Question

What is the net ionic equation for the reactions: CuNO3 (aq) + NaCl (aq) --> CuCl (s) + NaNO3 (aq)? Additionally, how would one solve this type of equation?

Answer 1

The net ionic equation for the reaction between CuNO3 and NaCl, producing CuCl and NaNO3 is Cu^{2+} (aq) + 2 Cl^{-} (aq) → CuCl2 (s), after cancelling the spectator ions Na^{+} and NO3^{-}.

Step-by-step explanation:

To derive the net ionic equation for the reaction between CuNO3 (aqueous) and NaCl (aqueous), which produces CuCl (solid) and NaNO3 (aqueous), you should first write the balanced molecular equation, then write the ionic equation by dissociating all strong electrolytes into their ions, and finally, cancel out the spectator ions to find the net ionic equation.

The balanced molecular equation is:

Cu(NO3)2 (aq) + 2 NaCl (aq) → CuCl2 (s) + 2 NaNO3 (aq)

The ionic equation is:

Cu^{2+} (aq) + 2 NO3^{-} (aq) + 2 Na^{+} (aq) + 2 Cl^{-} (aq) → CuCl2 (s) + 2 Na^{+} (aq) + 2 NO3^{-} (aq)

Here, the sodium (Na^{+}) and nitrate (NO3^{-}) ions are spectator ions as they appear on both sides of the equation without change. Cancelling these out, we are left with the net ionic equation:

Cu^{2+} (aq) + 2 Cl^{-} (aq) → CuCl2 (s)