In this video, we will learn how to
convert molecular equations into net ionic equations by identifying spectator ions
that don’t participate in the chemical change. Lead(II) nitrate and potassium
iodide are both white powders, and when we mix them with water, we’ll get two
colorless solutions. But if we mix these two solutions
together, the solution will turn bright yellow due to the formation of a yellow
solid. If we take a look at our balanced
chemical equation, we can see that the solid that forms, which we call a
precipitate, is lead(II) iodide. Our other product is potassium
nitrate. But let’s take a moment to think
about what’s going on here at the atomic level.
When we have ionic substances
dissolved in water, which is indicated by the state symbol aq or aqueous, the ionic
substance is broken up into the ions that make it up, so lead(II) nitrate would be
lead two plus ions and nitrate ions in solution. So if we were to take a look inside
our reaction vessel with microscopic eyes immediately after we mixed the two
solutions, we would see lead two plus ions, nitrate ions, potassium ions, and iodide
ions. We wouldn’t actually see any
lead(II) nitrate and potassium iodide left, although there might be a few
undissolved bits here and there. Of course, as soon as we mix our
solutions, the ions in the solution will begin to collide with each other and react
to form our products.
When the lead ions and the iodide
ions collide, they’ll react with each other to form the product lead(II) iodide. But our other product, potassium
nitrate, is still aqueous, meaning that it’s still broken up into potassium ions and
nitrate ions in the final solution. So if we compare our reaction
vessel before and after the reaction, the potassium ions and the nitrate ions didn’t
actually participate in the chemical reaction, where the lead ions and the iodide
ions reacted to form lead(II) iodide. So we call these two ions potassium
and nitrate spectator ions because they didn’t participate in the chemical change of
the reaction. They effectively just watched the
chemical change happen in the reaction vessel, just like spectators in the audience
of a game.
So this way of writing the chemical
equation isn’t perhaps the most helpful to show what’s going on in this
reaction. When we write a chemical equation
like this, it’s called a molecular equation. Molecular equations show all
chemical species intact, using their molecular formulas, even if they’re dissociated
into ions. There’s another way of writing the
chemical equation for a reaction, which is the ionic equation. In the ionic equation, we just
break up all aqueous chemical species into ions.
So let’s write the ionic equation
for this reaction. Our first chemical species,
lead(II) nitrate, would be broken up into lead two plus ions and nitrate ions. We shouldn’t forget the two here,
because there’s two units of nitrate in one unit of Pb(NO3)2. Then there’s potassium iodide,
which would be broken up into potassium ions and iodide ions. Then we have our products. Our first is lead(II) iodide, which
is a solid, so we won’t break it up into ions and we’ll leave it alone. Then we have our final product,
potassium nitrate, which is broken up into potassium ions and nitrate ions.
When we look at our ionic equation,
we can see that we have potassium ions and nitrate ions on both sides of the
chemical equation, which is what we saw earlier. We can also see that the lead ions
and the iodide ions are chemically combining to form lead(II) iodide. Now, if we remove the spectator
ions from our ionic equation, we’ll end up with the net ionic equation. Remember, our spectator ions are
those that appear on both sides of the chemical equation, the nitrate and potassium
ions. So removing those from our
reactants leaves us with lead two plus and 2I−, and removing them from our products
just leaves us with lead(II) iodide. So by removing the spectator ions,
we’re removing the ions that don’t participate in the chemical change. So the net ionic equation only
shows us the chemical change that’s happening in our reaction.
Writing net ionic equations will
always involve removing the spectator ions, but we should always remember to include
chemical species that are solids, liquids, and gases in our net ionic equation
because those will never be broken up into ions.
Now, I mentioned that the net ionic
equation is useful because it’s only showing us the chemical change that’s actually
occurring in the reaction. To see why this is useful, let’s
take a look at another reaction. In this reaction, we have lead(II)
acetate reacting with sodium iodide to form lead(II) iodide and sodium acetate. We can see that there’s some
aqueous chemical species in this reaction, so it would probably be useful to write
the net ionic equation.
Our first step will be to write the
ionic equation. Because our first chemical species
is aqueous, it will break up into the ions it’s made of, lead(II) plus ions and
acetate ions. Next we have sodium iodide, which
is also aqueous, so it will be broken up into sodium and iodide ions. Our first product is lead(II)
iodide. Since it’s a solid, it will stay
intact. We won’t break it up into ions. But our next product is aqueous, so
we’ll break it up into sodium and acetate ions.
Now it’s time to create the net
ionic equation by removing the spectator ions. Looking at our ionic equation, we
can see that we have two sodium ions on both sides of the equation and two acetate
ions on both sides of the equation. So these are our spectator
ions. We can create our net ionic
equation by removing these from the ionic equation. This will leave us with lead two
plus plus 2I− reacting to form lead(II) iodide.
So these two reactions have the
same net ionic equation, which means that the chemical change in both reactions is
the same, which is not something that we might have picked up on just looking at the
molecular equation for both of these reactions. So now that we know how to create
ionic equations and identify spectator ions so that we can create a net ionic
equation, let’s try some practice problems.
What is the ionic form of the
chemical expression CuSO4 aqueous?
We can see that the state symbol in
this chemical expression is aq or aqueous, meaning that this substance is dissolved
in water. When ionic substances are dissolved
in water, they break apart into the positive and negatively charged ions that make
them up. So if we want to create the ionic
form for the expression CuSO4 aqueous, we need to figure out what ions CuSO4 is made
of. Well, Cu is the symbol for
copper. Copper forms positively charged
ions, but it can form ions of multiple different charges. So we’ll need to figure out what
its charge is by looking at the other ion in this expression.
