solutions 2009-20101 everyone has problems, but chemists have solutions

Solutions 2009-2010 1

Everyone has Problems,

but Chemists have Solutions


A solution is defined as a homogeneous mixture of two or more substances.

Types of Solutions

Solute Solvent Example

Gas Gas Air

Gas Liquid Carbonated soda

Liquid Liquid 3% Hydrogen peroxide

Solid Liquid Salt water

Solid Solid Brass (Cu/Zn)

What you are dissolving

What you are dissolving into


Solubility is a measure of how much solute will dissolve in a solvent at a specific temperature.

The saying “like dissolves like” is helpful in predicting the solubility of a substance in a given solvent. What this expression means is that two substances with intermolecular forces of similar type and magnitude are likely to be soluble in each other.


Will bromine be more soluble in water or in carbon tetrachloride?

Bromine (Br2) is nonpolar, therefore it will be more soluble in the nonpolar carbon tetrachloride

H2O

CCl4

“Like Dissolves Like”


FYI-When eating hot peppers or hot spicy foods has left you "breathing fire“, drinking water will not relieve you, because the water cannot dissolve the oils which give the spicy taste. A chemist might suggest that the ideal would be to drink a chaser of a non-polar liquid. But most such liquids are toxic, and even gargling with benzene or toluene doesn’t sound very pleasant. A better solution is to eat greasy foods like meats which will dissolve the oils. Alternatively, food like pasta or bread, that can absorb the oils will help "put out the fire." The smell and taste of onions and garlic are also due to oils that are not easily washed away by water and are not absorbed well.


Two liquids are said to be miscible if they are completely soluble in each other in all proportions.

• Oil and water are immiscible.


Chemists characterize solutions by their capacity to dissolve a solute.

• An unsaturated solution contains less solute than it has the capacity to dissolve.

• A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature

• A supersaturated solution contains more solute than is present in a saturated solution. The supersaturated solution is not very stable. In time, some of the solute will come out of solution and form crystals. The remaining solution will then be _________.

saturated


Concentration UnitsChemists use several different

concentration units, each of which has advantages as well as limitations. The four most common units of concentration: percent by mass, mole fraction, molarity and molality.

The choice of concentration unit is based on the purpose of the measurement.


Mole Fraction (X)

XA = Moles of A

Moles of A + Moles of B + Moles of C …

•mole fraction is appropriate for calculating partial pressures of gases and for dealing with vapor pressures of solutions


XA = Moles of A

Moles of A + Moles of B + Moles of C …

Calculate the mol fraction ethanol (C2H5OH) if 25.0 g ethanol is mixed with 50.0 grams water.

XEthanol = .543

.543 + 2.77

25.0 g C2H5OH x 1 mol C2H5OH = .543 mol C2H5OH

46.07 g C2H5OH

50.0 g H2O x 1 mol H2O = 2.77 mol H2O

18.02 g H2O

= .164


Molarity (M)

Molarity = moles of solute

liters of solution

•the advantage of molarity is that it is generally easier to measure the volume of a solution using precisely calibrated volumetric flasks, than to mass the solvent


Molarity = moles of solute

liters of solution

Calculate the molarity of an aqueous solution containing 34.2 grams of potassium chlorate in 225 mL of solution.

.279 mol KClO3

.225 L

34.2 g KClO3 x 1 mol KClO3 = .279 mol KClO3

122.55 g KClO3

= 1.24 M


Molality (m)

molality = moles of solute

mass of solvent in kg

•molality is independent of temperature (the volume of a solution typically increases with increasing temperature, so that a solution that is 1.0 M at 25 oC may become 0.97 M at 45 oC because of the increase in volume)


What is the molality of a solution containing 14.5 g of ferric nitrate in 285 mL of water?

= .210 m .0599 mol Fe(NO3)3

.285 kg water

14.5 g Fe(NO3)3 x 1 mol Fe(NO3)3 = .0599 mol Fe(NO3)3

241.88 g Fe(NO3)3

Dwater

1 g = 1 mL = 1 cm3


Percent by Mass (also called percent by weight or weight percent)

Mass of soluteMass of solutionX

100%•like molality, percent by mass is independent of temperature

•we do not need to know the molar mass of the solute in order to calculate the percent by mass


What is the mass percent table salt in a solution containing 25.0 grams of table salt dissolved in 100.0 mL of water?

