This is an introductory course for students with limited background in chemistry; basic concepts involved in chemical reactions, stoichiometry, the periodic table, periodic trends, nomenclature, and chemical problem solving will be emphasized with the goal of preparing students for further study in chemistry as needed for many science, health, and policy professions..
Precipitation & Net Ionic Equations Part I
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En provenance du cours de Duke University
Introduction to Chemistry: Reactions and Ratios
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Solutions and Solubility Rules
<p>In the past five weeks of the course, we have learned concepts, rules, and skills, including but not limited to: the definitions of atoms, elements, molecules, and compounds; energy changes; Coulomb’s Law, molecular stoichiometry; atomic structures; periodic trends; the mole; compounds; atomic and molecular masses; molecular stoichiometry; balancing chemical equations.</p><p>In this coming week, we will be discussing aqueous solutions, solubility rules, precipitations and electrolytes! We are also going to learn the first type of chemical reaction equation in this course: Dissolutions.</p>
Rencontrer les enseignants
Prof. Dorian A. CanelasAssistant Professor of the Practice
Chemistry
In this lecture I'll be discussing how to apply solubility rules to
determine if a product will precipitate when two solutions are mixed.
From there I'll cover the process of determining the net ionic equation.
During this lecture we will be using the solubility rules extensively.
These rules are summarized on this page.
They are also found on the reference section of the home page.
Please feel welcome to pull out a copy of the solubility rules to
use them whenever you're doing these types of problems.
To begin with, let's review the species present in ionic solutions.
Here I have two aqueous solutions.
One made with sodium chloride and the other made with silver nitrate.
The cartoon depicts a large number of cations and
anions that dissociated in these solutions.
Since both of these solids dissolve extensively in
water the solubility product constant values for
these solutions must be much, much greater than 1.
Meaning there are a large number of ions dissolved.
If we mix these two solutions what would happen?
Recall that we can write the dissolution equations for both of these solutions.
In the first dissolution we started with solid sodium chloride.
And when we dissolve that in water, that gave us the dissociated sodium and
chloride ions.
For the silver nitrate solid, when we dissolved that in water,
the silver 1 cation and the nitrate anions disassociated.
The main point here is to recall that if a compound is
soluble the ions float apart and they are separate entities.
Which means that the sodium and
the chloride are no longer associated with each other.
They're dissociated.
And the silver and the nitrate are no longer associated with each other.
So if we mix these two solutions there would
be four separate species all floating around in the water.
Let's look at these four species and
determine if any of them will react to form a precipitate.
For example, in order for
something to react it must collide with another species.
If a sodium collided with a chloride would that cause a precipitate?
Of course it wouldn't.
Sodium and chloride are soluble in water.
What if a silver cation collided with a nitrate anion,
would that cause a precipitate?
Once again we know that silver nitrate is soluble in water so
no precipitate would result.
How about if a sodium cation bumped into a silver cation?
Would that cause any kind of reaction?
Of course it wouldn't.
The same is true if the two anions collided.
It would not cause any kind of reaction.
In fact, the anions repel each other don't they?
So what we really need to think about is the cation from one
of the original solutions reacting with the anion from one of the other solutions.
How about if a sodium bumped into a nitrate?
Would that cause a precipitate?
Well, we can either look at our solubility rules to determine that sodium nitrate is
soluble, or we can look up its Ksp value and find that that is much greater than 1.
Which means that when a sodium bumps into a nitrate in an aqueous solution,
no reaction will occur, and they will just float back apart and remain dissociated.
Solvated ions in the solution.
What if a silver cation bumps into a chloride anion?
In this case the Ksp value, if we were to look it up on a table we
would find to be approximately 10 to the minus 10.
In other words, silver chloride is not very soluble in water.
So if the silver cation bumped into a chloride anion a precipitate would result.
That gives us a reaction.
And we can write the net chemical equation for that reaction.
I'll do that up at the top.
The aqueous silver cation is going to collide with the aqueous chloride anion
and that results in solid silver chloride which would then precipitate.
What are the sodiums and the nitrates doing while all of this happens?
Well they're not doing anything actually.
They're just sort of watching the precipitation occur.
And for that reason the sodium and
the nitrate in this particular case are considered to be spectator ions.
The reaction that occurred is at the top.
I'll put it in a box to emphasize.
The reason it's called the net ionic equation is that
we are not including the spectator ions in the equation.
Because the spectator ions would be the same on the left and
the right side of the chemical reaction equation.
We just go ahead and leave those off.
Let's do another example of writing a net ionic equation.
In this case, let's mix a couple of different solutions.
First I'm going to make aqueous solutions of calcium chloride and
nickel(II) sulfate.
The calcium chloride dissolves very well.
And the nickel(II) sulfate dissolves very well.
So remember there are separate ions for these soluble species in solution.
I have four species present if I were to mix these two solutions.
Calcium 2 plus, chloride ion, nickel 2 plus, and sulphate anion.
With these four species in solution would you
predict that a precipitate would form if I mixed these things together?
You might not be able to predict that unless you looked at
your solubility rules.
Another way you could predict that would be to look up the solubility product
constants for the different species that could form.
We already know there's a solubility product constants for
the calcium chloride and the nickel(II) sulfate must be greater than 1,
because we were able to dissolve those to make the solution.
