2.5 Dissolution and the Solubility Product Constant Part II

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Introduction to Chemistry: Structures and Solutions

79 classificações

Duke University

Introduction to Chemistry: Structures and Solutions

79 classificações

This is an introductory course for students with limited background in chemistry; basic concepts such as atomic and molecular structure, solutions, phases of matter, and quantitative problem solving will be emphasized with the goal of preparing students for further study in chemistry.

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Solutions

<p>Welcome to week 6! We have reached the last week of video lectures and exercises for the course! Are you ready to make the final push through the rest of the material? After that, you should be ready to tackle the final exam which opens next week. I have enjoyed working on this course; and I hope you have too!</p> <p>This week, we will first release a couple of review videos related to solution and solubility from <i>Introduction to Chemistry: Reactions and Ratios</i>. Please review these important concepts before starting Lesson Two - Solutions. Then we will start with a review of dissolution and covers some concentration units. This is accompanied by solution-related demonstrations and another bonus interview with a chemist who also works at Duke. The interviews were conceived and filmed by Abdul Latif, an undergraduate in the humanities who is interested in the history of chemistry and how people choose their professions. Enjoy!</p>

2.1 Introduction to Solutions Part I 17:41

2.2 Introduction to Solutions Part II 11:45

2.3 [Demo] Electrolytes10:30

2.4 Dissolution and the Solubility Product Constant Part I 15:20

2.5 Dissolution and the Solubility Product Constant Part II11:58

2.6 Solubility Rules 13:20

Conheça os instrutores

Prof. Dorian A. Canelas
Assistant Professor of the Practice
Chemistry

Let's end this presentation by doing some review exercises.

Here's a couple of problems to consider.

First, write the dissolution equation for ammonium phosphate dissolved in water.

And then once you've done that, answer the second question.

If 0.075 moles of ammonium phosphate is dissolved in

water to make 500 milliliters of solution.

Then what is the molar concentration of ammonium phosphate, and

what is the molar concentration of ammonium ions?

Go ahead and pull out a piece of paper and try this one on your own now.

The first thing you have to be able to do to answer this question is

write down the chemical formula of ammonium phosphate.

So what is the chemical formula of ammonium phosphate?

Correct, so

ammonium phosphate is the ammonium cation dissolved with the phosphate anion.

Now, ammonium the charges NH4+ and

Phosphate the charge is three minus so to balance the charges,

I need to have three ammonium cation for every one phosphate anion.

I dissolve that in water, I'll use an equilibrium arrow and

I'll say I'm dissolving it in water, that results in three, this is a solid,

by the way, I should write, that does, results in three aqueous ammonium cations.

And one phosphate aqueous anion.

So that's the correct dissolution equation.

Now that we've done that you need to answer the next question,

what is the molar concentration of ammonium ions in solution?

This is number 2b.

Please go ahead and work that answer out now if you haven't all ready.

Let's finish this presentation by

solving the question that were asked on a previous slide.

If 0.075 moles of ammonium phosphate solid is dissolved in water,

then what is the molar concentration of ammonium phosphate and

what is the molar concentration of ammonia ions?

The first thing we need to do to solve this problem is figure out what

the molar concentration of ammonium phosphate is.

That's simple enough.

We have 0.075 moles of the solid and we're dissolving that in 500 milliliters.

We can do a little bit of quick calculation.

We can say 500 milliliters.

We know that there are 1000 milliliters in one liter, right?

So I multiply those numbers together and

see that the milliliters cancel and I get 0.500 liters.

Most of you can probably do that in your head.

That's fine.

So I'm going to take my 0.075 moles and divide it by 0.5 liters and

I see that I have an initial concentration this is the answer to part a of 0.15

molar ammonium phosphate The second part, part b is a little bit more complicated.

To do part b I'm actually going to re-write the equation.

So this is for part b.

I'm going to rewrite the chemical reaction.

[NOISE] So I've got ammonium phosphate, that's a solid.

It's dissolving in water,

usually solvent is actually put under the arrow, but you'll see people put

it above too I'm not particularly strict about that with myself.

Okay, so there's a chemical reaction repeated here's my

aqueous species ammonium and phosphate.

I'm going to solve this problem using something called a reaction table.

A reaction table has three components.

It has the initial amount of everything.

The amount the things changed by and then it has the amount of equilibrium.

Sometimes people call this the end but equilibrium isn't really an end.

It's just something we eventually get to where the amounts are no

longer changing because the rates are the same in both direction.