25125

X 100%

= 20.0 %

Dwater

1 g = 1 mL = 1 cm3


The effect of temperature on solubility

• The solubility of gases in water usually decreases with increasing temperature.

• In most, but not all cases, the solubility of a solid substance increases with temperature.


Solubility curves, which are used to illustrate how the solubility of a solute is affected by temperature, are determined experimentally


Because the solubility of solids generally decrease with decreasing temperatures fractional crystallization, the process of separating a mixture of substances into pure components on the basis of their differing solubilities, can be used to purify substances.


This occurs because supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.

Crystallization is the process in which dissolved solute comes out of solution and forms crystals.


A 90.0 g sample of KNO3 is contaminated with a small amount (less than 5.0 g) of NaCl. Explain how you would purify the KNO3.

The goal is not to necessarily get all 90.0 g of KNO3, but to get as much pure KNO3 as possible.


Add water to the crystals (100.0 g at least).

Heat the solution until all of the crystals dissolve (excess of __ oC).

Slowly cool the solution until it reaches 0 oC. At that temperature, about _____ of KNO3 is still soluble and ____ of the NaCl is still soluble.

That means that ____ of KNO3 will have formed a pure crystal free from NaCl.

50

13 g

5 g

77 g


Mixtures of ions can be separated by precipitation. Precipitation is when an insoluble solid forms and separates from a solution.

*Note that both precipitation and crystallization describe the separation of excess solid substance from a supersaturated solution.


The Effect of Pressure on the Solubility of Gases

For all practical purposes, external pressure has no influence on the solubilities of liquids and solids, but it does greatly affect the solubility of gases. The quantitative relationship between gas solubility and pressure is given by Henry’s Law, which states the the solubility of gas in a liquid is proportional to the pressure of the gas over the solution.


c = kP

Here c is the molar concentration (mol/L) of the dissolved gas; P is the pressure (in atm) of the gas over the solution (if several gases are present , P is the partial pressure of the specific gas of interest); and k is a constant (for that specific gas) that depends only on temperature. (For example, at a given temperature, CO2 has the same k value, AS LONG AS THE TEMP STAYS CONSTANT.) The constant k has the units mol/L . atm.


The solubility of pure nitrogen gas at 25 oC and 1 atm is 6.8 x 10-4 mol/L. What is the concentration of nitrogen dissolved in water under atmospheric conditions? The partial pressure of nitrogen gas in the atmosphere is 0.78 atm.

c = kP6.8 x 10-4 mol/L = k . (1 atm)

K = 6.8 x 10-4 mol/L . atm

Step 1: Solve for k (which remains constant for this gas at this temp)

Step 2: Solve for the solubility of nitrogen gas at 0.78 atmc = kP

c = (6.8 x 10-4 mol/L . atm)(0.78 atm)

c = 5.3 x 10-4 M

Now you can use this same k value at a different pressure, as long as the temp stays the same.


Explain how the effervescence of a soft drink when the cap is removed is a demonstration of Henry’s Law.

Before the bottle is sealed, it is pressurized with a mixture of air and CO2 and water vapor. Because of the high partial pressure of CO2, the amount dissolved in the soft drink is many times the amount that would dissolve under normal atmospheric conditions. When the cap is removed, the pressurized gases escape, and the excess CO2 comes out of solution causing the effervescence.


Most gases obey Henry’s law, but there are some important exceptions…

• Dissolved gases that react with water have higher solubilities

NH3 + H2O NH4+ + OH-

CO2 + H2O H2CO3

• Normally, oxygen gas is only sparingly soluble in water, however, its solubility in blood is dramatically greater because hemoglobin can bind up to four oxygen molecules


Colligative Properties

Colligative properties (or collective properties) are properties that depend only on the number of solute particles in solution and NOT on the nature of the solute particles. The colligative properties are vapor-pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.


All solutes that dissolve in water fit into one of two categories: electrolytes and nonelectrolytes.

• An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.

• A nonelectrolyte does not conduct electricity when dissolved in water

Electrolytes and Nonelectrolytes


A solution’s ability to conduct electricity depends on the number of ions it contains.

(b) A weak electrolyte solution contains a small number of ions, and the light bulb is dimly lit.

(a)A nonelectrolyte solution does not contain ions, and the light bulb is not lit.

The molar amounts of the dissolved solutes are equal in all three cases.

