General Chemistry Solutions & Their Behavior

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is concerning solutions and their behavior.0003

Here is the unit lesson overview.0012

We are first going to do the introduction into the different units of solution concentration,0015

basically what we call molarity, molality, weight percent, and parts per million.0023

After we get into the different concentrations of units of solutions, I then want to go0032

into what you may have learned in high school chemistry which the rule, like dissolves like.0038

We are going to see how solutions can interact with each other.0045

The third one is then solubility.0050

What is the factors that influence how well a solid or a gas dissolves into a solution?0054

We are going to learn that there is really two factors.0063

The first one is pressure; the second one is temperature.0066

After we talk about the factors that affect solubility, we will then go into what we call colligative properties.0071

The colligative properties that we are going to be discussing are more or less four of them.0079

The first one is vapor pressure.0085

The second and third are together, boiling point and freezing point.0087

Last but not least is osmotic pressure.0092

As always we will get into summary slide and sample problems.0096

Let's go ahead and now jump into the units of concentration.0105

Whenever you buy a beverage or something like that from the supermarket,0109

they usually give you maybe like the percentage of fruit juice suppose in your beverage.0114

That percentage is a unit of concentration.0122

It tells you how much of that ingredient is in the entire beverage.0124

Of course for chemistry, we also need a way of defining a solution concentration.0131

The most common solution concentration unit that we see in the chemistry laboratory is what we call molarity.0138

Molarity is formally defined as the following.0147

This is going to be equal to moles of solute for every liter of solution.0150

Again moles of solute for every liter of solution.0165

Molarity is typically symbolized with a capital M.0168

There is a couple of ways we can actually verbalize this too.0174

Let's say we had 0.67M NaCl.0177

If we were to translate this and verbalize it, we would say 0.67 molar NaCl solution.0184

Again this is the language that we use.0197

Anytime you see capital M, it is molar.0200

Again this is the most common unit of concentration we see in the chemistry lab for solutions.0204

The second one is related to molarity; it is just a different variation.0210

This is called molality; molality is going to be equal to the following.0215

This is going to be equal to moles of solute for every not liter of solution but kilogram of solution.0224

Moles of solute for every kilogram of solution.0238

Typically molality is symbolized with lowercase m.0241

0.67M NaCl, we would verbalize this as 0.67 molal NaCl solution.0246

Molarity followed by molality.0263

The next two we don't see too much.0267

But you do see this third one called weight percent.0270

You do see it quite often commercially.0274

For example, that beverage you buy in the supermarket.0279

What weight percent is is the following.0281

This is going to be equal to the mass of solute divided by the mass of solution times 100.0285

Weight percent is commonly abbreviated m/m percent; that is what we commonly see.0301

Again this is what we call weight percent, mass of solute divided by the mass of solution.0310

Finally the fourth and final concentration is ppm; ppm stands for parts per million.0315

Really we reserve the use of ppm for solutions that are very dilute or0327

the concentration of solute is incredibly small; used for very dilute solutions.0333

What parts per million is is basically the following.0347

This is going to be equal to milligrams of solute for every liter of solution0350

which is also equivalent to milligrams of solute divided by kilograms of solution.0360

You can see why this is reserved for very small quantities.0370

You see that the mass that we use for the solute is intentionally fixed to milligrams.0373

These are the four concentration units to express a concentration for a solution.0380

I now want to go into how solutions can interact with each other.0390

The way solutions can interact with each other is an old rule of thumb we have heard before.0395

It is called like dissolves like.0402

This refers to something that we have discussed previously; this refers to polarity.0404

Basically polar solutions will mix and be miscible.0416

Nonpolar solutions will also mix and be miscible.0436

However polar and nonpolar do not mix.0446

You can see the importance of when we talked about VSEPR theory and using VSEPR to determine molecular polarity.0459

You can see why this would come in handy.0466

For example, an alcohol and H₂O; both of these are polar.0470

These two will mix and be miscible with each other.0491

When you go to the supermarket and you purchase rubbing alcohol, rubbing alcohol is typically 70 percent.0496

It will say that on the bottle.0507

What that means is that it is 70 percent alcohol and the rest is just water, 30 percent.0508

Again that is going to be miscible.0513

The next one that we can see is the following.0521

We can take water and fat; water we know is polar already.0526

But the fats and oils are going to be nonpolar.0539

We know that when we combine water and oil, we get a heterogeneous layer, a heterogenous mixture.0550

The two do not mix; immiscible.0555

Again anytime you are predicting if two solutions are going to mix, just use the rule of thumb0562

like dissolves like to determine if it is going to be homogeneous or heterogeneous.0569

