Cu(NO3)2(aq) and K2S(aq)
Express your answer as a net ionic equation. Identify all of the phases in your answer. Enter noreaction if no precipitate is formed.
The balanced molecular equation follows:
Cu(NO3)2(aq) + K2S(aq) ==> CuS(s) + 2KNO3(aq)
Here is what you do. Separate the aq solutions into ions.This will give you the TOTAL ionic equation.
Cu(NO3)2(aq) + K2S(aq) ==> CuS(s) + 2KNO3(aq)
Cu^2+(aq) + 2(NO3)^-(aq) + 2K^+(aq) + S^2-(aq) ==> CuS(s) + 2K^+(aq) + 2(NO3)^-(aq)
Now you cancel those ions that appear in both sides of the equation. I see 2(NO3)^-(aq) and 2K^+(aq). That leaves the NET ionic equation.
Cu^2+(aq) + S^2-(aq) ==> CuS(s)
The three steps are as follows:
1. Write and balance the molecular equation.
2. Separate in ions showing (aq) for those in solution and (s) for a ppt. This gives you the total ionic equation.
3. Cancel materials that appear on both sides of the equation. That gives you the net ionic equation. Do you know what those canceled ions are called. Those are spectator ions; they don't enter nor are they needed to make the equation work. ANY Cu^2+ soluble salt (in this case it was nitrate ion) + ANY soluble sulfide (in this case K2S) will give you CuS.
Well, it seems we have a bit of a party going on here with Cu(NO3)2 and K2S mixing. Let's see what happens when these two get together!
Cu(NO3)2(aq) + K2S(aq)
Well, let's break this down a bit. Cu(NO3)2 contains copper ions (Cu^2+) and nitrate ions (NO3^-), while K2S contains potassium ions (K^+) and sulfide ions (S^2-).
When these guys mingle, the following reaction occurs:
Cu(NO3)2(aq) + K2S(aq) -> CuS(s) + 2 KNO3(aq)
Look at that, we got a precipitate! The copper ions (Cu^2+) from Cu(NO3)2 combine with the sulfide ions (S^2-) from K2S to form a solid copper sulfide (CuS), which is the precipitate. The potassium ions (K^+) from K2S pair up with the nitrate ions (NO3^-) from Cu(NO3)2 to form potassium nitrate (KNO3), which stays dissolved in the aqueous solution.
So, the net ionic equation for this reaction is:
Cu^2+(aq) + S^2-(aq) -> CuS(s)
And there you have it! The party resulted in the formation of some solid copper sulfide (CuS). Enjoy the show!
To find the net ionic equation for the reaction between Cu(NO3)2(aq) and K2S(aq), we first need to write out the balanced chemical equation using the chemical formulas.
Cu(NO3)2(aq) + K2S(aq) → CuS(s) + 2KNO3(aq)
The full ionic equation is:
Cu^2+(aq) + 2NO3^-(aq) + 2K^+(aq) + S^2-(aq) → CuS(s) + 2K^+(aq) + 2NO3^-(aq)
Now, we can eliminate the spectator ions (ions that appear on both sides of the equation unchanged) from the equation. In this case, the spectator ions are K^+ and NO3^-.
The net ionic equation is:
Cu^2+(aq) + S^2-(aq) → CuS(s)
The phases in the net ionic equation are:
Cu^2+(aq) + S^2-(aq) → CuS(s)
To write the net ionic equation, we need to first write the balanced chemical equation for the given reaction:
Cu(NO3)2(aq) + 2K2S(aq) → CuS(s) + 4KNO3(aq)
In this reaction, Cu(NO3)2(aq) and K2S(aq) are the reactants, while CuS(s) and 4KNO3(aq) are the products.
Next, we can write the ionic equation by representing the reactants and products as ions:
Cu2+(aq) + 2NO3-(aq) + 2K+(aq) + S2-(aq) → CuS(s) + 4NO3-(aq) + 4K+(aq)
Finally, to obtain the net ionic equation, we remove any ions that appear on both sides of the equation. In this case, the NO3- ions appear on both sides:
Cu2+(aq) + S2-(aq) → CuS(s)
The net ionic equation is: Cu2+(aq) + S2-(aq) → CuS(s)
The phases (states) of the substances in the equation are:
Cu(NO3)2(aq) - aqueous
K2S(aq) - aqueous
CuS(s) - solid