(c) A strong electrolyte solution contains a large number of ions, and the light bulb is brightly lit.


Vapor PressureWhen a liquid evaporates, its gaseous

molecules exert a vapor pressure. Vapor pressure is defined as the pressure of the vapor of a substance in contact with its liquid or solid phase. Vapor pressure changes with temperature; the higher the temperature, the higher the vapor pressure.


At the boiling point, bubbles form within a liquid. The pressure inside the bubble is due solely to the vapor pressure of the liquid. The pressure exerted on the bubble is largely atmospheric.

When the vapor pressure (internal) equals the external (atmospheric) pressure, the bubble rises to the surface of the liquid and bursts.


Phase Diagram• A phase diagram shows

the conditions at which a substance exists as a solid, liquid or gas. This is a phase diagram for water.

• At 1 atm (760 mm Hg) the vapor pressure of water (inside the bubble) is equal to the external atmospheric pressure at 100 oC, the point at which water boils. (See your Water Vapor Reference Sheet to obtain water’s vapor pressure at different temperatures.)


If you live at higher altitudes, the atmospheric pressure is lower, therefore, the liquid boils at a lowertemperature.

What would be true about water’s vapor pressure at the lower temp?

higher/lower


Do you need to boil your eggs for less time or more time if you live in Denver, Colorado? Explain.

It takes a certain amount of heat (total kinetic energy) to cook a hard-boiled egg. Because water boils at a lower temperature in Denver, (approx 95 oC) it is supplying less heat to the egg, therefore the egg needs to be boiled for longer.

*Remember, liquid water in Denver will not get hotter than 95 oC if that is its boiling point!!!


Vapor-pressure Lowering If a solute is nonvolatile (therefore the solute

does not have a measurable vapor pressure), the vapor pressure of the solution will always be less than that of the pure solvent. Thus, the relationship between solution vapor pressure and solvent vapor pressure depends on the concentration of the solute in the solution. This relationship is expressed by Raoult’s law

P1 = X1 . P0

1 , where

P1 = partial pressure of the solvent over a solutionP0

1 = vapor pressure of the pure solvent (torr, mm Hg, atm or kPa)

X1 = mole fraction of the solvent in the solution


Calculate the new vapor pressure when 10.0 mL glycerol (C3H8O3), a nonvolatile solute, is added to 500.0 mL water at 50 oC. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.We must first must calculate the mole fraction of the solvent (water).

500.0 mL x .988 g x 1 mol H2O = 27.4 mol H2O

1 mL 18.02 g

10.0 mL x 1.26 g x 1 mol C3H8O3 = 0.137 mol C3H8O3

1 mL 92.09 g

Xwater = 27.4 = .995

27.4 + 0.137 Now we can use Raoult’s law to calculate the new vapor pressure

Psolution = Xwater . P0water Psolution = (.995) (92.5 torr) = 92.0

torr

Notice, the vapor pressure has been lowered, just as we expected!


In a solution containing only one solute, X1 = 1 – X2, where X1 is the mole fraction of the solvent, and X2 is the mole fraction of the solute. The previous equation can therefore be rewritten substituting 1 – X2 for X1

P1 = (1 – X2) . P01

P1 = P01 – X2P0

1

-P01 + P1 = - X2P0

1

P01 - P1 =

X2P01P = X2P0

1

Using this formula will give you the CHANGE in vapor pressure.


Let’s solve the previous problem again, this time solving for change (decrease) in vapor pressure using this derived formula.

Calculate the vapor pressure lowering when 10.0 mL glycerol (0.137 mol) is added to 500.0 mL (27.4 mol) water at 50 oC. At this temperature, the vapor pressure of pure water is 92.5 torr

We must first must calculate the mole fraction of the solute (glycerol). Xglycerol = 0.137 = .00498

27.4 + 0.137

Now we can use Raoult’s law to calculate how much the vapor pressure was lowered by the addition of the glycerol

P = Xglycerol . P0water

P = (.00498) (92.5 torr) = 0.461 torr

P = X2P01

This shows that the vapor pressure would be lowered by .461 torr

(92.5 torr – 0.461 torr = 92.0 torr), the same answer we obtained two slides ago!


Why is the vapor pressure of a solution less than that of the pure solvent?