It turns out that the solubility product constant for
nickel combining with chloride is also greater than 1.
When those values are large remember Ksp has this formula.
It's the concentrations of the products, which here are these ions,
over the concentration of reactants.
And in the case of a dissolution reactant,
the reactants are always a solid and so that becomes 1.
What we see if we look at these solubility product values is
that the Ksp value for calcium sulfate is a very small number.
So if a calcium cation bumps into a sulfate anion in solution,
a precipitate results.
Let's write the net ionic equation that would result from that precipitation.
So again here is the formula for the solubility product constant for
calcium sulfate.
The reaction would be a calcium cation reacting with
a sulfate anion to give solid calcium sulfate product.
If we look at this reaction, we see that it looks sort of like the reverse of
a dissolution reaction, doesn't it?
In a dissolution reaction which, we talked about in week three,
the solid was dissolving to give the two ions.
In this case two ions are combining to make a solid.
And that solid falls to the bottom of the container.
We'll look at some demonstrations of that later in the video.
Let's do another example.
First, let's look at a container of sodium phosphate.
Now to do this problem,
you need to be able to write the chemical equation of sodium phosphate,
which means you need to be able to remember that phosphate is PO43 minus.
That means it must have three sodiums with it.
So there's the formula of sodium phosphate.
If we dissolve sodium phosphate in water to make an aqueous solution the products
from that dissolution are 3 sodium cations and 1 phosphate anion.
In both cases those species are now aqueous.
What's the solubility product constant for this dissolution reaction?
Well, equilibrium constants are always the concentrations of
the products divided by the concentrations of the reactants.
So for this specific reaction, the products are the sodium cation, and
the phosphate anion.
But remember we have 3 cations, so this 3 becomes
an exponent in the equation, here it is right here.
The reactant here is sodium phosphate, which is a solid.
You could think of water being a reactant too, that's a liquid.
Both of those things go to 1 because we assume that they are in excess.
So then the Ksp expression simplifies to the solubility product
constant equals the concentration of the sodium cation cubed
times the molar concentration of the phosphate anion.
Since sodium phosphate was very soluble in water,
the value for the Ksp must be much, much greater than 1.
That means that there are lots of ions in this particular aqueous solution.
So that's our first container.
We're going to mix that with another solution.
The other solution, let's prepare in container B.
In this container let's make a solution of silver nitrate.
Silver nitrate has a formula AgNO3, and if we dissolve that in water
it dissociates to make 1 silver cation and 1 nitrate anion.
We can write the expression for the solubility product constant.
Once again, it's reactants over products, which is the molar concentration of
silver cation times the molar concentration of nitrate
anion divided by the reactant here, which was silver nitrate solid.
Oh, that's right, solids become 1 in the Ksp expressions.
So I can re write the Ksp expression as the molar concentration of
silver cation times the molar concentration of nitrate anion.
[SOUND] We've done a lot of writing solubility product constant expressions.
I think we're getting the hang of it.
The next thing we're going to do, once we've decided that's much,
much greater than 1, which it must be if the silver nitrate dissolves well,
which it did, is we're going to mix these two containers.
So just to remind you, here were the two containers of solutions that we prepared.
I made a solution of sodium phosphate, hypothetically.
And I also made a solution of silver nitrate.
Both solutions are aqueous, in water.
If I mix those two containers together I then have
four different species present in solution.
I have more of the sodium cation than I do of the other ions, don't I?
Because I have 3 moles of sodium for every 1 mole of phosphate, silver, and
nitrate, assuming I dissolve all of the same amount of each solid.
But the question we really need to ask ourselves is,
well we know the sodium phosphate is soluble, how about sodium nitrate?
And if that's soluble, what about silver phosphate?
So the question is does sodium nitrate or silver phosphate precipitate?
Which one of those do you think precipitates?
Correct, it's the silver phosphate that precipitates.
Let's write the net ionic equation that occurs when the silver
phosphate precipitates.
Well the silver cation is going to combine with the phosphate anion.
Both of these species are aqueous of course, so
let's write aq for each of those.
And the product is going to be silver phosphate.
Because phosphate has a 3 minus charge, I need to have 3 silvers in my solid.
So, in order to balance this net ionic equation,
I need to put a 3 over by the silver.
What about the sodium and the nitrate?
What do we call those again?
Those are the spectator ions.
[SOUND]
Okay.
So hopefully you're feeling comfortable with net ionic equations at this point.
I'm labeling the spectator ions.
There we go.
They're all labeled.
Let's test our prediction with a real demonstration of the chemistry.
What I have here is an aqueous solution of sodium phosphate.
As you can see, it is clear, colorless, and homogeneous.
To the naked eye it just looks like water, because we can't see the dissolved ions.
I'm putting some of that solution into the test tube using a pipette.
Next, I'm going to take a sample of the aqueous solution of silver nitrate.
The silver nitrate solution is also clear, colorless and homogeneous.
It is stored in an amber bottle because it is photo sensitive.
Silver nitrate finds use in photographic film.
Before I mix them I'm showing you that both solutions look fairly identical.
When I mix them together, voila, a precipitate forms.
We have just identified that precipitate as silver phosphate.
I hope you've enjoyed this demonstration.
I'm going to leave you with a Steven Wright quote, which I hope you will enjoy.
Good luck on the quizzes this week.
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