Problems #11-25: Complete Molecular, Complete Ionic and Net Ionic

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Problem #11: Write the complete ionic and net ionic equation for this reaction in aqueous solution:

AgNO3 + CH3COOH ---> AgCH3COO + HNO3

Please include state symbols in both reactions.


1) Complete ionic:

Ag+(aq) + NO3-(aq) + CH3COOH(aq) ---> AgCH3COO(s) + H+(aq) + NO3-(aq)

Acetic acid is a weak acid, consequently it is written in molecular form. In aqueous solution, it is only a few percent ionized. Silver acetate is insoluble and you learn this from a solubility chart.

2) Net ionic:

Ag+(aq) + CH3COOH(aq) ---> AgCH3COO(s) + H+(aq)

Nitrate is the only spectator ion.

Problem #12: Write balanced molecular equation and net ionic equations for the following reactions.

a) A solution of potassium hydroxide reacts with a solution of sodium hydrogen phosphate.
b) A solution of magnesium nitrate reacts with a solution of ammonium carbonate.

Solution to a:

6KOH(aq) + 3NaH2PO4(aq) ---> Na3PO4(aq) + 2K3PO4(aq) + 6H2O(l)

6OH-(aq) + 3H2PO4- ---> 3PO43-(aq) + 6H2O(l)

This is an acid base neutralization.

Solution to b:

Mg(NO3)2(aq) + (NH4)2CO3(aq) ----> MgCO3(s) + 2NH4NO3(aq)

Mg2+(aq) + CO32-(aq) ---> MgCO3(s)

This is a double replacement.

Problem #13: Write balanced molecular, complete ionic and net ionic equations for this reaction:

aluminum nitrate reacts with potassium chloride to produce aluminum chloride and potassium nitrate



Al(NO3)3(aq) + 3KCl(aq) ---> AlCl3(aq) + 3KNO3(aq)

Complete ionic:

Al3+(aq) + 3NO3-(aq) + 3K+(aq) + 3Cl-(aq) ---> Al3+(aq) + 3Cl-(aq) + 3K+(aq) + 3NO3-(aq)

Net ionic:


NR stands for 'no reaction.' Eveything, on both sides, is soluble and stays in solution. Everything, on both sides, ionizes. No precipitate is formed. No gas is formed. No liquid water (a hallmark of the acid base neutralization) is formed. The key is that everything is a spectator ion so everything, on each side, gets eliminated in the net ionic.

Conclusion? No chemical reaction occured.

The reason I put this reaction in is because there will be a series of example reactions in whch something happens and then, on the test, a NR appears without its possibility ever being mentioned.

You have been warned!

If you take AP Chemistry, you will be asked to predict products and write net ionic equations. NR is not an acceptable answer for any of the problems presented.

Problem #14: Write balanced net ionic equations for the following reactions in aqueous solution:

a) sodium carbonate and vanadium(V) chloride
b) lithium phosphate and chromium(III) fluoride

Solution to a:

Since there are no acids or bases in the two reactants for the first reaction, we predict this to be a double replacement reaction. On that basis, we write this:
5Na2CO3(aq) + 2VCl5(aq) ---> V2(CO3)5(s) + 10NaCl(aq)

All three soluble substances are ionic, so they becomes ions in solution. The sodium ion and the chloride ion are spectator ions. The net ionic is:

2V5+ + 5CO32- ---> V2(CO3)5(s)

How do you know that V2(CO3)5 precipitates? The answer is that you usually can't figure it out from a solubility chart because vanadium is not usually included. However, most metallic carbonates precipitate, so it's a fairly reasonable guess that V2(CO3)5 is not soluble.

And, yes, this is an unusual problem.

Solution to b:

We also predict this to be a double replacement reaction:
Li3PO4(aq) + CrF3(aq) ---> CrPO4(s) + 3LiF(aq)

Heavy metal phosphates are almost always insoluble. The net ionic is this:

Cr3+(aq) + PO43-(aq) ---> CrPO4(s)

Now, a problem! It turns out that lithium phosphate is fairly insoluble. Here's an example from Yahoo Answers which mentions this. The problem is that many high school chemistry teachers may not know this (I didn't for a number of years!). So, this is a more chemically correct net ionic:

Cr3+(aq) + Li3PO4(s) ---> CrPO4(s) + 3Li+(aq)

The problem is that your teacher (or an answer in an online chemistry class) might expect the first net ionic I wrote above. My recommendation is to give the expected answer and move on. Don't try and argue the point.

Problem #15: What is the net ionic equation for copper(II) hydroxide reacting with dilute sulfuric acid? (Warning: this is a complicated answer!)


Molecular equation:

Cu(OH)2(s) + H2SO4(aq) --> CuSO4(aq) + 2H2O(l)

This is an acid base neutralization. It can also be viewed as a double replacement, but acid base is the most common answer.