One of the driving forces in physical and chemical processes is an increase in disorder – the greater the disorder, the more favorable the process. Vaporization increases the disorder of a system because molecules in a vapor are not as closely packed and therefore have less order than those in a liquid. Because a solution is more disordered that a pure solvent, the difference in disorder between a solution and a vapor is less than that between a pure solvent and a vapor. Thus solvent molecules in a solution have less of a tendency to leave the solution than to leave the pure solvent to become vapor, and the vapor pressure of a solution is less than that of the pure solvent.


Pure solvent

The pure solvent is highly ordered, therefore many of the particles will leave the solvent because it GREATLY increases disorder, this results in high vapor pressure.

Nature tends towards disorder (entropy)

This solution is already somewhat disordered, therefore less of the solvent particles will leave the solvent, this results in lower vapor pressure.

Nonvolatile solute


If both components of a solution are volatile (that means they both have measurable vapor pressure), the vapor pressure of the solution is the sum of the individual partial pressures.

Raoult’s law holds equally well in this case.

PT = XAP0A + XBP0

B

Where PT is equal to the total pressure (PA + PB)


Fractional Distillation

Solution vapor pressure has a direct bearing on fractional distillation, a procedure for separating liquid components of a mixture based on their different boiling points.


Fractional distillation is somewhat analogous to fractional crystallization.

When you boil a mixture containing two substances with appreciably different boiling points (80.1 oC and 110.6 oC), the vapor formed will be somewhat richer in the more volatile compound (the one that vaporizes/boils at the lower temp). If the vapor is condensed in a separate container and that liquid is boiled again, a still higher concentration of the more volatile substance will be obtained in the vapor phase. By repeating this process many times, it is possible to obtain a pure sample of the more volatile substance.


Boiling-point elevation

Because the presence of a nonvolatile solute lowers the vapor pressure of a solution, it must also affect the boiling point of the solution.


Phase Diagram of a pure substance. “Normal” boiling and freezing points are indicated.


For each temperature the solute decreases the vapor pressure of the solution.

Notice that the solution boils at a higher temperature than the pure substance (boiling occurs when the vapor pressure is equal to atmospheric pressure – in this case 1 atm - so now the solution needs to be hotter to reach a vapor pressure of 1 atm.)

Phase diagram of a pure substance and curves representing the effect of adding a solute.


Boiling-point elevation (Tb) is defined as the CHANGE in the boiling point (boiling point of the solution minus the boiling point of the pure solvent).

The value of Tb is proportional to the concentration (molality) of the solution. That is,

Tb = kbm

where kb is the molal boiling point elevation constant. The units of kb are oC/m.

(You can find a table of molal boiling-point elevation and freezing-point depression constants of several common liquids on your reference sheet.)


What is the new boiling point when 175 grams of sucrose, C12H22O11, are added to 285 mL of water? Tb = kbm

= 1.79 m .511 mol C12H22O11

.285 kg water

175 g C12H22O11 x 1 mol C12H22O11 = .511 mol C12H22O11

342.30 g C12H22O11

Tb = (.52 oC/m)(1.79 m) = .93 oC

.93 oC + 100.00 = 100.93 oC


Freezing-point depressionFreezing involves a transition from the disordered state

to the ordered state. For this to happen, energy must be removed from the system. Because a solution has greater disorder than the solvent, more energy needs to be removed from it to create order than in the case of the pure solvent. Therefore, it needs to get colder to freeze a solution, giving the solution a lower freezing point than the solvent.


Freezing-point depression (Tf) is defined as the CHANGE in the freezing point (freezing point of the pure solvent minus the freezing point of the solution). Note** Tf is ALWAYS positive.

The value of Tf is proportional to the concentration (molality) of the solution. That is,

Tf = kfm

where kf is the molal freezing point depression constant.


An aqueous solution of glucose freezes at –2.56 oC. How many grams of glucose must have been added to 150.0 mL of water to form this solution?

X = .207 mol

1.38 mol C6H12O6

1 kg water

Tf = (1.86 oC/m)(m) = 2.56 oC

m = 1.38

X mol C6H12O6

.150 kg water=

.207 mol C6H12O6 x 180.15 g C6H12O6 = 37.3 g C6H12O6

1 mol C6H12O6


Explain why salt is placed on frozen roads and sidewalks.

Salt depresses the freezing point of water, which means that the ice on the roads and sidewalks will melt into water, even at temperatures below 0 oC.


