Complete Molecular, Complete Ionic and Net Ionic: Ten Problems

Here are two NR that I just decided to put here:

NH₄NO₃ + Pb(ClO₄)₂ ---> Pb(NO₃)₂ + NH₄ClO₄

AlCl₃ + K₂CO₃ ---> Al₂(CO₃)₃ + KCl

Every compound is soluble and every compound ionizes 100%. No precipitates, no gas formation.

No reaction.

Problem #1: Write the following equation as a complete ionic equation and a net ionic equation:

2Al(s) + 6HBr(aq) ---> 2AlBr₃(aq) + 3H₂↑

Solution:

Complete ionic:

2Al(s) + 6H⁺(aq) + 6Br¯(aq) ---> 2Al³⁺(aq) + 6Br¯(aq) + 3H₂↑

You have to know that HBr is a strong acid and, as such, will ionize 100% in solution. You have to recognize AlBr₃ as ionic and H₂ as molecular.

Net ionic:

2Al(s) + 6H⁺(aq) ---> 2Al³⁺(aq) + 3H₂↑

Problem #2: Write the following equation as a complete ionic equation and a net ionic equation:

Br₂(ℓ) + 2NaI(aq) ---> I₂(s) + 2NaBr(aq)

Solution:

Complete ionic:

Br₂(l) + 2Na⁺(aq) + 2I¯(aq) ---> I₂(s) + 2Na⁺(aq) + 2Br¯(aq)

Net ionic:

Br₂(ℓ) + 2I¯(aq) ---> I₂(s) + 2Br¯(aq)

Problem #3: Write the complete molecular, complete ionic and net ionic equations for:

solutions of sodium sulfide and hydrochloric acid react to form sodium chloride and hydrogen sulfide

Solution:

This is a double replacement reaction.

complete molecular:

Na₂S(aq) + 2HCl(aq) ---> 2NaCl(aq) + H₂S(g)

complete ionic:

2Na⁺(aq) + S²¯(aq) + 2H⁺(aq) + 2Cl¯(aq) ---> 2Na⁺(aq) + 2Cl¯(aq)(aq) + H₂S(g)

Recognize sodium sulfide as a soluble, ionic substance. HCl is a strong acid and ionizes 100% in solution. NaCl is the quintessential ionic substance.

net ionic:

2H⁺(aq) + S²¯(aq) ---> H₂S(g)

Yes, you do need to recognize that H₂S is a gas and, as such, is a molecular substance. When a substance is a gas, it is always molecular, never ionic. Be careful, because you could see HCl(g) in an example. It needs to be written as a formula. HCl(aq) must be written as ions. HCl(g) are gas molecules floating freely in space. HCl(aq) is the HCl dissolved in water. It does make a difference!

Problem #4: Write the complete molecular, complete ionic and net ionic equations for:

solutions of lead(II) chloride and potassium sulfate react to form a precipitate of lead(II) sulfate and aqueous potassium chloride

Solution:

This is a double replacement reaction.

complete molecular:

PbCl₂(aq) + K₂SO₄(aq) ---> PbSO₄(s) + 2KCl(aq)

You know that PbSO₄ precipitates from using a solubility chart.

complete ionic:

Pb²⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + SO₄²¯(aq) ---> PbSO₄(s) + 2K⁺(aq) + 2Cl¯(aq)

net ionic:

Pb²⁺(aq) + SO₄²¯(aq) ---> PbSO₄(s)

Problem #5: Write the complete ionic and net ionic equations for:

HNO₃(aq) + KOH(aq) ---> KNO₃(aq) + H₂O(ℓ)

Solution:

This is a neutralization reaction.

complete ionic:

H⁺(aq) + NO₃¯(aq) + K⁺(aq) + OH¯(aq) ---> K⁺(aq) + NO₃¯(aq) + H₂O(ℓ)

net ionic:

H⁺(aq) + OH¯(aq) ---> H₂O(ℓ)

Problem #6: Write the complete ionic and net ionic equations for:

H₂SO₄(aq) + 2NaOH(aq) ---> Na₂SO₄(aq) + 2H₂O(ℓ)

Solution:

This is a neutralization reaction.

complete ionic:

2H⁺(aq) + SO₄²¯(aq) + 2Na⁺(aq) + 2OH¯(aq) ---> 2Na⁺(aq) + SO₄²¯(aq) + 2H₂O(ℓ)

net ionic:

2H⁺(aq) + 2OH¯(aq) ---> 2H₂O(ℓ)

which reduces to:

H⁺(aq) + OH¯(aq) ---> H₂O(ℓ)

Here lies a potential problem when you go to acid base topics: in terms of moles, sulfuric acid and sodium hydroxide react in a one to two molar ratio, NOT a one to one ratio as the net ionic seems to say. In reality, every one molecule of H₂SO₄ that reacts requires 2 formula units of NaOH. Do not be deceived by the net ionic.

