Chemical reactions usually take place in mixtures dissolved in water. In this unit we will learn the definition and nature of solutions, the factors affecting the rate at which substances dissolve, calculations and units of concentration. We will also discuss the increase in boiling point and freezing point of solutions, and experimental methods of separating substances in solution.
A solution is a homogeneous mixture. The substance that does the dissolving is called the
solvent and the substance that is dissolved is called the solute. The substance in the largest amount is usually the solvent but not always. For example, it is possible to dissolve 180 g of sugar in 100 g of water. Solvents and solutes can be any state of matter. Below is a table listing many combinations that make up a solution.
salt water flavored drink (like
Kool-ade) rubbing alcohol
(70% isopropanol) carbonated water water water isopropanol salt sugar, dye, and flavoring water unpolluted air water nitrogen (72%) carbon dioxide gas oxygen (22%), argon
(1%), carbon dioxide
(0.1%), and water vapor (varies < 5%) zinc or tin brass copper pewter tin silver, lead bronze copper
Every part of a solution has the same composition with the solute particles evenly distributed.
This is why the first cup of flavored drink has the same flavor and sweetness as the last cup.
Moreover, solutions with solvents of liquid and gas are clear since the solutes are uniformly mixed and the particles of the solutes are now widely separated within the solvent.

When ionic compounds are dissolved in water the cation and anion separate and each is surrounded by water molecules. In a water molecule, the electron density is concentrated around the oxygen, so the molecule has a partial negative charge over oxygen and a partial positive charge near the hydrogens. The negative oxygens on water will surround each cation to dissolve it. Likewise, the positive hydrogens of water will surround each anion. See the animation of the process at http://mw2.concord.org/public/student/solution/dissolve.html
& there is a lesson at http://molit.concord.org/database/activities/186.html
.
Water is a polar molecule, with positive and negative ends http://commons.wikimedia.org/wiki/File:
Watermolecule.png
NaCl dissolves in water and sodium is surrounded by water, solvated. http://commons.wikimedia.org/wiki/File:Na%
2BH2O.svg
See the animation of the process at http://mw2.concord.org/public/student/solution/dissolve.html
& there is a lesson at http://molit.concord.org/database/activities/186.html
.
Polar molecules behave much the same with the positive end of the molecule attracted to the negative oxygen of water and the negative end of the polar molecule attracted to the positive hydrogens of water. However, nonpolar molecules do not dissolve, since the attraction between water molecules is strong (including H-bonding) the nonpolar molecules are excluded.
The solute must be evenly distributed in the solvent to be a solution. Muddy water that separates after a long period of time is called a suspension of clay and water. Whipped cream is a mixture with evenly distributed milk fats and sugar in water. This mixture is called a colloid.
Colloids are very much like solutions except the dissolving particles are so large that the mixture is not clear like in a solution.
There are four factors affecting the rate of dissolving and three of them are ways you already know increase the rate of dissolving. 1) Agitation. Agitation means stirring, swirling, or shaking a solution. The more the solvent and the solute are mixed actively and not allowed to remain still, the faster the solution will form. 2) Increasing heat. Most solid and liquid solutes

will dissolve when heated (which increases molecular motion). 3) Decrease surface area. By reducing the size of the pieces of solid being dissolved the process is faster.
All these factors affect how fast the ions, molecules, or atoms in a solute are broken apart and surrounded by the solvent. To see how this works picture a solid ionic compound arranged in its crystal lattice where each ion is connected to the surrounding ions, except for those ions on the outer surfaces where there are fewer connections. Water, the most common solvent, moves randomly and starts to connect to the different ions. At some point the water molecules, which can surround either positive or negative ions, provide a more stable system for the ion than the surrounding ions, because some of these ions are the same charge that repel the ion and some are opposite charges that attract the ion. The ion is surrounded by water and leaves the area. New water molecules move in and surround the next ion.
When a solution is agitated (stirred, shaken, swirled, etc), the process of moving the surrounded ions away from the undissolved substance proceeds faster and the solid is dissolved more rapidly. If the solution is heated the molecules of water and the particles of the solid move more rapidly, both of which help break up of the crystal lattice and move the
solvated ions away from the dissolving substance.
