Complete Molecular, Complete Ionic and Net Ionic: Twenty-Five Problems

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Problem #26: Complete the reaction & write the net ionic equation:

Iron(III) chloride + hydrochloric acid --->


If you tried to do double replacement, you'd get the same products as reactants and so you would write this reaction as NR (no reaction).

There is a trick to this one. It turns out there is an iron(III) chloride complex, formula = FeCl4-

The question wants this for an answer:

FeCl3 + HCl ---> FeCl4- + H+

This problem is illustrative of the main problem students face in doing net ionic problems: you have to know a large amount of seemingly random bits of information (like the fact that iron(III) chloride forms a complex).

Problem #27: 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)

On the left-hand side, 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)

Problem #28: Write the net ionic equation for the following reaction:

BaCl2(aq) + Mg(NO3)2(aq) ---> MgCl2(aq) + Ba(NO3)2(aq)


This is an example of no reaction (commonly signified as NR). Notice that all four substances are ionic and soluble. That means that each substance ionizes 100% to give this total ionic equation:

Ba2+(aq) + 2Cl-(aq) + Mg2+(aq) + 2NO3-(aq) ---> Mg2+(aq) + 2Cl-(aq) + Ba2+(aq) + 2NO3-(aq)

Everything is identical on each side of the arrow, so everything is eliminated for being a spectator ion. Since nothing is left, we call it NR.

Comment: when the question writer intends NR to be the answer, you will often see the reaction asked with products not present:

BaCl2(aq) + Mg(NO3)2(aq) --->

What you have to be able to do is (1) predict the correct products and (2) identify them as soluble, ionic substances (that will then dissociate 100% in solution).

Problem #29: Write the net ionic equation for the following reaction:

3CuCl + (NH4)3PO4 ---> 3NH4Cl + Cu3PO4

Please include state symbols in the answer.


Cu3PO4 is insoluble. Everything else is soluble. Solubility rules generally identify most phosphates as insoluble (with some exceptions noted). Copper(I) phosphate is not one of those exceptions.

Net ionic:

3Cu+(aq) + PO43-(aq) ---> Cu3PO4(s)

Problem #30: Write the net ionic equation for the following reaction:

CH3COOH(aq) + Ca(OH)2(aq) --->


Acetic acid is a weak acid and, as such, is written unionized in the net ionic equation. Calcium acetate precipitates.

2CH3COOH(aq) + Ca2+(aq) + 2OH-(aq) ---> Ca(CH3COO)2(s) + 2H2O(ℓ)

Note that calcium hydroxide is shown fully ionized in solution. Ca(OH)2 is only slightly soluble, but what does dissolve, ionizes 100%.

Problem #31: Write the net ionic equation for:

AsCl3 + 3H2O(ℓ) ---> 3HCl(aq) + As(OH)3(aq)


1) Comments on each compound:

AsCl3 is a molecular compound. Hence, it is written in molecular form.

H2O is a molecular compound. Hence, it is written in molecular form.

HCl is a strong acid which completely dissociates in water. Hence, it is written in ionic form, i.e. H+ and Cl¯.

As(OH)₃ in a weak acid with pKa = 9.2, and most of As(OH)3 in aqueous solution exists as molecules. Hence, it is written in molecular form.

2) Therefore, the net ionic equation is :

AsCl3(ℓ) + 3H2O(ℓ) ---> 3H+(aq) + 3Cl¯(aq) + As(OH)3(aq)

3) The difficulty is that you might think that's not the correct answer. That's because you might think the formula As(OH)3 is a hydroxide and, thus, think that the hydrogen ion reacts with the arsenic hydroxide (which is the wrong name for the compound) to give this wrong answer:

AsCl3(ℓ) + 3H2O(ℓ) ---> As3+(aq) + 3Cl¯(aq) + 3H2O(ℓ)

and conclude that no reaction took place, that the AsCl3 simply dissolved in water and ionized. However, As(OH)3 is actually arsenious acid. There is no arsenic(III) hydroxide. Another acid formula you need to be aware of is the one for boric acid, B(OH)3.