More discussion below the next problem.

Problem #7: Write the complete ionic and net ionic equations for:

H₃PO₄(aq) + 3NaOH(aq) ---> Na₃PO₄(aq) + 3H₂O(ℓ)

Solution:

This is a neutralization reaction.

complete ionic:

H₃PO₄(aq) + 3Na⁺(aq) + 3OH¯(aq) ---> 3Na⁺(aq) + PO₄³¯(aq) + 3H₂O(ℓ)

Notice that the phosphoric acid was written in a molecular way. This is because it is a weak acid and, in solution, ionizes only to a small extent (somewhere in the 1% to 5% amount, it varies due to the specific acid and its concentration).

In previous problems, I used HCl and HNO₃. Both of these are stong acids and ionize 100%, so I wrote them as ions.

However, weak acids should be written as molecules, not as ions. Typically, weak acids are one to five percent ionized, meaning that, in solution, they exist mostly as molecules. Even though some ions are present, the most common procedure is to write weak acids as molecules.

The take-a-way is that you need to know which acids are strong and which are weak. By the way, these comments also apply to weak bases. For example, an aqueous solution of ammonia should be written NH₃(aq), not NH₄⁺(aq) + OH^-(aq) or as NH₄OH(aq).

net ionic:

H₃PO₄(aq) + 3OH¯(aq) ---> PO₄³¯(aq) + 3H₂O(ℓ)

Notice that writing it in this fashion preserves the 1 to 3 molar ratio of acid to base.

Problem #8: Write the complete ionic and net ionc equations for:

NH₃(g) + HCl(aq) ---> NH₄Cl(aq)

Solution:

This is a neutralization reaction. Notice that water is not formed, yet we still consider this neutralization. The acid is HCl and the base is NH₃. My thinking on showing this is that you run a non-zero risk of being taught only neutralizations that produce water and then being asked a question like this on a test.

complete ionic:

NH₃(g) + H⁺(aq) + Cl¯(aq) ---> NH₄⁺(aq) + Cl¯(aq)

net ionic:

NH₃(g) + H⁺(aq) ---> NH₄⁺(aq)

If the ammonia was an aqueous solution, the net ionic would be this:

NH₃(aq) + H⁺(aq) ---> NH₄⁺(aq)

The only thing I changed was (g) to (aq).

Another possibility is this:

NH₃(g) + HCl(g) ---> NH₄Cl(s)

The above reaction does not take place in solution. Gaseous NH₃ and HCl react directly to form solid ammonium chloride. The above reaction is not normally used in teaching about net ionic equations, but some teacher out there may just very well be silent in class and then give you the above equation on a test. If that happens, the above reaction should be considered the net ionic equation.

Problem #9: Write balanced, molecular, ionic and net ionic equations for potassium hydroxide reacting with zinc sulfate.

Solution:

This is a double replacement reaction.

The partial equation looks like this:

KOH(aq) + ZnSO₄(aq) --->

balanced molecular:

2KOH(aq) + ZnSO₄(aq) ---> Zn(OH)₂(s) + K₂SO₄(aq)

You have to recognize that the reaction is double replacement, predict the products and that zinc hydroide is insoluble. And, yes, predicting the products is hard because you have to know a large numer of bits of information.

complete ionic:

2K⁺(aq) + 2OH¯(aq) + Zn²⁺(aq) + SO₄²¯(aq) ---> Zn(OH)₂(s) + 2K⁺(aq) + SO₄²¯(aq)

net ionic:

Zn²⁺(aq) + 2OH¯(aq) ---> Zn(OH)₂(s)

Problem #10: Write the balanced molecular, complete ionic and net ionic equation for the double preciptation that occurs when the following solutions are mixed:

barium hydroxide and magnesium sulfate

Solution:

This is a double replacement reaction. Did you pick up on 'double preciptation?' This is an unusual reaction.

balanced molecular:

Ba(OH)₂(aq) + MgSO₄(aq) ---> BaSO₄↓ + Mg(OH)₂↓

complete ionic:

Ba²⁺(aq) + 2OH¯(aq) + Mg²⁺(aq) + SO₄²¯(aq) ---> BaSO₄↓ + Mg(OH)₂↓

net ionic:

Ba²⁺(aq) + 2OH¯(aq) + Mg²⁺(aq) + SO₄²¯(aq) ---> BaSO₄↓ + Mg(OH)₂↓

The complete ionic and the net ionic are the same equation. There are no spectator ions. I decided to use down arrows just because. For a double precipitation reaction, use either (s) or ↓, don't use one of each.

