what is the net ionic reaction for coppper(II)chloride and sodium hydroxide?
To determine the net ionic reaction for copper(II) chloride (CuCl2) and sodium hydroxide (NaOH), we need to first write the balanced molecular equation, followed by the complete ionic equation, and finally, the net ionic equation.
First, let's write the balanced molecular equation:
CuCl2 + 2NaOH → Cu(OH)2 + 2NaCl
Next, let's write the complete ionic equation by breaking down all soluble compounds into their respective ions:
Cu2+(aq) + 2Cl-(aq) + 2Na+(aq) + 2OH-(aq) → Cu(OH)2(s) + 2Na+(aq) + 2Cl-(aq)
Now, let's simplify the equation by removing the spectator ions. Spectator ions are those that appear on both sides of the equation and do not participate in the actual reaction:
Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)
The net ionic equation is the simplified equation, which shows only the ions that directly participate in the reaction. In this case:
Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)
So, the net ionic reaction for copper(II) chloride and sodium hydroxide is Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s).
Write the molecular equation. I assume CuCl2 and NaOH are in solution. They won't react as dry compounds.
CuCl2(aq) + 2NaOH(aq) ==> Cu(OH)2 + 2NaCl(aq)
Now separate each compound into ions using the following guidelines.
Aqueous solutions of strong electrolytes are written as ions.
Aqueous solution of weak electrolytes are written as the molecule.
Insoluble compounds are written as the molecule.