Sometimes people call this an ice box.

Can you see why people call this an ice box?

I have i, c, and e for initial change in equilibrium.

The initial amount of sodium phosphate before I dissolved it was 0.075 moles.

What was the initial amount of ammonium cation?

Before the ammonium phosphate was dissolved,

how much of the aqueous species were present?

Well, before I dump the solid into the water,

of course I had none of these ions present.

So zero and zero for the initial amounts of each of those.

Now in this particular reaction,

during the dissolution process all of the ammonium phosphate dissolves because for

ammonium phosphate K is much, much greater than one.

We can look that up or we can just look up that it's soluble.

I'll talk more about solubility rules in the next presentation.

So after all the solid has dissolved at equilibrium because this

compound is so soluble.

I should have zero moles of sodium phosphate solid left.

So after the reaction occurs.

After all the salt has dissolved.

I should have zero moles of ammonium phosphate left.

Which means that the change of the ammonium phosphate had to be

minus 0.075 moles.

And if we look at the stoichiometry of course,

one mole of sodium phosphate is going to create three mols of ammonium cations.

If I look at the stoichiometry,

one mole of ammonium phosphate is going to create three moles of ammonium cation.

So here where I have the change for ammonium cation,

it's going to go up by three times 0.075 moles.

What about for phosphate?

Well, I'm going to look at the reaction stoichiometry.

One mole of ammonium phosphate is making one mole of phosphate right?

So if 0.075 moles of ammonium phosphate

dissolves that should produce 0.075 moles of phosphate.

So, in equilibrium how many moles of aluminum and

phosphate are present once all of the solid has dissolved?

Well, I would have three times 0.075

or 0.225 moles of ammonium.

And 0.075 moles of phosphate.

Finally, part b was asking me about the molar concentration of ammonium ions.

I still only have a half a liter of solution, so

I need to take my 0.225 moles of ammonium cations.

And divide that by my half of a liter of solution, and

I see that that gives me 0.45 molar concentration from my ammonium cations.

For the first two examples in this lecture,

we wrote the solubility product constant expression for

very simple ionic compounds: sodium chloride, and silver chloride.

But now we're looking at a more complicated ionic compound,

ammonium phosphate.

What would the solubility product constant look like for this type of ionic compound?

The rules that we used before about it being products over reactants still apply.

But there's a little twist in

the rules that deals with the reaction stoichiometry.

Let's go ahead and write the solubility product constant expression for

the dissolution of ammonium phosphate.

So to write that expression I'll start by writing Ksp.

And I know that it's going to

be the molar concentrations of the ions over the reactant,

in this case it's the ammonium phosphate solid which becomes one.

So, let's start by writing that,

so I have the molar concentration of ammonium cation.

And remember, I am using square brackets here to show molarity and

I need to multiply that by the molar concentration of the phosphate ion.

I'm not quite finished with the numerator though at this point.

I have to add something to the Ksp expression to

accommodate for the three that's the coefficient.

To show that there were three times as many ammonium cations formed as there

were in the previous examples that had the sodium and the silver.

So what I am going to do then,

and this is the way that the convention was developed and it's a human construct,

is that three coefficient becomes an exponent here in the expression.

So if it were to be a two, then this would be ammonium ion concentration squared, but

since it's a three, it's the ammonium ion concentration cubed.

And that's just how people decided to do it.

These KSP values are a human construct.

I'm going to divide all that by the concentration of ammonium phosphate,

but right as I'm writing it I realize oh, that's right, that's a solid.

So that all is going to go to be one.

So the real expression for

Ksp then simplifies to mean the molar concentration of ammonium cation, a cube,

cubed times the molar concentration of the phosphate ion.

And I can look that value up on a table of Ksp values.

One of those tables is linked in the reference information again.

It's pretty simple to write Ksp expressions isn't it.

Here's the answer to this one.

This concludes the introduction to equilibrium constants.

Remember the equilibrium constant tell us the ratio of the products to reactants

following a set of rules that have been made up by scientists in the past.

For ionic compounds dissolving in water, a larger value for

KSP means that there are more ions in solution.

And we can either calculate the solubility product constant by doing an experiment.

Or we can look up the KSP value on a table and

use that value to do some other types of calculations.

In either case the solubility product constant gives us a rough guide to

determining relative solubilities if we're comparing different ionic compounds.

The larger Ksp, the more soluble that compound is.

That means there's more ions in that aqueous solution.

I'll talk more about aqueous solubility rules in the next lecture.

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