Question

Complete and balance the molecular equation, including phases, for the reaction of aqueous copper(II) chloride, CuCl2, and aqueous potassium phosphate, K3PO4.

CuCl2(aq) + K3PO4(aq)--->

Enter the balanced net ionic equation, including phases, fir this reaction.

Answer 1

You need 3 components for this question: Molecular Equation(M.E), Complete Ionic Equation(C.I.E), and Net Ionic Equation (N.I.E).

Let's do the first part of the first step:
M.E = \[CuCl _{2}(aq) + K _{3}PO _{4}(aq)\]
Let's use the model: AB + CD -> AD+CB (double replacement equation)

So using that model, we will determine the second part of the equation:
\[Cu _{3}(PO _{4})_{2} + KCl\]
Notice how the states are missing. We will get to that later. For now, let's understand why we got what we got:
So remember, usually copper will exist in +1 and +2 oxidation states(charges) but sometimes copper can also exist in +3 just like Iron can but only in really complex solutions. For this particular ionic compound, since Cu2+ is attached to phosphate, the charges of the compound can be determined like this:
\[Cu ^{2+} PO _{4}^{3-}\]
+ +
\[Cu ^{2+} PO _{4}^{3-}\]
+
\[Cu ^{2+}\]
\[+6 Cu = -6PO _{4}\]
Since the charges are balanced, and for every 3 atoms of Cu2+, there will always be 2 atoms of PO4 3- in the compound. As for KCl ionic compound, since K is in group IA and it always exists in nature with a +1 charge and since Cl always exists with a -1 charge being in group 17 and since the charges are already balanced, we can just leave the new product compound as KCl and not K3Cl2.

Now let's determine the states of these products. In order to do this, we have to consider the solubility rules(look them up in your textbook):
The rules we will use in this are:
1) Cu2+ is most commonly considered an insoluble cation.
2) PO4 3- is also insoluble but with 2 exceptions: 1) PO4 3- combined with IA group will result as soluble and 2) PO4 3- combined with NH4+ (considered a cation in combination with Phosphate) will result as soluble.
3) Potassium is in group I and all group I cations are are soluble.
4) All halides(group 17) are considered soluble but with exceptions of cations: Ag+, Cu+, Pb+, Hg2 2+. Since Cl is a halogen, it is soluble.

After considering those rules, we can now apply the states:
\[Cu _{3}(PO_{4})_{2}(s) + KCl(aq)\]
So our complete M.E is:
\[CuCl _{2}(aq)+K _{3}PO _{4}(aq) \rightarrow Cu _{3}(PO _{4})_{2}(s) + KCl(aq)\]

Answer 2

Now we will balance the molecular equation by adding coefficients:
\[3CuCl _{2}(aq)+2K _{3}PO _{4}(aq) \rightarrow Cu _{3}(PO _{4})_{2}(s)+6KCl\]
Atoms of elements in original/complete molecular eqn:
Cu - 1 Cu - 3
Cl - 2 Cl - 1
K - 3 K - 1
P - 1 P - 2
O - 4 O - 8

Atoms of elements in balanced molecular eqn:
Cu - 3 Cu - 3
Cl - 6 Cl - 6
K - 6 K - 6
P - 2 P - 2
O - 8 O - 8

Now the second step is to determine the complete ionic equation. The way to determine this equation is by breaking up all the aqueous solutions into ions and polyatomic ions. You do not break up solids and liquids in complete ionic equations. Let's begin

C.I.O:
\[3Cu ^{2+}(aq) + 6Cl ^{-}(aq) + 6K ^{+} + 2PO _{4}^{3-}(aq) \rightarrow Cu _{3}(PO _{4})_{2}(s) + 6K ^{+}(aq) + 6Cl ^{-}(aq)\]

Basically just add multiply the coefficients attached to first compound by the atoms of atoms of elements within that same compound. So first would be 3x1 Cu = 3Cu2+ and 3x2 = 6 Cl- and then do the same with the second compound, 2x3 = 6K+ and 2x1=PO4 and not 8PO because PO4 is a polyatomic ion by itself and you keep the same coefficient of 2 for all polyatomic ions. Now continue doing the same for the other side of the equation.

Answer 3

Now the final step is to determine the net ionic equation. The way to do this is to get rid of all the spectator ions in the complete ionic equation previously done. Spectator ions are ions that are present in the reaction but they are unchanged by the reaction. These ions do appear in the molecular and overall ionic equation but not in net ionic equation because they do not participate in the actual reaction.

Now the N.I.E (actual reaction) is:
\[3Cu ^{2+}(aq) + 2PO _{4}^{3-}(aq) \rightarrow Cu _{3}(PO _{4})_{2}(s)\]
The above is a precipitate reaction.

So the spectator ions in the complete ionic equation were:
\[6Cl ^{-}(aq) + 6K ^{+}(aq) \rightarrow 6K ^{+}(aq) + 6Cl ^{-}(aq)\]
The above is not a real reaction.

Just remember to cross out those spectator ions on both sides of the equation and what is left in the complete ionic equation is your net ionic equation.

Hope that helped! I'm currently taking Chemistry I myself so this was good practice for me. My personal thanks to you.

Answer 4

I think I forgot to add (aq) state to K+ in C.I.E. above.

Answer 5

and correction: multiply the coefficient and not 'add.' pretty late at night. My apologies.