Increasing the surface area of a solid involves adding more surfaces to the solute. Breaking or cutting the solid creates new surfaces, so smaller pieces have greater surface area than a larger piece of the same mass. These smaller pieces provide more places for the solvent to interact with the solute. This increases the rate of dissolving since more areas are exposed to and engage in solute particles being surrounded by and dissolved by the solvent.
Some nice simulations of the dissolving process are shown at: http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/molvie1.swf
& http://molo.concord.org/database/activities/186.html
Pressure the Fourth Factor
Besides agitation, temperature, and surface area, the fourth factor affecting the rate of dissolving is pressure. But pressure fundamentally affects only gases, because solids and liquids are not compressible, so pressure has very little affect on them. For a gas, an increase in pressure will increase the solubility of the gas. This is why bottles of soda are so stiff. The high pressure inside the bottle is necessary to maintain the high solubility of the carbon dioxide in the carbonated water and the fizz you expect from a fresh bottle of soda. Once the bottle has been opened and the pressure is released, the amount of carbon dioxide decreases.
If the cap is left off the bottle the amount of carbon dioxide continues to decrease until the soda is flat, indicating most of the carbon dioxide has left the solution.
Concentration
Solutions are described in terms of concentration. The laundry detergent is Ultra because it is more concentrated, the bug spray is more effective because it has a higher concentration of repellant, the water is more pure because the concentration of dissolved solids is low, and your allergies are triggered by a higher concentration of pollen in the air. Concentration is used for solutions because the total amount of solute is dissolved in either a little bit of solvent

or a lot of solvent and that matters. Streams dissolve a lot of water and carry it down to the ocean to make it salty, but there is no salty taste in streams because the stream only dissolves a small bit of salt at a time. It is concentrated in the ocean, where the amount of salt increases and the amount of water remains essentially constant.
Concentration can be described in a variety of ways: parts per million or billion (or trillion), mass per volume, volume per volume, percent of solute to solution, moles of solute per liter of solution, or moles of solutes per kilogram of solution. ppm or ppb
The EPA (Environmental Protection Agency) monitors our water and air carefully and our water services have elaborate purification systems that make sure that the water we get in our taps is clean and safe. In fact the water from most municipalities only has small amounts of solutes like iron ions, Fe 3+ , calcium ions, Ca 2+ , sodium ions, Na + , and chlorine ions, Cl – , that are measured in parts per million, ppm, or parts per billion, ppb.
The picture of 100 dots is 10 times 10. The one black dot represents 1 part per hundred. Below each cluster of dots is 10 blue dots, so this is a picture of 1000 particles. The one black dot represents 1 part per thousand (the black dot is 1.5 times bigger than the other dots to emphasize its location). To get to 1 part per million, there would have to be 1000 similar pictures of 1000 dots all blue except for the 1 black dot already shown (imagine stacking 1000 papers of the picture on top of each other). To represent 1 part per billion, it would take
1,000,000 similar pictures of 1000 dots.
The parts per million and billion are ratios of the mass of the solute to total mass of the solvent.
The density of water is assumed to be 1.00 g/mL just like pure water since very little solute is in the solution.
The EPA regulates the amount of substances that are allowed in pure air and water. For air the amounts of pollutants are closely watched because many people have difficulty breathing even when traces of certain substances are in the air. The limit for contamination in air can be very small and are reported as parts per billion. Ozone is one of the primary pollutants and at a level of 115 parts per billion the air is unhealthy for many people and all children especially if they are active. When ozone reaches a level of 400 parts per billion the air is unhealthy for all people. Another pollutant from cars, buses, trucks, and many other oil powered systems is nitrogen dioxide. Nitrogen dioxide is responsible for the brown-red color in a city sunset.
This gas is hazardous when it reaches 1.6 parts per million.
