Here's how to write the complete ionic equation for the reaction between lead(II) nitrate and sodium sulfate:

1. Write the balanced molecular equation:

Pb(NO₃)₂(aq) + Na₂SO₄(aq) → PbSO₄(s) + 2NaNO₃(aq)

2. Identify the soluble ionic compounds:

* Lead(II) nitrate (Pb(NO₃)₂) is soluble in water.

* Sodium sulfate (Na₂SO₄) is soluble in water.

* Lead(II) sulfate (PbSO₄) is insoluble in water (it forms a precipitate).

* Sodium nitrate (NaNO₃) is soluble in water.

3. Separate the soluble ionic compounds into their ions:

* Pb(NO₃)₂(aq) → Pb²⁺(aq) + 2NO₃⁻(aq)

* Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq)

* PbSO₄(s) remains as a solid.

* 2NaNO₃(aq) → 2Na⁺(aq) + 2NO₃⁻(aq)

4. Combine the ions to form the complete ionic equation:

Pb²⁺(aq) + 2NO₃⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → PbSO₄(s) + 2Na⁺(aq) + 2NO₃⁻(aq)

This is the complete ionic equation.