Once two solutions mix, the next question is how well do they mix?0579

In other words, what are the factors affecting solubility--how well two components combine together?0586

The first factor is the effect of pressure; this is what we call Henry's law.0608

Henry's law tells us that solubility is going to be directly proportional to the following--something we call Henry's constant and P.0618

Capital P is going to be the vapor pressure of the solute.0637

Let's imagine the following; let's say we had a can of soda.0653

In this can of soda was H₂O and CO₂ gas.0660

Let's say that this was open versus the same can of soda closed.0667

Now the question is in which of these cans is the CO₂ going to be more in the water?0682

The answer of course is going to be where the can is closed.0690

When the can is closed, the vapor pressure of CO₂ is greater than when0693

the can is open because in the can that is open, CO₂ can escape.0701

Basically we see the relationship that solubility of the CO₂ is directly proportional to its pressure.0709

Henry's constant helps make this proportionality into the equation.0720

Henry's constant is going to be unique for each substance.0725

Once again Henry's law tells us that solubility is directly proportional to vapor pressure.0737

That is the first factor.0747

The second factor is going to be the effect of temperature.0749

The effect of temperature is different on a gas versus a solid.0753

Let's go ahead and look at that can of soda again.0759

Let's have both of these cans open.0765

Once again we have H₂O, CO₂, H₂O, and CO₂.0771

Let's make the left can hot; let's make the right can cold.0778

Now the question is the following.0786

In which cans of soda is the CO₂ going to escape more?0788

In other words, is a hot can or cold can of soda going to fizz more?0795

It turns out that the warmer the can, the more fizz you get.0801

But what is this is fizz that you and I hear?0815

This fizz that we hear is CO₂ gas escaping which means that the CO₂ gas is not as soluble in the water.0819

It is less soluble; it is escaping instead.0834

The effect of temperature on gas solubility is an inverse relationship.0838

That is as the temperature increases, the solubility of a gas goes down.0845

That is as the temperature increases, the solubility of a gas goes down.0854

Less of it stays in solution.0859

More of it escapes which explains why a warmer can of soda fizzes more than a colder can.0861

The effect of temperature on solid solubility is now we are going to find is going to be opposite; opposite to gas solubility.0872

This one we know; it is a little more intuitive.0885

Let's say we are brewing a tea bag.0890

Which of the situations are you going to get the tea to brew faster, in cold water or hot water?0894

It is in hot water; you see that happen before your eyes.0899

That is because as temperature goes up, solids dissolve more easily.0902

Just think about it.0914

When are we able to clean better, our dishes that is, in cold water or hot water?0917

It is hot water because all the fats and oils are going to dissolve more easily in the water and in the soap.0921

As temperature goes up, solids dissolve much more easily.0928

The effect of temperature on solid solubility is completely opposite to that of gas solubility.0936

Again the examples that we did was a tea bag brewing faster in warmer water.0946

Another sign from everyday life is the appearance of your tap water.0962

Doesn't tap water appear cloudy when it is warmer?0967

The answer is yes because there is more dissolved ions and minerals coming out.0971

Warm tap water is not as clear.0977

It is cloudier due to more due to more dissolved ions and minerals that are present and abundant in tap water.0984

Again these are the factors that affect solubility; it is pressure and temperature.1002

We now move on to what are known as colligative properties.1011

Colligative properties of a solution are defined as the following.1016

These are properties that depend on the relative quantity of solute particles and not on the chemical identity per say.1020

We are interested in solute amount; again that is going to be relative to solution.1029

There is a law that helps to quantify the relationship; this is called Raoult's law.1045

Raoult's law is P is equal to x times P⁰ where P is the vapor pressure of the solute in solution.1052

P⁰ is the vapor pressure of the pure solute just by itself.1074

x is going to be the mole fraction of solute.1088

As you can see, as x increases, that is as solute amount goes up, so does the vapor pressure.1100

That makes sense; you toss more of a solute in.1123

It is going to have just a higher vapor pressure right above the surface of the liquid.1126

What is x again?--x is going to be the mole fraction; net mole fraction of solute.1136

This is going to be the moles of solute divided the moles of solute plus moles of solvent1145

which is also can be rewritten as moles of solute divided by the moles of total solution.1159

Of course the mole fraction of solute is going to be less than or equal to 1.1171

Once again this is Raoult's law.1179

Basically the vapor pressure of the solute is directly proportional to what the quantity is in solution.1183

The next two colligative properties also depend on amount.1195

This is a boiling point elevation and freezing point depression.1200

Basically as solute is added to a solution, the freezing point of the solution decreases.1206

