I misread the question i posted earlier.
The equation is
Pb(OH)2 (s) + NaOH (aq)---> ?
I know that the complex ion formed is Pb(OH)4^2- , but would that be written in a balanced net-ionic equation?
correction: *how would that be written in a balanced net-ionic equation?
Pb(OH)2(s) + 2OH^-(aq) ==> Pb(OH)4^-2(aq)
To determine the balanced net-ionic equation for the reaction between Pb(OH)2 (s) and NaOH (aq), we first need to write the complete ionic equation and then cancel out the spectator ions to obtain the net-ionic equation.
The complete ionic equation can be written as follows:
Pb(OH)2 (s) + 2Na+(aq) + 2(OH)-(aq) → Pb(OH)2(s) + 2Na+(aq) + 2OH-(aq)
In this equation, both Pb(OH)2 and NaOH are in their ionic form with their respective charges. The (s) denotes the solid state of Pb(OH)2, and (aq) represents the dissolved state of Na+ and OH- ions in water.
Now, let's cancel out the spectator ions. In this case, Na+ and OH- ions appear on both sides of the equation without undergoing any change. Therefore, they can be eliminated:
Pb(OH)2 (s) + 2OH-(aq) → Pb(OH)2(s)
The resulting equation is the net-ionic equation. It shows only the species that participate directly in the reaction, which in this case is the Pb(OH)2 (s) compound.
So, the balanced net-ionic equation for the reaction between Pb(OH)2 (s) and NaOH (aq) is:
Pb(OH)2 (s) + 2OH-(aq) → Pb(OH)2(s)