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How to Write Ionic Equations is an extensive discussion of the topic. It has a lot of good information, said in a different sequence than I do below. If you go over there, make sure you ignore the late-1990's style and concentrate on the information.
I. Complete Molecular Equations
These equations can also be called 'complete formula equations,' 'total formula equations,' or simply 'formula equations.' There is no standard name. I once saw an answer on Yahoo Answers that referred to the complete molecular equation as the empirical equation. I have also seen it referred to as the non-ionic equation.
You have to know them all because you never know what a particular teacher/textbook might use. Or, perhaps, a future lab partner who learned it one way, while you had learned it a different way.
This type of equation shows the full formula for each substance involved (or the full name of each substance), without reference to a substance being ionic or molecular. Many times, problems of this type will start out with an equation in words. Here are two examples:
barium chloride solution reacts with sodium sulfate solution to make solid barium sulfate and aqueous sodium chloride
aqueous solutions of hydrochloric acid and sodium hydroxide react to produce sodium chloride and water
The wording in problems like the above can vary somewhat. The word 'aqueous' could be used as well as 'precipitate.' Sometimes aqueous is used for the reactants, but assumed for the products. Sometimes aqueous is never used and the question writer simply assumes you know everything is happening in aqueous solution. There is no standard way to arrange a chemical equation using the names of the substances. (However, when using formulas, the examples used tend be done in a similiar style world-wide.)
I started out with names for the complete molecular equations because your first answer in a given problem is often to translate the names into a complete molecular equation equation. This means that you must be able to (1) write correct chemical formulas from the names and (2) balance chemical equations.
Here are the word equations above, repeated using formulas:
BaCl2(aq) + Na2SO4(aq) ---> BaSO4(s) + 2NaCl(aq)
HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O(ℓ)
Only full formulas (never words or ions) are involved in a complete molecular equation. Ions (never words) will be used for the complete ionic equation and the net ionic equation, which will follow just below. You can use words for an ionic equation (either complete or net), it just isn't usually done.
I have chosen to use "ℓ" to indicate the liquid state, as opposed to the more common "l". Keep in mind that "ℓ" and "aq" mean different things.
II. Complete Ionic Equations
The word total can also be used, as in 'total ionic equation' or even simply 'total equation.' There is no standard term for this type of equation.
To transform a complete molecular equation to a complete ionic equation, you need to know the difference between an ionic compound and a molecular compound. The basics are as follows:
Ionic compounds are between metals and nonmetals or between metals and polyatomic ions. Examples are sodium chloride (NaCl), magnesium nitrate [Mg(NO3)2] and ammonium sulfate [(NH4)2SO4]. When an ionic substance is dissolved in aqueous solution, it ALWAYS ionizes and the ions always have a charge. Remember to never split apart a polyatomic ion.
Sometimes a student wonders if a compound between two polyatomics, say NH4NO3, is an ionic substance. Yes, it is. It has an ionic bond between the two polyatomics and thus qualifies as an ionic substance. It ionizes into NH4+ and NO3¯.
Molecular compounds involve only nonmetals. Examples include sulfur trioxide (SO3), carbon dioxide (CO2), water (H2O) and glucose (C6H12O6). Molecular substances never ionize in solution, they exist as complete molecules and they NEVER have a charge. By the way, molecular compounds are also called covalent compounds. All the bonds in molecules like the examples above are covalent. Not one ionic bond exists in a covalent compound.
There next thing you need to know:
ionic substances which are insoluble are always written as the full formula, never as ions
This needs a bit of explanation. First of all, a tiny amount of an insoluble substance does dissolve and ionize. However, unless it is made clear by context, you always ignore the tiny amount that dissolves (and ionizes). Your default choice is to write an insoluble ionic compound as the full formula and not as ions. You might think this is bit strange, but keep in mind that almost all of an insoluble substance never dissolves, so it never has a chance to ionize. An ionic substance is considered to only be broken down into separate ions when it is in aqueous solution.
I just made reference to context. To illustrate, consider this:
Magnesium hydroxide is insoluble but the state symbol tells you to consider it as being dissolved. What you do is ignore the insoluble concept and consider only the tiny amount of magnesium hydroxide that does dissolve. Since it is ionic, it ionizes 100% and you would write this in an ionic equation:
Mg2+(aq) + 2OH¯(aq)
In order to know which substances are soluble and which are insoluble, you must know your solubility chart. There are many of them available across the Internet.