That is what we call a depression.1227

The boiling point of solution is going to go up.1230

That is what we call elevation.1236

If we were to look at a phase diagram of pressure versus temperature, we are going to get something like this.1239

For example, let's go ahead and do this for water.1250

For water, this is 0 degrees Celsius; this is 100 degrees Celsius.1256

Let me go ahead and do a blue line.1262

The blue line is now going to represent the water this time with salt added.1264

This is with NaCl.1271

What happens is as you can see the freezing point now decreases.1274

The boiling point now increases.1280

This is bp_new which is greater than 100 degrees Celsius.1286

This is freezing point_new; that is going to be less than 0 degrees Celsius.1291

Freezing point depression, boiling point elevation; let's go ahead and explain why.1298

Basically this is telling me the following; that more energy needed for vaporization to occur.1305

As I toss sodium chloride in the water, we are introducing more attractions, more attractive forces.1322

This is really due to more attractive forces between solute and solvent.1330

If all of a sudden the attractive forces shot up, we have to supply more energy in the1343

form of heat to overcome those attractive forces to induce vaporization, hence a higher boiling point.1348

For the freezing point, we actually need a cooler temperature.1362

The colder temperature needed because what happens is the following.1368

When we reach a colder temperature and try to induce freezing, like molecules tend to interact with each other.1377

What I mean by that is the following.1398

That water is going to want to interact with itself.1399

NaCl is going to want to interact with itself.1406

In order for the solute and solvent to separate out like this, we need to1411

remove the thermal energy so that interactions like that will be minimized.1417

A cooler temperature helps to minimize what we call unwanted interactions.1424

Water sticks with water; sodium chloride sticks with sodium chloride.1440

Freezing process can occur where we eventually result in solidification.1444

Cooler temperature minimizes unwanted interactions so that solidification can occur.1451

Again this is boiling point elevation and freezing point depression.1469

We actually have two equations that can help us through this.1475

They are the following.1480

The change in the freezing point is going to be equal to some constant K times lowercase m times i.1483

I am going to call this K_F.1496

The change in boiling point is equal to some constant K_B times m times i.1499

What these are is the following.1507

i is what we call the van't hoff factor.1510

This is going to be proportional to number of ions from solute.1518

m is just the molality.1529

Finally K's are just going to be constants unique to the solute.1535

But what you see is that the temperatures are directly related to the amounts.1542

Temperature is directly related to the amount.1551

The more solute you have in solution, the greater the change.1555

The more solute you have, the lower the freezing point.1560

The more solute you have, the higher the boiling point, the greater the ΔT.1563

Once again these are called boiling point elevation and freezing point depression.1569

The final colligative property, that is the final property that is dependent on relative solute amount is what we call osmotic pressure.1575

Before we get into osmotic pressure, let's first go ahead and define osmosis.1588

Osmosis is the flow of solvent through a semipermeable membrane into a more concentrated solution.1593

In other words, solvent naturally flows from a dilute area to a more concentrated area.1599

The typical diagram to illustrate this is the following.1632

Let's say we had a container; in the container is this barrier.1637

This barrier is what we call a semipermeable membrane; what this is is the following.1642

It allows only certain sized particles through.1652

Once again it allows only certain sized particles through.1663

In this case, it is going to be solvent molecules only because we are going to assume that a solvent1667

like water is going to be relatively smaller to a much bigger and heavier solute compound such as sodium chloride.1673

Let's say initially that these two regions which are separated by a semipermeable membrane have equal levels of liquid.1685

However let's say that one side was just H₂O.1698

On the other side, we had Na⁺, Cl^-, and H₂O.1702

That means this side represents my concentrated side.1709

This side represents my dilute side.1717

If we allow this to proceed, after some time, we are actually going to get a change in water levels because of osmosis.1721

Solvent is going to flow from the dilute area to the more concentrated area.1734

In other words, I am going to get solvent going this way.1739

What that results in is my dilute side is going to drop in volume.1742

My concentrated side is going to increase in volume because of the presence of more solvent molecules.1752

This is what is known as osmosis.1762

The colligative property is what we call π; this is osmotic pressure.1767

Pretty much what osmotic pressure is, it is the pressure needed to be1776

applied to prevent the flow of osmosis; to prevent osmosis from occurring.1785

π is equal to i times M times R times T where i is once again that van't hoff factor that we talked about.1801

Again that is going to be proportional to the number of solute ions.1815

Big M is the molarity.1823

R is going to be our universal gas constant, 0.08206 liters atmosphere K mole.1827