I want you to notice the (s) after the copper(II) hydroxide. This is because copper(II) hydroxide is insoluble, consequently (aq) is not used. Two points: (1) usually, insoluble stuff appears on the product side, not often on the reactant side and (2) your teacher may demand that (aq) be used rather than (s).

The rationale for (aq) is that the Cu(OH)2 that does react dissolves (and ionizes, as we shall see) first and so it reacts as aqueous rather than solid.

Complete ionic based on solid for Cu(OH)2:

Cu(OH)2(s) + H+(aq) + HSO4-(aq) --> Cu2+(aq) + SO42-(aq) + 2H2O(l)

It is important to note that sulfuric acid is a strong acid, but only to the extent of the dissociation of the first H+. The HSO4- ion that results is a weak acid, and is not dissociated.

Since there are no spectator ions, nothing is eliminated and the net ionic equation is the same as the complete ionic equation.

Again, there could be a problem (or two). One problem is that your instructor will insist that sulfuric acid is fully dissociated in BOTH hydrogens. Another possible problem is that the copper(II) hydroxide will be treated as soluble and written as the ions rather than the solid. In that case, this is the net ionic tha results:

H+(aq) + OH-(aq) ---> H2O(l)

Two problems from Yahoo Answers:

Problem #16

Problem #17

Problem #18: When a solution of sodium hydroxide is added to a solution of ammonium carbonate, H2O is formed and ammonia gas, NH3, is released when the solution is heated. Write a complete molecular, complete ionic and net ionic equations for this reaction.


The reactants for the molecular equation are these:

(NH4)2CO3(aq) + NaOH(aq) --->

The products are these:

(NH4)2CO3(aq) + 2NaOH(aq) ---> Na2CO3(aq) + 2NH3(g) + 2H2O(l)

The above is the balanced molecular equation. Note that sodium carbonate is also a product, one that was not mentioned in the problem text.

By the way, this is a (more-or-less) double replacement, with a bit of decomposition thrown in. The ammonia and water come from NH4OH, a "compound" which is unstable, decomposing immediately to ammonia and water.

Complete ionic:

2NH4+(aq) + CO32-(aq) + 2Na+(aq) + 2OH-(aq) ---> 2Na+(aq) + CO32-(aq) + 2NH3(g) + 2H2O(l)

Net ionic:

2NH4+(aq) + 2OH-(aq) ---> 2NH3(g) + 2H2O(l)

which reduces to:

NH4+(aq) + OH-(aq) ---> NH3(g) + H2O(l)

Problem #19: Write the complete molecular, complete ionic and net ionic equations for ammonium carbonate reacting with barium hydroxide.



(NH4)2CO3(aq) + Ba(OH)2(aq) ---> BaCO3(s) + 2H2O(l) + 2NH3(g)

complete ionic:

2NH4+(aq) + CO32-(aq) + Ba2+(aq) + 2OH-(aq) ---> BaCO3(s) + 2H2O(l) + 2NH3(g)
net ionic equation:
2NH4+(aq) + CO32-(aq) + Ba2+(aq) + 2OH-(aq) ---> BaCO3(s) + 2H2O(l) + 2NH3(g)

Note the last two equations are the same. Everything has changed between reactants and products, there are no spectator ions. This example is a bit reminiscent (at least to the ChemTeam!) of the double precipitation in problem #10.

Problem #20: Zinc chloride solution is poured into a solution of ammonium carbonate.


Note: ammonium does not always break down into ammonia gas.


ZnCl2(aq) + (NH4)2CO3(aq) ---> ZnCO3(s) + 2NH4Cl(aq)

complete ionic equation:

Zn2+(aq) + 2Cl-(aq) + 2NH4+(aq) + CO32-(aq) ---> ZnCO3(s) + 2NH4+(aq) + 2Cl-(aq)

net ionic:

Zn2+(aq) + CO32-(aq) ---> ZnCO3(s)

When ammonium is reacted with a base, ammonia is produced. In the above problem, there is no base.

Problem #21: Pb(NO3)2(aq) + Na2S(aq) --->



Pb(NO3)2(aq) + Na2S(aq) ---> PbS(s) + 2NaNO3(aq)

full ionic:

Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + S2-(aq) ---> PbS(s) + 2Na+(aq) + 2NO3-(aq)

net ionic:

Pb2+(aq) + S2-(aq) ---> PbS(s)

Sodium ion and nitrate ion were the spectator ions removed.

Problem #22: ammonium phosphate + calcium chloride --->



2(NH4)3PO4(aq) + 3CaCl2(aq) ---> 6NH4Cl(aq) + Ca3(PO4)2(s)

You know the calcium phosphate precipitates by knowing the solubility table.

net ionic:

3Ca2+(aq) + 2PO43-(aq) ---> Ca3(PO4)2(s)

Rearranged to put the cation first on the reactant side.