Curious. Look here for air pollutants and then look at the calculator that converts the Air
Quality Index, AQI, to units of concentration: http://www.epa.gov/air/urbanair/ & http://www.airnow.gov/index.cfm?action=resources.aqi_conc_calc
.
Aqueous (or water) solutions can also have particles in small amounts that are listed best using ppm. For example, many ions in bottled water are listed in concentrations of ppm.
While many bottled waters are filtered tap water (Aquifina, Dasani) others use the water

purified by cities for their bottled water. Other bottled waters are can be nearly mineral free spring waters. Mineral water is water whose unusual taste is favored by many people to compliment their food instead of simply for hydration. Perrier, from Vergeze, France, contains
150 ppm calcium, 320 ppm bicarbonate (hydrogen carbonate, HCO
3–
), and 4.2 ppm magnesium (plus other ions); Avita, from Northern Michigan, contains 46 ppm calcium, 151 ppm bicarbonate, and 10.3 ppm magnesium; and Spa Reine, from Belguim, contains 3.5 ppm calcium, 11 ppm bicarbonate, and 1.3 ppm magnesium. All these different waters taste different and people favor one bottle over another because of the small amounts of ions that are dissolved in the different waters.
Our tap water is regulated by the EPA so many tap water systems provide very similar concentrations of ions even though the taste can be quite different. Many of these substances are measured in concentration units of ppm, but sometimes the concentration unit used is µg per L (a microgram, µg, is one millionth of a gram). Here is a list of drinking water contaminate limits published by the EPA: http://www.epa.gov/safewater/contaminants/index.html
.
Lead and arsenic are commonly known to be dangerous in drinking water. The maximum amount of lead allowed in drinking water is 0.015 µg Pb per liter of solution. The maximum amount of arsenic is 0.010 µg As per liter of solution.
Some consumer products are sold with the concentration listed as volume of solute per volume of solution. Isopropanol is listed as 70% isopropanol, which means that its concentration is 70 mL of isopropanol for each 100 mL of solution. Lugol’s solution, which is an iodine solution used to disinfect cuts and scraps, is 2.2% solution of iodine/iodide.
In chemistry, the amount we are interested in is the mole, so chemicals are often sold in concentrations listing the moles per liter of solution. The strong acid, hydrochloric acid, is sold in 12 moles per liter solution. The unit for moles per liter is called molar and has a label M. So hydrochloric acid is sold in 12 M concentration. Sulfuric acid, one of the top chemicals in the world is sold as 18 M concentration. Calcium hydroxide a strong base, but it is also insoluble, so its highest concentration is 0.00105 M.
Calculating Concentration
Parts per million is a ratio calculation that determines the amount in a million based on the given mass of solute and solution (if the solvent is water, the density of the solution is 1.000 g/mL, since there is very little solute so the density is the density of water). For example,
“Determine the ppm of lead in a 1 L solution that contains 0.0015 g of lead.” {Note that this solution is 100,000 times greater than the EPA amount mentioned above}
1 L • 1.00 g/mL = 1000 mL • 1 g/mL = 1000 g of solution
Known Unknown
0.0015 g Pb = x gives x = 1.5 ppm
1000 g solution 1,000,000
Another example, “Find the maximum amount of PCB’s, polychlorinatedbiphenyls, (widely used chemicals that were banned in 1979) in 2 L, if the EPA limit is 0.5 ppb in drinking water.”

Known Unknown
0.5 g PCB = x gives x = 0.000001 g PCB
1,000,000,000 2,000 g solution
A ratio can be used to calculate the percent solution too, by recognizing that a percentage is the amount over 100. For example, “A solution of salt, ice, and water has a temperature of -21.0
°C, what is the percent salt of this solution if 2.00 grams of salt is used with 9.00 g of solution
(water, ice, and salt)?”
Known Unknown
2.00 g salt = x gives x = 22.2 % salt solution
9.00 g solution 100 or
“Find the amount of sugar in 75 g of maple sap if maple sap is only 2% sugar (this is just like a percent problem in math so solve it that way if you can).”
Known Unknown
2 g sugar = x gives x = 1.5 g sugar in the sap.