Finally temperature is going to be the kelvin temperature.1836

But as you can see, that π is directly proportional to M and to i.1841

Basically if you have an area of great concentration where you have a1851

lot of solute, osmosis is going to happen very easily and very readily.1857

I am going to need much more pressure to stop that process from occurring.1862

Again this is called osmosis.1868

Let's go ahead and summarize our presentation on solutions and their behavior.1873

Solution concentration we found is often expressed in four ways--molarity, molality, weight percent, and as ppm.1882

Again weight percent is really what we see commercially.1891

Ppm is really for dilute solutions, very very dilute solutions.1897

We saw the rule that like dissolves like.1903

In other words, polar solutions are going to be miscible with other polar solutions.1908

Nonpolar solutions are going to be to miscible with other nonpolar solutions.1913

Simply put, polar and nonpolar do not mix, forming heterogeneous mixtures.1918

What we saw was the traditional portrayal of fat oil plus water giving us a heterogeneous mixture.1923

We also discussed factors that influence the solubility.1932

The solubilities of gases and solids in a solution we found are influenced by pressure and by temperature.1936

Remember that for temperature, it is going to be opposite for solids and gases.1943

It is an opposite effect.1952

Finally we also discussed colligative properties.1956

We saw that colligative properties are not really dependent on the identity of the compound itself but really just the relative solute amount.1958

Now that that is our summary for solution behavior, let's get into a pair of sample problems.1970

Calculate the vapor pressure of water at 20 degrees Celsius in a solution1978

prepared by dissolving ten grams of sucrose in 100 grams of water.1981

You are told that the vapor pressure of pure water at this temperature is 17.5 torr.1986

Let's go ahead and write out the equation.1992

This is pressure is equal to mole fraction times P⁰.1994

This is going to be the vapor pressure of solution.1999

This is the mole fraction of solute.2004

This is the vapor pressure of the pure solvent.2014

The vapor pressure of the pure solvent is given to us to be 17.54 torr.2025

x is the mole fraction... I am sorry... not of the solute.2033

But it is going to be the mole fraction of the solvent.2038

What we see is the following.2042

That as the mole fraction of solvent, as x goes down, the vapor pressure also drops.2043

As we toss more and more sucrose into this water, the water is going to be less and less volatile.2055

Its vapor pressure is going to drop.2062

All we have to do is plug it in, fill in the equation.2065

What we need therefore is the mole fraction of sucrose and the mole fraction of water.2069

x of sucrose is going to be equal to the following.2075

The moles of sucrose divided by the moles of sucrose plus moles of H₂O.2082

Let's get the moles of sucrose.2097

That is going to be 10 grams of sucrose times 1 mole divided by the molar mass of sucrose.2098

Sucrose is given to us right here, C₁₂H₂₂O₁₁.2106

That is just going to be molar mass of sucrose.2113

That is going to be divided by the total moles.2118

That is going to be divided by 10.0 grams over molar mass of sucrose plus the moles of water2121

which is going to be 100.0 grams of water divided by the molar mass of water, 18.016 grams per mole.2131

That is going to give us the mole fraction of sucrose.2141

To get the mole fraction of the water therefore, every time I have a fractional counterpart, the sum has to equal to 1.2146

The mole fraction of water is just 1 minus the mole fraction of sucrose.2156

That is it.2162

All I do, I plug that directly into P is equal to xP⁰ to get the vapor pressure of the water.2163

As you can see, it is going to drop because of the added sucrose.2173

P is less than P⁰.2179

That is going to be sample problem one, vapor pressure lowering.2182

Let's now move on to the final sample problem.2189

What is the molality of C₆H₁₂O₆ in the solution prepared by dissolving 90.5 grams in 250 grams of water?2192

Remember molality is lowercase m.2201

That is equal to the moles of solute divided by kilograms of the solution.2203

This is going to be equal to the solute is the C₆H₁₂O₆.2212

The water is the solvent.2219

The moles of solute, let's go ahead and get that.2223

That is 90.5 grams of C₆H₁₂O₆ times 1 mole divided by its molar mass which is approximately 180 grams.2226

I am going to take that. I am going to divide it by the kilograms of solvent.2238

The total amount of solvent we have is 90.5 plus the 250.0 grams.2242

I want kg; I want kilograms.2251

I am going to take this; I am just going to multiply by 10^-3.2252

When all is said and done, we get our answer in blank molal of C₆H₁₂O₆.2259

That was sample problem two and simply using the molality equation to calculate the concentration.2270

This concludes our lecture and presentation on solution behavior.2278

I will see you next time on Educator.com.2283