Write the net ionic equation for the following reaction: aqueous solutions of copper(ii) nitrate and sodium hydroxide are mixed to form solid copper(ii) hydroxide and aqueous sodium nitrate. 1. cu2+ + 2 no− 3 + 2na+ + 2oh− → cu2+ + 2 oh− + 2 na+ + 2 no− 3 2. cu2+ + 2 oh− → cu(oh)2(s) 3. cu(no3)2(aq) + 2 naoh(aq) → cu(oh)2(s) + 2 nano3(aq) 4. 2 na+ + 2 no− 3 → 2 nano3(aq)

3. Cu(NO₃)₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaNO₃(aq)

Explanation:

Copper(II) is the ion Cu⁺² and nitrate is the ion NO₃⁻, so the compound will be formed by the change of the charge without the signals: Cu(NO₃)₂. Sodium has the ion Na⁺, and hydroxide OH⁻, so the compound is NaOH.

Then, copper (II) hydroxide is Cu(OH)₂, and sodium nitrate is NaNO₃.

The reaction is:

Cu(NO₃)₂(aq) + NaOH(aq) → Cu(OH)₂(s) + NaNO₃(aq)

To balance the equation, the elements must have the same amount at both sides of it, so we must multiply NaOH and NaNO₃ by 2:

Cu(NO₃)₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaNO₃(aq)