Problem #32: Write the net ionic equation for the following reaction:

Ag2CO3(s) + 2HNO3(aq) ---> 2AgNO3(aq) + H2O(ℓ) + CO2(g)


Ag2CO3(s) + 2H+(aq) ---> 2Ag+(aq) + H2O(ℓ) + CO2(g)

Problem #33: Complete the reaction & write the net ionic equation:

HClO4(aq) + Mg(OH)2(s) --->


Note the presence of solid magnesium hydroxide. Since the solid state is considered to NOT be dissociated, it is written as the full formula. Perchloric acid is a strong acid; it ionizes 100% in solution.

The complete ionic equation is this:

2H+(aq) + 2ClO4¯(aq) + Mg(OH)2(s) ---> Mg2+(aq) + 2ClO4¯(aq) + 2H2O(ℓ)

and the net ionic equation is this:

2H+(aq) + Mg(OH)2(s) ---> Mg2+(aq) + 2H2O(ℓ)

Problem #34: Write the net ionic equation for this reaction:

Ca(OH)2(s) + 2CH3COOH(aq) ---> Ca(CH3COO)2(aq) + 2H2O(ℓ)


The net ionic would not eliminate anything, however there would be one change from the molecular equation above:

Ca(OH)2(s) + 2CH3COOH(aq) ---> Ca2+(aq) + 2CH3COO¯(aq) + 2H2O(ℓ)

The one change is because calcium acetate is a strong electrolyte and, as such, should always be written as ions when in solution.

Note the acetic acid, a weak electrolyte, is only ionized in solution to a small extent and, consequently, is written in the molecular way and not as ions.

Problem #35: Write the net ionic equation for this reaction:

Cr(NO3)3 9H2O + 3NH3 ---> Cr(OH)3 + 3NH4NO3 + 9H2O


The lack of state symbols is deliberate. Let us suppose this reaction takes place in aqueous solution. With state symbols, we have this:

Cr(NO3)3 9H2O(aq) + 3NH3(aq) ---> Cr(OH)3(s) + 3NH4NO3(aq) + 9H2O(ℓ)

Cr(NO3)3 9H2O is one entire formula, so the "aq" occurs at the end of the formula.

Here is the complete ionic equation:

Cr3+(aq) + 3NO3¯(aq) + 9H2O(ℓ) + 3NH3(aq) ---> Cr(OH)3(s) + 3NH4+(aq) + 3NO3¯(aq) + 9H2O(ℓ)

Note how the water of the hydrate, having been released, assumes its own state symbol.

Let's remove spectator ions:

Cr3+(aq) + 3NH3(aq) ---> Cr(OH)3(s) + 3NH4+(aq)

However, the above equation is not correct, since it is not balanced. So, we balance it:

Cr3+(aq) + 3NH3(aq) + 3H2O(ℓ) ---> Cr(OH)3(s) + 3NH4+(aq)

The three waters added back in balance the change from ammonia to ammonium as well as the three hydroxides on the chromium(III) hydroxide.

Now, suppose the chromium(III) nitrate is reacting as a solid, giving this equation with state symbols:

Cr(NO3)3 9H2O(s) + 3NH3(aq) ---> Cr(OH)3(s) + 3NH4NO3(aq) + 9H2O(ℓ)

In that case, nothing can be eliminated and the ammonium nitrate would be written in the ionized state. Doing that is left to the reader.

Problem #36: Hydrogen sulfide gas reacts with iron(III) bromide. The reaction produces iron(III) sulfide and aqueous hydrogen bromide. What are the molecular and net ionic equations?


1) molecular:

3H2S(g) + 2FeBr3(aq) ---> Fe2S3(s) + 6HBr(aq)

2) net ionic:

3H2S(g) + 2Fe3+(aq) ---> Fe2S3(s) + 6H+(aq)

Problem #37: Solid sodium hydroxide reacts with an aqueous solution of hydrogen chloride to form water and an aqueous solution of sodium chloride.


molecular: NaOH(s) + HCl(aq) ---> NaCl(aq) + H2O(ℓ)

net ionic: NaOH(s) + H+(aq) ---> Na+(aq) + H2O(ℓ)

Problem #38: What is the net ionic equation for dissolving gaseous HCl?


