When sodium hydroxide and copper (II) sulphate are forms. The unbalanced equation for this sulphas is: (10 marks) NaOH+CuSO_(4)7Na_(2)SO_(4)+Cu(OH)_(2) Write the; a. Balanced equation (including status of matter of products and reactants) b. Net ionic equation

a. Balanced equation:

$$2NaOH + CuSO_4 \rightarrow Na_2SO_4 + Cu(OH)_2$$

In this reaction, sodium hydroxide (NaOH) reacts with copper (II) sulfate (CuSO_4) to form sodium sulfate (Na_2SO_4) and copper (II) hydroxide (Cu(OH)_2), which is a blue precipitate.

b. Net ionic equation:

To write the net ionic equation, we need to identify the species that undergo a change during the reaction. In this case, the copper (II) ion (Cu^2+) and the hydroxide ion (OH^-) are the species that change.

The net ionic equation is:

$$Cu^{2+} + 2OH^- \rightarrow Cu(OH)_2$$

In the net ionic equation, we only include the species that undergo a change, omitting the spectator ions (in this case, the sodium ions and the sulfate ions).