100 g solution 75 g solution or
“If a wine is 8% alcohol (ethanol) by volume, then how many mL of alcohol are in a 200 mL glass of wine?”
Known Unknown
8 mL alcohol = x gives x = 16 mL of alcohol (about 3 tsp)
100 mL solution 200 mL solution
Molarity
Chemical calculations of molarity are usually solved using dimensional analysis, but the ratio calculation will work too. Molarity, which is moles per liter, is the most common concentration problem and it will show up in a wide variety of problems. It is particularly important in experimental calculations including titration (which was introduced in the acidbase unit).
Example 1. “Find the molarity of a sugar solution if 1.5 grams of sugar, C
12
H
22
O
11
, are dissolved in 70 mL of solution.”
Known Unknown
1.5 g sugar • 1 mol C
342 g C
12
12
H
H
22
22
O
O
11
11
• 1 (a place holder) = x mol/L x = 0.0627 M
0.070 L solution
M or M are unit symbols for molarity
Example 2. “The molarity of barium hydroxide solution is highest at about 0.1 M. Find the mass of barium that will dissolve in 250 mL of water. Use 3 significant figures;”
Known Unknown
0.250 L solution • 0.10 mol Ba(OH)
2
• 171 g Ba(OH)
2
= x g x = 4.28 g
1 L of solution 1 mol Ba(OH)
2
Experimental data often uses molarity, but, like a titration, there is a change in substances that requires stoichiometry to solve for the unknown.

Example 3.  A titration of 0.15g of the solid acid, KHP, which has a molar mass of 204 g/mol, is done with an unknown concentration of potassium hydroxide, KOH. If 33.0 mL of the solution are used than what is the concentration of the base solution? The balanced equation is
KHP + KOH  H
2
O + K
2
P (KHP is an abbreviation not a chemical formula).”
Known Unknown
0.15 g KHP • 1 mol KHP • 1 mol KOH • 1 = x mol/L ; x = 2.2X10
-3 M
204 g KHP 1 mol KHP 0.033 L KOH
To find concentration the answer must have units of mol/L—two units. In the last step, all that is
needed is to divide the mol KOH by L of KOH. The 1 is a placeholder in the calculation and both units, mol/L, are part of the answer.
Example 4a. “A titration between 0.10 M Ba(OH)
2
and an unknown concentration of HBr is conducted with 17.7 mL of Ba(OH)
2
is added to 10 mL of HBr. Find the mass of HBr, hydrobromic acid, that reacts with the base. The balanced chemical equation is:
Ba(OH)
2
 +  2HBr  2 H
Known
2
O + BaBr
2
Unknown
0.0177 L Ba(OH)
2
• 0.10 mol Ba(OH)
2
• 2 mol HBr • 80.9 g HBr = x g ; x = 0.29 g
1 L Ba(OH)
2
1 mol Ba(OH)
2
1 mol HBr
Example 4b. “A titration between 0.10 M Ba(OH)
2
and an unknown concentration of HBr is conducted with 17.7 mL of Ba(OH)
2
is added to 10 mL of HBr. Find the unknown
concentration of HBr, hydrobromic acid, that reacts with the base. The balanced chemical equation is: Ba(OH)
2
 +  2HBr  2 H
Known
2
O + BaBr
2
Unknown
0.0177 L Ba(OH)
2
• 0.10 mol Ba(OH)
2
• 2 mol HBr • 1 = x M ; x = 0.354 M
1 L Ba(OH)
2
1 mol Ba(OH)
2
0.010 L
Another concentration calculation that requires units of mol/L, so the 1 is a placeholder.
Molality
Molality is used in calculations of the properties of a solvent. For this reason the denominator of molality is kg of solvent and is not about the mass or volume of the solution. Molality is not used in stoichiometry calculations, so the examples are simple calculations.
Example. “Determine the molality of a 1 L solution that contains 4.0 grams of sodium chloride, NaCl (table salt). Assume the density of the solution is 1.00 g/mL.”
