MnSO4(aq)+Na3PO4(aq)→Mn3(PO4)2(s)+Na2SO4(aq)
Step 2: Dissociate all aqueous compounds into their ions.
- MnSO₄ (aq) → Mn²⁺ (aq) + SO₄²⁻ (aq)
- Na₃PO₄ (aq) → 3Na⁺ (aq) + PO₄³⁻ (aq)
- Na₂SO₄ (aq) → 2Na⁺ (aq) + SO₄²⁻ (aq)
Step 3: Cancel spectator ions.
- The sodium ions (Na⁺) and sulfate ions (SO₄²⁻) are spectator ions and can be canceled out.
Step 4: Write the net ionic equation.3Mn2+(aq)+2PO43−(aq)→Mn3(PO4)2(s)3\text{Mn}^{2+}(aq) + 2\text{PO}_4^{3-}(aq) \rightarrow \text{Mn}_3\text{(PO}_4)_2(s)3Mn2+(aq)+2PO43−(aq)→Mn3(PO4)2(s)
Explanation:
In these reactions, the spectator ions (such as Na⁺ and SO₄²⁻) do not participate in the chemical change, so they are removed from the equation. The remaining ions in the reaction form precipitates, which are represented as solid compounds in the net ionic equations. This highlights the key species involved in the reaction that undergo chemical changes, leaving out the ions that remain in solution unchanged.