Osmotic Pressure

The net movement of solvent molecules through a semipermeable membrane from a pure solvent of a dilute solution to a more concentrated solution is called osmosis. The osmotic pressure (p) of a solution is the pressure required to stop osmosis. This pressure can be measured directly from the difference in the final fluid levels.


Like boiling point elevation and freezing point depression, osmotic pressure is directly proportional to the concentration of solution. If two solutions are of equal concentration (hence, equal osmotic pressure), they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more dilute solution is described as hypotonic.


Using colligative properties to determine molar mass

The colligative properties of nonelectrolyte solutions provide a means of determining the molar mass of a solute. Theoretically, any of the four colligative properties is suitable for this purpose. In practice, however, only freezing-point depression and osmotic pressure are used because they show the most pronounced changes.

58

A 7.85 g sample of a compound is dissolved in 301 g of benzene. The freezing-point of the solution is 1.05 oC below that of pure benzene. What is the molar mass of this compound?

•Our first step is to calculate the molality of the solution.

Molality = Tf = 1.05 oC = 0.205 m

kf 5.12 oC/m

•Since there is 0.205 mole of the solute in 1 kg of solvent, (0.250 m) , the number of moles of solute in 301 g, or 0.301 kg of solvent is

•Finally, because we know that 7.85 g of the substance is equal to .0617 mol, we can calculate the molar mass of the solute as follows

7.85 g = 127 g/mol

0.0617 mol

0.205 mole =

1 kg solvent .301 kg solvent

.0617 mol


The colligative properties of electrolytes require a slightly different approach than the one used for the colligative properties of nonelectrolytes. The reason is that electrolytes dissociate into ions in solution, and so one unit of an electrolyte compound separates into two or more particles when it dissolves. (Remember, it is the number of solute particles that determines the colligative properties of a solution.)


For example, each unit of NaCl dissociates into two ions – Na+ and Cl-. Thus the colligative properties of a 0.1 m solution of NaCl should be twice as great as those of a 0.1 m solution containing a nonelectrolyte, such as sucrose. Similarly, we would expect a 0.1 m Ba(NO3)2 solution to depress the freezing point by three times as much as a 0.1 m sucrose solution because Ba(NO3)2 produces three ions. To account for this effect we must modify the equations for colligative properties as follows:

Tb = ikbm

Tf = ikfm

The variable i is the van’t Hoff factor which is equal to the number of particles in solution after dissociation. (Remember, only electrolytes dissociate!)


What is the new freezing point when 25 grams of ammonium nitrate are added to 285 mL of water?

Tb = ikbm

.312 mol NH4NO3

.285 kg water

25 g NH4NO3 x 1 mol NH4NO3 = .312 mol NH4NO3

80.05 g NH4NO3

Tb = (2)(1.86 oC/m)(1.10 m) = 4.09 oC

New freezing point = -4.09 oC

= 1.10 m


ColloidsThe solutions we discussed so far are true

homogeneous mixtures. Now consider what happens if we add fine sand to a beaker of water and stir. The sand particles are suspended at first but then gradually settle to the bottom. This is an example of a heterogeneous mixture. Between these two extremes is an intermediate state called a colloidal suspension, or simply, a colloid. A colloid is a dispersion of particles of one substance (the dispersed phase) throughout a dispersing medium made of another substance.


Types of ColloidsDispersing Medium

Dispersed phase

Name Example

Gas Liquid Aerosol Fog, mist

Gas Solid Aerosol Smoke

Liquid Gas Foam Whipped cream

Liquid Liquid Emulsion Mayonnaise

Liquid Solid Sol Milk of magnesia

Solid Gas Foam Plastic foams

Solid Liquid Gel Jelly, butter

Solid Solid Solid gel Certain alloys (steel), gemstones


One way to distinguish a solution from a colloid is by the Tyndall effect. When a beam of light passes through a colloid, it is scattered by the dispersed medium. No such scattering is observed with ordinary solutions because the solute molecules are too small to interact with visible light.

Another demonstration of the Tyndall effect is the scattering of sunlight by dust or smoke in the air.


The End

solutions 2009-20101 everyone has problems, but chemists have solutions

Documents

better solution

unsaturated solution

remaining solution

mol c2h5oh

mol h2o

moles of amoles

mol fraction ethanol

g c2h5oh x