SO4 is the polyatomic ion sulphate,
which has the charge two minus. So that means in order for this
compound to have an overall neutral charge, copper must have a charge of two plus,
meaning that this compound is copper(II) sulfate. Of course, we shouldn’t forget our
state symbols since these ions are still in water. So the ionic form of the chemical
expression CuSO4 aqueous is Cu2+ aqueous plus SO42− aqueous.
So now that we’re confident in
creating the ionic form for chemical expressions, let’s put this skill to the test
and identify some spectator ions in an equation.
A redox reaction is described by
the equation Mg solid plus CuSO4 aqueous reacting to form MgSO4 aqueous plus Cu
solid. Including the state symbol,
identify the spectator ion in this reaction.
The reaction in this question,
where solid magnesium metal is reacting with copper(II) sulfate to form magnesium
sulfate and copper metal, is an example of a redox reaction that involves the
transfer of electrons. But the fact that this reaction is
a redox reaction isn’t important for answering this question, since all we’ve been
tasked with is identifying a spectator ion. A spectator ion is an ion that does
not participate in the chemical change in a reaction.
To identify these spectator ions,
we’ll first have to write out the ionic form of all the chemical species involved in
this reaction. We’ll notice that some of the
chemical species involved in this reaction, copper sulfate and magnesium sulphate,
are aqueous. When we have an ionic substance
that’s aqueous, that means it’s split up into the ions that make it up. So let’s go through our chemical
equation and split up everything into ions.
Our first reactant is magnesium
metal. Since it’s a solid, we don’t have
to do anything with it. Next is copper(II) sulfate, which
would be split up into the copper two plus ion and the sulfate two minus ion. Our first product is magnesium
sulfate, which would be split up into the magnesium two plus ion and the sulfate two
minus ion. And our final product is solid
copper. Again since this is a solid, we
just leave it alone. Now that we’ve written our chemical
equation with everything split up into ions that can be, which is called an ionic
equation, we can identify our spectator ion. Since a spectator ion doesn’t
participate in the chemical change, it will be the same on both the reactant side
and the product side of our reaction.
We can see that the sulfate ion is
on the reactant side and the product side. So this must be our spectator
ion. So the spectator ion is sulphate or
SO42−. And we shouldn’t forget the state
symbol, which is aqueous.
What is the net ionic equation for
the neutralization between ammonium hydroxide and hydrochloric acid? NH4OH aqueous plus HCl aqueous
reacts to form NH4Cl aqueous plus H2O liquid. (A) H+ aqueous plus OH− aqueous
reacts to form H2O liquid. (B) NH4+ aqueous plus H+ aqueous
plus OH− aqueous reacts to form NH4+ aqueous plus H2O liquid. (C) NH4+ aqueous plus OH− aqueous
reacts to form NH3+ aqueous plus H2O liquid. (D) NH4+ aqueous plus Cl− aqueous
reacts to form NH4Cl aqueous. (E) H+ aqueous plus H2O liquid
reacts to form H3O+ aqueous.
The net ionic equation for a
reaction is a way of expressing the chemical equation with all of the spectator ions
removed. Spectator ions are ions that are
involved in a reaction that don’t actually participate in the chemical change. To identify them, we’re first going
to need to break apart all of these aqueous reactants and products into the ions
that make them up, since when an ionic species is aqueous, it’s broken up into the
ions that it’s made of. I’ll remove the answer choices so
that we have room to write everything out.
Our first chemical species,
ammonium hydroxide, is aqueous. It will break apart into NH4 or
ammonium ions and OH− or hydroxide ions when it’s in solution. Our next reactant is hydrochloric
acid. Since it’s aqueous, it will break
apart into hydrogen ions and chloride ions when it’s dissolved in water. Now let’s move on to the product
side of the reaction. Our first product, ammonium
chloride, is also aqueous, so it will be split up into NH4 ions and Cl− ions. Our final product is water. Since water is a liquid, it doesn’t
break up into ions, so we’ll leave it alone. This equation that we’ve ended up
with is called the ionic equation, since it shows all of the chemical species that
are involved in the reaction as broken up into the ions that make them up.
Now that we have our ionic
equation, we can identify the spectator ions in this reaction. Since spectator ions don’t
participate in the chemical change, they’ll appear on both the reactant side and the
product side of the reaction. Looking through our reactants and
products, we can see that ammonium appears on both the reactant side and the product
side, as does the chloride ion. So if we remove these spectator
ions, we’ll end up with a net ionic equation for this reaction. On the reactant side, removing the
spectator ions just leaves us with OH− and H+ ions. And on the product side, all we’re
left with is water. So this is the net ionic equation
for the reaction. OH− aqueous plus H+ aqueous reacts
to form H2O liquid. So now if we bring back our answer
choices, we can see that the net ionic equation that we came up with matches answer
Now let’s summarize everything
we’ve learned with the key points. Spectator ions are ions involved in
a chemical reaction that don’t participate in the chemical change. We can identify our spectator ions
in a molecular equation, which uses molecular formulas for all reactants and
products even if they are actually broken up into ions, by creating an ionic
equation which shows all of the soluble ionic substances as broken up into the ions
that they’re made of. If we remove these spectator ions
from the ionic equation, we’re left with the net ionic equation, which shows the
chemical change that’s actually occurring in the reaction.