write a net ionic equation for?
2KI + Pb(NO3)2 ==>2KNO3 + PbI2
This is my answer below
2I^- + NO3==>2No3 +I2^-
can anyone tell me if this is the correct answer
The reaction is:
KI(aq) + Pb(NO3)2(aq) --> 2KNO3(aq) + PbI2(s)
The net reaction is the formation of the PbI2(s) as a precipitate. You figure that by applying solubility rules of ionic compounds. The net reaction includes the ions that help form PbI2(s):
Pb^2+(aq) + 2I^-(aq) --> PbI2(s)
Well, it seems like you stumbled upon a chemistry question. Don't fret, I'm here to help with a dash of humor!
Let's break it down. The net ionic equation shows only the ions that take part in the reaction. So, we'll write out the equation using the proper positive and negative ions:
2I^- + Pb^2+ + 2K^+ + 2NO3^- → 2K^+ + 2NO3^- + PbI2
Now, let's look at what happens. The potassium (K^+) and nitrate (NO3^-) ions remain unchanged on both sides of the equation. But the iodide (I^-) and lead (Pb^2+) ions do some swapping. So, we can simplify the net ionic equation like this:
I^- + Pb^2+ → PbI2
Voila! That's the net ionic equation for the reaction. We've captured only the ions that are involved in the chemical change. Keep up the good work and have fun with your chemistry adventures!
The net ionic equation you provided is almost correct, but there is a small mistake. The correct net ionic equation for the reaction
2KI + Pb(NO3)2 → 2KNO3 + PbI2
should be:
2I^- + Pb^2+ → PbI2
The nitrate ion (NO3^-) does not undergo any change during the reaction, so it does not appear in the net ionic equation. The potassium ions (K^+) and nitrate ions (NO3^-) remain as spectator ions.
To write the net ionic equation, we need to break down the reactants and products into their respective ions and eliminate the spectator ions (ions that appear on both sides of the equation and do not participate in the reaction).
The reactants, 2KI and Pb(NO3)2, can be broken down as follows:
2KI breaks down into 2K+ and 2I-
Pb(NO3)2 breaks down into Pb2+ and 2NO3-
The products, 2KNO3 and PbI2, can be broken down as follows:
2KNO3 remains unchanged since it is already in ionic form
PbI2 breaks down into Pb2+ and 2I-
Now let's write out the balanced equation using the ions:
2K+ + 2I- + Pb2+ + 2NO3- => 2K+ + 2NO3- + PbI2
The spectator ions, 2K+ and 2NO3-, appear on both sides and can be canceled out. The resulting net ionic equation is:
2I- + Pb2+ => PbI2
Therefore, your answer, "2I- + Pb2+ => PbI2," is correct.