Two possible ways to answer:

HCl(g) ---> H+(aq) + Cl-(aq)

HCl(g) + H2O(ℓ) ---> H3O+(aq) + Cl-(aq)

The second method is more reflective of the actual chemical process. The hydrogen in the HCl is transfered (as an ion) to the water, making H3O+, which is called the hydronium ion.

The first equation can be considered as a shorthand for the second way and it is probable your teacher would prefer the second answer. The first answer is often considered to be a shorthand for the second equation.

Problem #39: What is the net ionic equation for dissolving gaseous NH3?


You have to recognize NH3 as a base. It is not an acid. This is the correct net ionic:

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

If you were to treat NH3 like HCl, this would be wrong:

NH3(g) + H2O(ℓ) ---> H3O+(aq) + NH2-(aq)

That sure does look like a plausible chemical reaction! In fact, a question could be worded so as to require the above equation as an answer. For example:

Ammonia does not act as an acid in aqueous solution. However, assume that it does and write the net ionic equation for that reaction.
The reasons why it does not happen are beyond the scope of this lesson.

Problem #40: What is the net ionic equation for dissolving solid glucose?


C6H12O6(s) ---> C6H12O6(aq)

Glucose does not ionize in solution.

This type of question is not commonly asked.

Problem #41: What is the balanced chemical equation for: liquid phosphoric acid reacting with aqueous barium hydroxide to produce a precipitate of barium phosphate and liquid water.


complete: 2H3PO4(ℓ) + 3Ba(OH)2(aq) ---> Ba3(PO4)2(s) + 6H2O(ℓ)

net ionic: 2H3PO4(ℓ) + 3Ba2+(aq) + 6OH-(aq) ---> Ba3(PO4)2(s) + 6H2O(ℓ)

Notice that there are no spectator ions to be eliminated.

If the phosphoric acid were in aqueous solution, this would be the net ionic:

2H3PO4(aq) + 3Ba2+(aq) + 6OH-(aq) ---> Ba3(PO4)2(s) + 6H2O(ℓ)

Since phosphoric acid is a weak acid, it is written in the molecular way when dissolved in aqueous solution.

Problem #43: Write the net ionic equation for this reaction:

NH3(aq) + CH3COOH(aq) ---> NH4CH3COO(aq)


NH3(aq) + CH3COOH(aq) ---> NH4+ + CH3COO¯(aq)

Problem #44: (a) What is the balanced equation of sodium acetate and barium nitrate? (b) What is the net ionic equation?


2CH3COONa(aq) + Ba(NO3)2(aq) ---> Ba(CH3COO)2(aq) + 2NaNO3(aq)

All four substances are soluble and all ionize 100% in solution.

Conclusion: NR (no reaction)

Notice how the question asks you what the net ionic equation is. That's a bit of a trap because you're thinking about what it would be, but the net ionic doesn't exist because the "reaction" is actually NR.

Another NR:

Ba(CH3COO)2(aq) + CaCl2(aq) ---> Ca(CH3COO)2(aq) + BaCl2(aq)

All four substances are soluble in solution and all four substances ionize in solution. Writing a full molecular equation looks like this:

Ba2+(aq) + 2CH3COO¯(aq) + Ca2+(aq) + 2Cl¯(aq) ---> Ca2+(aq) + 2CH3COO¯(aq) + Ba2+(aq) + 2Cl¯(aq) All the ions in the above full ionic equation would be removed, as they are all spectator ions. The net ionic equation would be NR.

And, another NR:

CoCl2(aq) + Na2SO4(aq) ---> ???

It gives the appearance of a double replacement, so you write the reaction:

CoCl2(aq) + Na2SO4(aq) ---> CoSO4(aq??) + 2NaCl(aq)

You know NaCl is soluble. Since you're not sure about cobalt(II) sulfate, you look it up and find it to be soluble.

Nothing precipitates. Everything ionizes 100%.

No reaction.

Here's another NR: Which net ionic equation best represents the reaction that occurs when as aqueous solution of potassium nitrate is mixed with an aqueous solution of sodium bromide?