A second double precipitation (only the molecular equation will be given):

3Ba(OH)₂(aq) + Fe₂(SO₄)₃(aq) ---> 3BaSO₄↓ + 2Fe(OH)₃↓

A third double precipitation (only the net ionic equation will be given):

3Sr²⁺(aq) + 6OH¯(aq) + 2Co³⁺(aq) + 3SO₄^2-(aq) ---> 3SrSO₄(s) + 2Co(OH)₃(s)

A fourth double precipitation (only the molecular equation is shown):

Ca(OH)₂(aq) + MgSO₄(aq) -----> CaSO₄(s) + Mg(OH)₂(s)

A fifth double precip: Write the molecular equation, ionic equation and net ionic equation for the reaction of aluminum bromide and silver hydroxide. Look at solubility rules to determine the precipitate.

AlBr₃(aq) + 3AgOH(aq) ---> Al(OH)₃(s) + 3AgBr(s)

Silver hydroxide is actually an insoluble substance, but a tiny bit does dissolve. So, I chose to write it with an (aq). I didn't write the question, so it is possible that the writer wanted the AgOH to be solid:

AlBr₃(aq) + 3AgOH(s) ---> Al(OH)₃(s) + 3AgBr(s)

In other words, you pour an aluminum bromide solution over some solid AgOH. In the context of a double precipitation, however, I don't think that's the case.

Problem #19 (in this file) has a reaction that isn't a double precipitation but it does result in a situation where you cannot eliminate any spectator ions as you go from the complete ionic to the net ionic.

A bit of a variation on the double precipitation theme: If solutions of Ca(OH)₂ and CoSO₄ are mixed, how many precipitation reactions will occur? (It could be zero, one, or two.)

What we need to do is look at the products and see if they are insoluble. CaSO₄ is one product and the other is Co(OH)₂. It turns out one is soluble and one is insoluble, so the answer to the question is one. This reaction turns out to not be a double precipitation.

By the way, if the answer was zero, then it would be an NR.

Bonus problem #1: For the following equation:

K₃PO₄(aq) + CaCl₂(aq) ---> KCl(aq) + Ca₃(PO₄)₂(s)

Write:

1. the coefficients of balancing
2. the solvation equation of the first compound
3. the solvation equation of the second compound
4. the overall ionic equation for the reaction
5. the net ionic equation for the reaction.

Solution:

2K₃PO₄(aq) + 3CaCl₂(aq) ---> 6KCl(aq) + Ca₃(PO₄)₂(s)

K₃PO₄(aq) ---> 3K⁺(aq) + PO₄³¯(aq)

CaCl₂(aq) ---> Ca²⁺(aq) + 2Cl¯(aq)

6K⁺(aq) + 2PO₄³¯(aq) + 3Ca²⁺(aq) + 6Cl¯(aq) ---> 6K⁺(aq) + 6Cl¯(aq) + Ca₃(PO₄)₂(s)

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

Question parts 2 and 3 are not commonly asked.

Bonus problem #2: Write the net-ionic equation for:

NaHCO₃(aq) + Ba(NO₃)₂(aq) ---> ???

Spoiler alert: it's a 'no reaction.'

Solution:

1) Recognize both reactants as soluble and fully ionized in solution:

NaHCO₃ because it is a sodium-containing salt.
Ba(NO₃)₂ because it is a nitrate-containing salt.

2) Identify possible products:

Ba(HCO₃)₂
NaNO₃

3) Consider sodium nitrate:

It is soluble and fully ionized in solution.

Sodium ion and nitrate ion remain unchanged and would be considered spectator ions and would be deleted from the net-ionic.

4) Consider barium hydrogen carbonate:

It does not form a precipitate, it remains in solution as barium ions and hydrogen carbonate ions.

Both barium ion and hydrogen carbonate ion would be stricked from the net-ionic.

5) Noting remains for the net-ionic. This "reaction" is an NR: no reaction.

6) Some more information on barium hydrogen carbonate remaining in solution:

In the Wikipedia article on calcium hydrogen carbonate, we find this statement:

"Very few solid bicarbonates other than those of the alkali metals and ammonium ion are known to exist."

We also find this in the Wiki article:

"The term [ChemTeam says: referring to the formula Ca(HCO₃)₂] does not refer to a known solid compound; it exists only in aqueous solution . . . . "

This source says:

A compound, Ba(HCO₃)₂, which is only stable in solution.