Question

The following chemical reaction takes place in aqueous solution: CuCl 2

​
(aq)+2NaOH(aq)→Cu(OH) 2
​
(s)+2NaCl(aq) Write the net ionic equation for this reaction.

Answer 1

To write the net ionic equation, first, break down all strong electrolytes (soluble salts, strong acids, strong bases) into their constituent ions. Then, remove the spectator ions, which are those ions that appear unchanged on both sides of the equation.

The full ionic equation for the reaction between copper(II) chloride (CuCl2) and sodium hydroxide (NaOH) in aqueous solution is:

`\[ \text{Cu}^(2+)(aq) + 2\text{Cl}^(-)(aq) + 2\text{Na}^(+)(aq) + 2\text{OH}^(-)(aq) \rightarrow \text{Cu}(\text{OH})_2(s) + 2\text{Na}^(+)(aq) + 2\text{Cl}^(-)(aq) \]`Now, to write the net ionic equation, remove the spectator ions, which are the sodium (Na+) and chloride (Cl-) ions because they appear on both sides of the equation and do not participate in the actual chemical change:

`\[ \text{Cu}^(2+)(aq) + 2\text{OH}^(-)(aq) \rightarrow \text{Cu}(\text{OH})_2(s) \]`

This is the net ionic equation for the reaction, showing only the ions and compounds that are directly involved in the chemical change. Copper(II) ions react with hydroxide ions to form solid copper(II) hydroxide.

Answer 2

The net ionic equation for the reaction of
`CuCl_2` and NaOH is:

`Cu^(2+)(aq)\ +\ 2OH^-(aq) \rightarrow Ca(OH)_2(s)`

How to write the net ionic equation?

Net ionic equation is simply an equation showing the net ion of a reaction in which the spectator ions are not considered.

The net ionic equation for the reaction of
`CuCl_2` and NaOH can be written as illustrated below:

`CuCl_2(aq)\ +\ NaOH(aq)\ \rightarrow\ \\\\But,\\\\CuCl_2(aq) = > Cu^(2+)(aq)\ +\ Cl^-(aq)\\\\NaOH(aq) = > Na^(+)(aq)\ +\ OH^-(aq)\\\\Thus,\ we\ have\\\\Cu^(2+)(aq)\ +\ Cl^-(aq)\ +\ Na^(+)(aq)\ +\ OH^-(aq) \rightarrow Ca(OH)_2(s)\ +\ Cl^-(aq)\ +\ Na^(+)(aq)`

Removing the spectator ions (
`Cl^-\ and\ Na^(+)`), we have:

`Cu^(2+)(aq)\ +\ 2OH^-(aq) \rightarrow Ca(OH)_2(s)`