The last item I will mention is this:
weak acids and bases are always written in a molecular way, never in an ionic way
Here is an example:
HF(aq) + NaOH(aq) ---> NaF(aq) + H2O(ℓ)
The above is written as a complete molecular equation. Now, the complete ionic equation:
HF(aq) + Na+(aq) + OH¯(aq) ---> Na+(aq) + F¯(aq) + H2O(ℓ)
Notice that HF, a weak acid, is not shown ionized, as is the NaOH and the NaF. Now, for the net ionic:
HF(aq) + OH¯(aq) ---> F¯(aq) + H2O(ℓ)
Compare this to the net ionic equation that will result from the reaction of HCl (a strong acid) and NaOH, just below.
There are lists of strong acids and strong bases. Here is one. many others are available on the Internet.
Acids and bases not on the list are considered to be weak. All weak substances are written as complete, unionized molecules.
And, of course, there are always exceptions. Mg(OH)2 is insoluble, so the state symbol (s) is used with it. However, if (aq) is used, as I discussed above, we consider it to ionize 100% because it's an ionic substance. Any amount of an ionic substance ionizes 100% when in solution. Since Mg(OH)2 is insoluble, it does not usually appear on a list of strong bases.
Above, in section I, were two complete molecular equations. Just below, I have written then as complete ionic equations:
Ba2+(aq) + 2Cl¯(aq) + 2Na+(aq) + SO42¯(aq) ---> BaSO4(s) + 2Na+(aq) + 2Cl¯(aq)
H+(aq) + Cl¯(aq) + Na+(aq) + OH¯(aq) ---> Na+(aq) + Cl¯(aq) + H2O(ℓ)
Please note that barium sulfate is an insoluble substance and that water is a molecular substance.
Sometimes, the word 'complete' is deleted and the text (or teacher) assumes you know that 'ionic equation' is NOT the same as "net ionic equation.' It is best to assume 'complete ionic equation' is what is meant by 'ionic equation.' In other words, the word 'net' is always included when one means to identify the 'net ionic equation.'
Here is an example of a mistake students sometimes make:
Ba2+(aq) + Cl¯(aq) + Na+(aq) + SO42¯(aq) ---> BaSO4(s) + Na+(aq) + Cl¯(aq)
The mistake is to not show the proper coefficients when writing the complete ionic equation from the molecular equation. For example, when BaCl2 ionizes, it forms one barium ion and two chloride ions in solution, not one of each.
Here is a Yahoo Answers Q&A showing this mistake.
III. The Spectator Ion
The spectator ion is an important idea. Here is the definition:
spectator ions are present on the reactant side and on the product side in exactly the same form
In both of the example equations above, the sodium ion and the chloride ion are the spectator ions.
Here's why identifying the spectator ions is important:
to go from a complete ionic equation to a net ionic equation, all spectator ions are eliminated from the equation
The problem students face is that the ability to identify the spectator ions is bound up in knowing (1) how to write correct formulas, (2) how ionic substances ionize in solution and (3) which substances are soluble and which are insoluble.
That can be a problem because there are many different bits of information to know before being able to answer the types of questions that are the subject of this tutorial. In addition, some of the bits you need to know wind up getting taught after covering this area. Cute!
IV. Net Ionic Equations
Using my two example equations, when we strike out the spectator ions, we are left with the following net ionic equations:
Ba2+(aq) + SO42¯(aq) ---> BaSO4(s)
H+(aq) + OH¯(aq) ---> H2O(ℓ)
Remember, in a net ionic equation, all spectator ions are completely removed. For all intents and purposes, they are not considered part of the reaction.
V. No Reaction
A 'no reaction' situation happens when all of the products are soluble and ionize in solution, just like the reactants.
Take a look at this molecular reaction:
CaCl2 + Mg(NO3)2 ---> Ca(NO3)2 + MgCl2
The reactants are both soluble and ionize in solution, giving this on the left-hand side of the complete ionic equation:
Ca2+(aq) + 2Cl¯(aq) + Mg2+(aq) + 2NO3¯(aq) --->
The products are both soluble and ionize in solution, giving this on the right-hand side of the complete ionic equation:
---> Ca2+(aq) + 2NO3¯(aq) + Mg2+(aq) + 2Cl¯(aq)
Put them together:
Ca2+(aq) + 2Cl¯(aq) + Mg2+(aq) + 2NO3¯(aq) ---> Ca2+(aq) + 2NO3¯(aq) + Mg2+(aq) + 2Cl¯(aq)
Everything on the left-hand side is exactly like everything on the right-hand side. Following the rule that says to eliminate all spectator ions, we get this for the net ionic equation:
In other words, there is no net ionic equation. This is called 'no reaction' and is often indicated like this: NR.
Warning: what teachers like to do is present all the information about how to do net ionic equations and never mention the existence of NR. Then, as you might expect, there is a 'no reaction' on the test.
Here is an NR question on Yahoo Answers. The questioner was confused by the idea that everything canceled out, a behavior that is the hallmark of NR.
