Question

Nickel(II) sulfate solution reacts with sodium hydroxide solution to produce a precipitate of nickel(II) hydroxide and a solution of sodium sulfate. Write the molecular equation for this reaction. Then write the corresponding net ionic equation.

Short answer

Molecular: \( \text{NiSO}_4 + 2\text{NaOH} \rightarrow \text{Ni(OH)}_2 + \text{Na}_2\text{SO}_4 \); Net ionic: \( \text{Ni}^{2+} + 2\text{OH}^- \rightarrow \text{Ni(OH)}_2 \).

Step-by-step solution

Write the balanced molecular equation

First, identify the reactants and products. Nickel(II) sulfate is \( \text{NiSO}_4 \), sodium hydroxide is \( \text{NaOH} \), nickel(II) hydroxide is \( \text{Ni(OH)}_2 \), and sodium sulfate is \( \text{Na}_2\text{SO}_4 \). The balanced molecular equation is:\[\text{NiSO}_4 (aq) + 2\text{NaOH} (aq) \rightarrow \text{Ni(OH)}_2 (s) + \text{Na}_2\text{SO}_4 (aq)\]

Write the complete ionic equation

The complete ionic equation involves splitting all soluble ionic compounds into their respective ions. However, nickel(II) hydroxide, being a precipitate, remains as a compound. The equation is:\[\text{Ni}^{2+} (aq) + \text{SO}_4^{2-} (aq) + 2\text{Na}^+ (aq) + 2\text{OH}^- (aq) \rightarrow \text{Ni(OH)}_2 (s) + 2\text{Na}^+ (aq) + \text{SO}_4^{2-} (aq)\]

Write the net ionic equation

To write the net ionic equation, cancel out the spectator ions, which appear on both sides of the complete ionic equation. These are the sodium ions \( \text{Na}^+ \) and the sulfate ions \( \text{SO}_4^{2-} \). The net ionic equation is:\[\text{Ni}^{2+} (aq) + 2\text{OH}^- (aq) \rightarrow \text{Ni(OH)}_2 (s)\]

Verify the Net Ionic Equation

Ensure the equation is balanced in terms of both mass and charge. Both sides have:- Nickel: 1 atom- Hydroxide: 2 ions- Total charge: \( 0 \)Thus, the net ionic equation is balanced.

Key concepts

Molecular Equation

When dealing with chemical reactions, the molecular equation is the first step in clearly understanding what happens during the process. In this case, we're looking at a reaction between nickel(II) sulfate \( \text{NiSO}_4 \) and sodium hydroxide \( \text{NaOH} \). A molecular equation expresses the complete formulas of the reactants and the products as they exist in solution. It's like looking at the whole picture.

The molecular equation for the reaction is:

• \( \text{NiSO}_4 (aq) + 2\text{NaOH} (aq) \rightarrow \text{Ni(OH)}_2 (s) + \text{Na}_2\text{SO}_4 (aq) \)

In this equation:

• The \( \text{(aq)} \) symbol means that the compound is dissolved in water, forming an aqueous solution.
• The \( \text{(s)} \) symbol indicates that nickel(II) hydroxide is a solid, or a precipitate, which forms out of the solution.

The balanced molecular equation shows that the number of atoms of each element is conserved during the reaction. Molecular equations give us the starting point for further analysis, such as writing ionic and net ionic equations.

Ionic Equation

The ionic equation provides a deeper look into the solution's chemical nature by breaking down the soluble compounds into their respective ions. Ionic compounds that dissolve in water dissociate into ions.

For the reaction between nickel(II) sulfate and sodium hydroxide, the complete ionic equation represents each species in the form that they exist in the solution:

• \( \text{Ni}^{2+} (aq) + \text{SO}_4^{2-} (aq) + 2\text{Na}^+ (aq) + 2\text{OH}^- (aq) \rightarrow \text{Ni(OH)}_2 (s) + 2\text{Na}^+ (aq) + \text{SO}_4^{2-} (aq) \)

• Note how the ions separated due to dissolution in water, while nickel(II) hydroxide, being insoluble, stays as a compound in the equation.

Understanding ionic equations helps to visualize which ions participate in forming the solid product (precipitate) and which remain dissolved, setting the stage for identifying spectator ions.

Net Ionic Equation

The net ionic equation strips away the ions that don't change during the course of the reaction, known as spectator ions, to focus on the heart of the reaction.

Spectator ions, in this case, are the sodium \( \text{Na}^+ \) ions and the sulfate \( \text{SO}_4^{2-} \) ions. After removing these, we are left with the net ionic equation:

• \( \text{Ni}^{2+} (aq) + 2\text{OH}^- (aq) \rightarrow \text{Ni(OH)}_2 (s) \)

This equation succinctly shows the formation of the precipitate, nickel(II) hydroxide, from its constituent ions. It highlights the essence of the chemical reaction—ions coming together to form a new substance that is no longer dissolved in water.

Net ionic equations are particularly useful for focusing on the substances that truly react, making it easier to understand the underlying chemical principles at work.