Notice how it is phrased to give you the impression that there is a net ionic equation when the truth is that this is a "no reaction" reaction.

Here's the "reaction:"

KNO3(aq) + NaBr(aq) ---> NaNO3(aq) + KBr(aq)

I like this: "Which . . . best represents" The correct answer is that the complete absence of a net ionic equation best represents which net ionic equation to use.

And another NR: What are the balanced molecular and net ionic equations for ammonium nitrate + potassium sulfide reacting?

Once again, note how the NR is written as if a reaction is expected to take place, when no reaction actually does take place.

NH4NO3(aq) + K2S(aq) ---> KNO3(aq) + (NH4)2S(aq)

All 4 substances are soluble and all four ionize 100%. That makes for an NR.

Here's another NR: Manganese(II) nitrate + sodium iodide ---> managanese(II) iodide + sodium nitrate

You might guess that MnI2 is insoluble. After all, some iodides (PbI2, AgI) are insoluble. It turns out that (a) MnI2 is soluble and (b) you may have to look it up individually, since many solubility charts do not include Mn. Here's one that does.

Another NR: Predict the products of KI and HCl reacting in aqueous solution.

You write a "double replacement" reaction:
KI(aq) + HCl(aq) ---> KCl(aq) + HI(aq)

Both reactants are soluble and strong electrolytes (they ionize 100% in solution). The products are also both soluble and strong electrolytes.

NR. I wrote "double replacement" because there really is no reaction.

Problem #45a: When a precipitation reaction occurs, the ions that do not form the precipitate:

A) evaporate
B) are cations only
C) form a second insoluble compound in the solution
D) are left dissolved in the solution
E) none of these

The correct answer is D.

Problem #45b: Aqueous potassium chloride will react with which one of the following aqueous solutions to produce a precipitate?

A) Calcium nitrate
B) Sodium bromide
C) Lead (II) nitrate
D) barium nitrate
E) Sodium chloride

The correct answer is C.

Problem #46: Write the net-ionic equation for this reaction:

CuCl2(s) + 2NaNO3(aq) ---> Cu(NO3)2(aq) + 2NaCl(aq)


1) Let's write the full ionic for this:

CuCl2(s) + 2Na+(aq) + 2NO3¯(aq) ---> Cu2+(aq) + 2NO3¯(aq) + 2Na+(aq) + 2Cl¯(aq)

2) Remove the spectator ions (sodium ion and nitrate ion):

CuCl2(s) ---> Cu2+(aq) + 2Cl¯(aq)

And, we're done. That's the answer.

3) But wait, there's more! The above reaction is showing the dissolving of solid copper(II) chloride into water. That is NOT considered a chemical reaction. So, the correct answer to this problem is:


Problem #47: Based on the solubility rules, which of the following will occur when solutions of CuSO4(aq) and MgCl2(aq) are mixed?

(a) MgCl2 will precipitate; Cu2+ and SO42¯ will be spectator ions
(b) CuSO4 will precipitate; Mg2+ and Cl¯ will be spectator ions
(c) MgSO4 will precipitate; Cu2+ and Cl¯ will be spectator ions
(d) CuCl2 will precipitate; Mg2+ and SO42¯ will be spectator ions
(e) No precipitate will form


1) This certainly appears to be a double replacement reaction:

CuSO4(aq) + MgCl2(aq) ---> MgSO4 + CuCl2

I deleted the state symbols from the products.

2) The question now becomes: Are either of the two products insoluble in aqueous solution?

3) The answer is no, neither MgSO4 nor CuCl2 are insoluble. In fact, both are quite soluble and each also ionizes 100% in solution. Oh, and both reactants are soluble and ionize 100% in solution.

4) This is an example of NR, so answer choice e is the correct choice.

5) And another NR for you to examine:

cobalt(II) chloride + copper(II) sulfate --->

It's a double replacement. I will leave you to determine the "products."

6) NR?

NaI + CaCl2 ---> CaI2 + NaCl

The reactants are both soluble and ionize 100% in solution.
The products are both soluble and ionize 100% in solution.

There is no chemical reaction. You just have a solution with sodium ions, iodide ions, calcium ions, and chloride ions. Nothing precipitates, no gas is formed. Yup, it's NR.