Solution mass = 1000 mL • 1.00 g/mL = 1000 g = 1.00 kg = mass salt + mass water
0.004 kg + y
Known
58.4 g NaCl 0.996 kg
Unknown
4.0 g NaCl • 1 mol NaCl • 1 = 6.88 X 10 –3 m
Need the kg of solvent, so change solution mass to solvent mass but make sure it has units of kg.
Example. “How many grams of calcium chloride, CaCl
2
, are needed to make a 0.20 m solution in 4.0 L of water?” use 1 g/mL as the density of water to change 1.0 L to 1.0 kg

Known Unknown
4.0 kg water • 0.20 mol CaCl
2
• 111 g CaCl
1.0 kg water 1 mol CaCl
2
2
= 89 g CaCl
2
Freezing Point Depression and Boiling Point Elevation
When a substance is placed in a pure solvent the freezing and the boiling point changes. This is the affect of antifreeze in the radiator of a car. The antifreeze is added to water and keeps the car’s cooling system from boiling over in summer or when the car is working hard. But it also keeps the water in the engine and radiator from freezing, which would cause either to break (see what happens to cast iron when water freezes: http://jchemed.chem.wisc.edu/JCESoft/CCA/CCA0/MOVIES/ICEBOMB.html
).
In the winter, salt is added to ice on the roads to melt the ice. This is another situation where the solution has a lower freezing point than the pure substance. The snow and ice are frozen near 0°C, but when the salt dissolves in the water, that inevitably forms on the surface of the ice (because of equilibrium between the solid and liquid states of water), the solution doesn’t freeze. This continues until all the ice has melted and only a salt water solution remains. Salt solutions will raise the boiling point also, but there isn’t a common need for extremely salty solutions. Sugar solutions, however, are used to make syrups and candy. These sugar solutions have the boiling points greater than pure water.
Any solute will change the boiling point and the freezing point. While a common
misunderstanding is to believe both the freezing and boiling point increased or both decreased, the change in freezing point and boiling point widens the gap between the two. The freezing point is lowered and the boiling point is raised. The full name of each effects helps to remind you of the direction of the change: freezing point depression and boiling point elevation.
These two properties and changes in vapor pressure and changes in osmotic pressure are called colligative properties. Colligative properties are physical properties of a solvent that are changed by the addition of a solute.
To understand boiling point elevation and freezing point depression it is important to understand how melting and freezing are in equilibrium at the freezing point and how boiling and condensing are in equilibrium at the boiling point (in a closed container at constant pressure). Let’s focus on the freezing point to describe what is happening with particles of solvent and solute. Equilibrium occurs when two processes, which are the reverse of each other like melting and freezing, have equal, unchanging rates of change in concentration.
During equilibrium the amounts of the two states of matter remain constant, but the amounts do not have to be equal. For example, a glass of ice water held at 0°C has both melting and freezing taking place at equal rates. Thus, the masses of the ice and liquid water remain constant (but not necessarily equal); even while, melting and freezing continue, so individual molecules of water change from solid to liquid and visa versa during equilibrium. To picture the movement of molecules and the process of equilibrium, imagine a busy restaurant that always has a full number of people sitting down to eat: 20. People outside (who must be a constant number: 100) are passing by and some will wait to be seated and others will go on by.
The number of people who are seated remains the same, but those individuals who are sitting

is always changing. The rate of people leaving the restaurant and those who can come into the restaurant to sit down for a meal is the same as is the number of people wandering around.
Now a colligative effect, like the freezing point depression, occurs when a solute is added to this equilibrium. With a solute in the liquid, the molecules of water have difficulty organizing themselves to form the solid. This slows down the freezing process. Now the rates of freezing and melting are no longer equal, melting is favored and faster, so more melting will occur until any frozen particles must melt, because of the higher rate, before more particles can freeze. If the temperature remains constant the addition of the solute will cause all the ice to melt.