Another way to ask an NR question is this:
What are the products when Al2(SO4)3 and NH4Cl react in aqueous solution?
The answer is NR because the products are (1) soluble and (2) ionize 100%. Both these points apply to the reactants. So, the four ions (aluminum, sulfate, ammonium, chloride) all stay in solution and are unchanged. If we wrote the balanced complete ionic equation, we would see this:
2Al3+(aq) + 3SO42¯(aq) + 6NH4+(aq) + 6Cl¯(aq) ---> 2Al3+(aq) + 6Cl¯(aq) + 6NH4+(aq) + 3SO42¯(aq)
Since everything is exactly the same on both sides, it's NR.
Here's a link to another NR question on Yahoo Answers.
You will find additional NR examples at 13, 23, 28, and 44 in my problem sets. There are several examples at #44. Also, an interesting one at #46.
And yet another NR on YA.
And another NR:
Write the balanced net ionic equation for HCl(aq) reacting with H2SO4(aq)
The problem is that these two compounds (both acids) do not react. However, notice how the question is phrased to indicate that the two compounds do, in fact, react. The question-writer is lying to you!!!
One more example (not NR), then some final comments:
Write the complete molecular, complete ionic and net ionic equations for this reaction:
solutions of sodium chloride and silver nitrate react to form a precipitate of silver chloride and aqueous sodium nitrate
complete molecular: NaCl(aq) + AgNO3(aq) ---> AgCl(s) + NaNO3(aq)
You know AgCl is insoluble from using a solubility chart.
complete ionic: Na+(aq) + Cl¯(aq) + Ag+(aq) + NO3¯(aq) ---> AgCl(s) + Na+(aq) + NO3¯(aq)
In this example, the sodium ion and the nitrate ion are the spectator ions. They occur on both sides of the arrow in exactly the same form.
net ionic: Ag+(aq) + Cl¯(aq) ---> AgCl(s)
An old school method to indicate a precipitate is to use a downwards-pointing arrow:
net ionic: Ag+(aq) + Cl¯(aq) ---> AgCl↓
An upwards-pointing arrow is used to indicate a gas is formed. Hydrogen gas, for example, would be H2↑ rather than H2(g)
VI. Last Comments
There are three major (and one minor) areas where net ionic equations are used. A major use is in the area called reduction oxidation (short name = redox). I have an entire section on redox (and it will probably be covered later on in your course.) Some of the equations I will use below are redox equations, but they will not be identified as such. The equations I use will be fairly simple and can be balanced by sight and will not require the techniques taught in the redox section.
The two other major types that you will see in the problems below are these:
(1) double replacement reactions (also called double displacement, also called metathesis). Yes, you should know all three names.
(2) acid base neutralizations.
The double replacement reactions are where you will need a solubility chart. Acid base neutralizations will result in water being formed from hydrogen ion and hydroxide ion.
Finally, although there are a number of single replacement reactions that can be written in net ionic form, I'm calling it a minor area. The main reason for single replacement being minor is that it is used much less as a source for net ionic type problems. A very large majority tend to be the double replacement and the neutralization reactions.
Here are two (unbalanced) single replacement examples:
Al(s) + HBr(aq) ---> AlBr3(aq) + H2↑
Br2(ℓ) + NaI(aq) ---> I2(s) + NaBr(aq)
I'll make them the first two problems in the problems numbered 1 to 10 (see #1-10 link below).
Here are two (unbalanced) double replacement examples:
Pb(NO3)2(aq) + Na2S(aq) --->
ammonium phosphate + calcium chloride --->
The answers to these two are at #21 and #22 (see #11-25 link below). Try them on your own before looking at the answers and see how well you do.
Here's another example: write the molecular equation, the full ionic equation and the net-ionic equation for: A solution of lead(II) nitrate is mixed with a solution of potassium iodide to produce a precipitate of lead(II) iodide and aqueous potassium nitrate. The answer:
molecular equation ---> Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq)
full ionic equation ---> Pb2+(aq) + 2NO3¯(aq) + 2K+(aq) + 2I¯(aq) ---> PbI2(s) + 2K+(aq) + 2NO3¯(aq)
net ionic equation ---> Pb2+(aq) + 2I¯(aq) ---> PbI2(s)
Last, last comment: another type of problem you sometimes see on a test without it being discussed in class is a double precipitation, where both products (as opposed to only one) of a double replacement reaction will precipitate. Here is an example of a double precipitation:
Sr(OH)2(aq) + CuSO4(aq) ---> SrSO4(s) + Cu(OH)2(s)
I will leave the net ionic equation unwritten. I will point out, however, that there are not any spectator ions to be eliminated. There are several more examples of double precipitation reactions at problem #10 here.
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