7) NR for sure:

Na3PO4 + K2SO4 ---> Na2SO4 + K3PO4

I left it unbalanced. All four substances are soluble and ionize 100% in solution.

Problem #48: Write the net ionic equation for the reaction between Borax and HCl.

Na2B4O7 10H2O(aq) + 2HCl(aq) ---> 4H3BO3(aq) + 2NaCl(aq) + 5H2O(ℓ)


1) I'll work backwards through the equation:

(a) water is a molecular compound. It will not ionize.
(b) sodium chloride is an ionic compound. It will ionize 100%.
(c) boric acid is weak. It will not ionize.
(d) HCl is strong. It will ionize 100%.
(e) Borax is soluble and ionic. It will ionize in solution.

In addition, the water of hydration will be released and become part of the aqueous solvent.

2) Based on the above, here is the complete ionic equation:

2Na+(aq) + B4O72¯(aq) + 10H2O(ℓ) + 2H+(aq) + 2Cl¯(aq) ---> 4H3BO3(aq) + 2Na+(aq) + 2Cl¯(aq) + 5H2O(ℓ)

Note what happened to the water of hydration. It simply became part of the aqueous solution.

3) The net ionic:

B4O72¯(aq) + 5H2O(ℓ) + 2H+(aq) ---> 4H3BO3(aq)

4) Here's one small change in the original equation:

Na2B4O7 10H2O(s) + 2HCl(aq) ---> 4H3BO3(aq) + 2NaCl(aq) + 5H2O(ℓ)

The Borax now has (s) behind it rather than (aq).

5) That changes the net ionic to this:

Na2B4O7 10H2O(s) + 2H+(aq) ---> 4H3BO3(aq) + 2Na+(aq) + 5H2O(ℓ)

Only two chloride ions get eliminated.

Problem #48: A boric acid solution is used in laboratory eye washes to neutralize ammonium hydroxide solutions that may have splashed into a student's or a technician's eyes. Write the non-ionic, total ionic, and net-ionic equations for this reaction.

Solution #1:

1) Ammonium hydroxide does not actually exist. I'll use it anyway. Here's the non-ionic:

H3BO3(aq) + 3NH4OH(aq) ---> (NH4)3BO3(aq) + 3H2O(ℓ)

2) Boric acid is a weak acid. NH4OH is a weak base. The total ionic is this:

H3BO3(aq) + 3NH4OH(aq) ---> 3NH4+(aq) + BO33¯(aq) + 3H2O(ℓ)

3) The above is also the net ionic. This is because there are no spectator ions.

Solution #2:

1) In solution, ammonia exists almost entirely as molecular NH3. Using NH3, here is the non-ionic:

H3BO3(aq) + 3NH3(aq) ---> (NH4)3BO3(aq)

2) The total ionic:

H3BO3(aq) + 3NH3(aq) ---> 3NH4+(aq) + BO33¯(aq)

3) The net-ionic:

H3BO3(aq) + 3NH3(aq) ---> 3NH4+(aq) + BO33¯(aq)

As you can see, it's the same as the total ionic. However, the following solution is the preferred answer because ammonium hydroxide is not a compound that exists.

Problem #50: What is the ionic equation of solid barium carbonate reacting with hydrogen ions from hydrochloric acid?


1) Carbonates react with acids to produce a salt, water, and carbon dioxide. Let's start by writing a complete molecular equation:

Na2CO3(aq) + 2HCl(aq) ---> 2NaCl(aq) + H2O(ℓ) + CO2(g)

2) Write the complete ionic equation:

2Na+(aq) + CO32¯(aq) + 2H+(aq) + 2Cl¯(aq) ---> 2Na+(aq) + 2Cl¯(aq) + H2O(ℓ) + CO2(g)

3) Eliminate spectator ions to get the net ionic:

CO32¯(aq) + 2H+(aq) ---> H2O(ℓ) + CO2(g)

4) Some comments:

There's a bit of a hint in the wording of the problem, telling you that chloride is not involved in the final answer.

However, nothing tells you to eliminate sodium ion until you actually do the problem.

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