What is the net ionic equation form the reaction between Na2S(aq) and Pb(HCO3)2(aq)?
Just wanting to make sure this correct:
2Na+(aq) + S2-(aq) + Pb2+(aq) + 2HCO3-(aq) --> 2Na+(aq) + 2HCO3-(aq) + PbS(s)
The question asks for the NET ionic equation. What you have written (correctly) is the MOLECULAR equation. To turn the molecular equation into the net ionic equation you cancel ions common to both sides. That means 2Na^+ and 2HCO3^- cancel leaving
Pb^2+(aq) + S^2-(aq) ==> PbS(s)
To determine the net ionic equation, we need to first write the balanced molecular equation for the reaction between Na2S and Pb(HCO3)2:
Na2S(aq) + Pb(HCO3)2(aq) → 2NaHCO3(aq) + PbS(s)
Now, let's break down the soluble compounds into their ions:
Na2S(aq): 2Na+(aq) + S2-(aq)
Pb(HCO3)2(aq): Pb2+(aq) + 2HCO3-(aq)
2NaHCO3(aq): 2Na+(aq) + 2HCO3-(aq)
From this, we can see that the net ionic equation is:
S2-(aq) + Pb2+(aq) → PbS(s)
To determine the net ionic equation for the reaction between Na2S(aq) and Pb(HCO3)2(aq), we need to balance the equation and identify the spectator ions.
First, let's write out the balanced molecular equation:
Na2S(aq) + Pb(HCO3)2(aq) → 2NaHCO3(aq) + PbS(s)
Now, let's break down the compounds into their ions:
Na2S(aq) → 2Na+(aq) + S2-(aq)
Pb(HCO3)2(aq) → Pb2+(aq) + 2HCO3-(aq)
Next, we need to determine which ions remain unchanged throughout the reaction. These ions are called spectator ions. In this case, the spectator ions are Na+ and HCO3-. They appear on both sides of the equation and do not participate in the actual reaction.
Therefore, we can rewrite the equation by removing the spectator ions:
Net Ionic Equation: S2-(aq) + Pb2+(aq) → PbS(s)
So, the correct net ionic equation for the reaction between Na2S(aq) and Pb(HCO3)2(aq) is:
S2-(aq) + Pb2+(aq) → PbS(s)