Problem #23: Cobalt(II) nitrate reacts with sodium chloride. Write the complete molecular, complete ionic and net ionic equations.


molecular (just reactants):

Co(NO3)2(aq) + NaCl(aq) --->

This is a double replacement reaction, so we write this for the full molecular:

Co(NO3)2(aq) + 2NaCl(aq) ---> 2NaNO3(aq) + CoCl2(aq)

complete ionic:

Co2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2Cl-(aq) ---> 2Na+(aq) + 2NO3-(aq) + Co2+(aq) + 2Cl-(aq)

Note that both products are soluble and both ionize.

net ionic: After elimination of all spectator ions, we are left with nothing. So:

N.R. (no reaction)

Problem #24: Write the molecular and net ionic equations for: TlNO3(aq) + KI(aq) --->

Comment: thallium compounds are not commonly asked in these types of questions nor are thallium compounds commonly included in a solubility table. That being said, thallium is a heavy metal (that's a hint about the solubility).



TlNO3(aq) + KI(aq) ---> TlI(s) + KNO3(aq)

net ionic:

Tl+(aq) + I-(aq) ---> TlI(s)

Comment: how do you know that TlI precipitates if it is not commonly included on solubility charts? The answer is that, in general, heavy metal iodides are insoluble (AgI, PbI2 and HgI2 are examples). This is one of the things that one learns as one studies the issues of what is soluble, what is not and what exceptions to the rules exist.

Problem #25: Ammonium chloride and sodium dihydrogen phosphate, NaH2PO4, are mixed in water. Write a balanced net ionic equation for the acid-base reaction that could, in principle, occur.


Let us write a partial molecular first:

NH4Cl(aq) + NaH2PO4(aq) --->

If you treat the above as a double replacement reaction, you can see that the sodium ion and the chloride ion are the spectator ions. Write a partial net ionic equation:

NH4+(aq) + H2PO4-(aq) --->

The key now is to recognize that the ammonium ion can only be an acid, it has no capacity to accept a proton (which is what a base would do). That forces the dihydrogen phosphate into the base role, that it, to accept a proton. By the way, it helps that the question text tips off that this reaction should be treated as an acid-base reaction.

net ionic:

NH4+(aq) + H2PO4-(aq) ---> NH3(g) + H3PO4(aq)

Another possibility is this:

NH4+(aq) + H2PO4-(aq) ---> NH4H2PO4(s)

It turns out that ammonium dihydrogen phosphate is quite soluble, but, evidently, it does precipitate out when the solution is very acidic. However, a different reaction is used rather than the one immediately above. See here:

ammonium dihydrogen phosphate

This question was asked and answered on Yahoo Answers.

First Bonus Problem: Write the molecular, complete ionic and net ionic equation for the reaction between sodium hydrogen sulfite and hydrobromic acid.



NaHSO3(aq) + HBr(aq) ---> NaBr(aq) + H2O(l) + SO2(g)

complete ionic equation:

Na+(aq) + HSO3-(aq) + H+(aq) + Br-(aq) ---> Na+(aq) + Br-(aq) + H2O(l) + SO2(g)

net ionic equation:

HSO3-(aq) + H+(aq) ---> H2O(l) + SO2(g)

Do NOT write H2SO3(aq).

From the Wikipedia page:

"There is no evidence that sulfurous acid exists in solution, but the molecule has been detected in the gas phase."

Second Bonus Problem: Write the complete ionic and net ionic equations for the following molecular equation:

2KAl(OH)4(aq) + H2SO4(aq) ---> 2Al(OH)3(s) + K2SO4(aq) + 2H2O(ℓ)


1) Here is the complete ionic equation:

2K+(aq) + 2Al3+(aq) + 8OH-(aq) + 2H+(aq) + SO42-(aq) ---> 2Al(OH)3(s) + 2K+(aq) + SO42-(aq) + 2H2O(ℓ)

Note that the sulfuric acid is treated as fully dissociated. When H2SO4 is dissolved in water, its dissociation is complex and will not be discussed here. However, in this example, the sulfuric acid will react completely, so we treat it as fully dissociated.

2) Here is the net ionic equation (after removal of all spectator ions):

2Al3+(aq) + 8OH-(aq) + 2H+(aq) ---> 2Al(OH)3(s) + 2H2O(ℓ)

3) This is the wrong answer! Why?

Notice that there are hydrogen ions and hydroxide ions on the left-hand side of the arrow. They will react, as follows:

2Al3+(aq) + 6OH-(aq) + 2H2O(ℓ) ---> 2Al(OH)3(s) + 2H2O(ℓ)

Two hydrogen ions and two hydroxide ions reacted to form two water molecules.

4) A second round of removing spectators gives the final answer:

2Al3+(aq) + 6OH-(aq) ---> 2Al(OH)3(s)

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