However, if the temperature is lowered, then the rate of freezing is increased (in a glass of pure ice and liquid water any temperature below 0°C will freeze all the water to ice, because the rate of freezing is faster than melting and there is no equilibrium of ice and water). An increase in the rate of freezing counteracts the decrease in the rate of freezing caused by adding solute particles and at a certain temperature the equilibrium between melting and freezing is reestablished. This temperature is the freezing point of the solution.
Here is a table of the freezing point depressions and boiling point elevations of different solutions of table salt, NaCl
Molality of
NaCl
Solution
Freezing Point
Depression (∆T)
Freezing Point of Solution
Boiling Point
Elevation
(∆T)
Boiling Point of
Solution
0 0 °C 0 °C 0 °C 100 °C
0.25 0.9 °C 0.9°C +0.3 °C 100.3°C
1.2 -4.5 *C -4.5 *C +1.2 *C 101.2 *C
2.7 -
10.0 °C
-
10.0 °C +2.8 °C 102.8 °C
Note that freezing point depression and boiling point elevation are changes in temperature not the actual freezing point or boiling point.
Types of Compounds as Solutes
All substances can be used as solutes, but NaCl, salt, will have a greater effect than sugar, and
CaCl
2
, calcium chloride, will have a greater effect than NaCl. This is because ionic compounds break into both cations and anions when they are dissolved, but covalent compounds only break into individual molecules. Here’s the changes in freezing and boiling points for water solutions (aqueous solutions).
Substance
(1.0 m )
Freezing Point
Depression (∆T)
Freezing Point of Solution
Boiling Point
Elevation (∆T)
Boiling Point of Solution
C
6
H
12
O
6
, dextrose
1.9 °C 1.9 °C +0.5 °C 100.5 °C

Substance
(1.0 m )
Freezing Point
Depression (∆T)
Freezing Point of Solution
Boiling Point
Elevation (∆T)
Boiling Point of Solution
NaCl
(Na + Cl
–
)
CaCl
2
(Ca 2+ , 2Cl
–
)
-
3.7 °C
5.6 °C -
3.7 °C
5.6 °C
+1.0 °C
+1.6 °C
101.0 °C
101.6 °C
Fe(NO
3
)
3
(Fe 3+ , 3NO
3
–
)
7.4 °C 7.4 °C +2.1 °C 102.1 °C
There is a formula for calculating the freezing point depression and boiling point elevation
(remember this is a change in temperature not the freezing point or boiling point) for any solution:
∆T f
= i•K f
•m ∆T b
= i•K b
•m
∆T f
is the change in temperature of the freezing point; & ∆T b
is change for boiling point
i = number of particles the substance breaks into when dissolved, and called the Van’t Hoff factor;
K f
is a freezing point constant that depends on the solvent (remember it is the solvent that is changing it’s properties); & K b
is the boiling point constant
m, is the molality of the solution.
Example. “What is the freezing point of water, when 1 grams of ethanol, C
2
H
5
OH, is added to
1 L of water? Assume the volume remains constant and the density of the solution is 1.00 g/mL.”
K f
= 1.86 °C/m for water (must be given or found in a table);
m = mol/kg = 1 g • 1 mol C
2
H
5
OH • 1 = 0.022 mol/kg = 0.022 m
46.08 g C
2
H
5
OH 1.0 kg water
Variables | Formula | Work (filled in formula) |Answer & Units
∆T f
= x | ∆T f
= i•K f
•m | x = 1•1.86°C/m•0.022m | x = 0.041 °C
i = 1 (molecular)
K f
= 1.86°C/m
m = 0.022 m
{Now, answer the question} freezing point = -0.041°C
Example. “What is the boiling point of water, when 1 grams of potassium bromide, KBr, is added to 2 L of water? Assume the volume remains constant and the density of the solution is
1.00 g/mL.”
K b
= 0.52 °C/m for water (must be given or found in a table);
m = mol/kg = 1 g • 1 mol KBr • 1 = 0.0042 mol/kg = 0.0042 m
119 g KBr 2.0 kg water
Variables
∆T f
= x
| Formula
| ∆T f
= i•K f
•m
| Work (filled in formula) |Answer & Units
| x = 2•0.52°C/m•0.0042 | x = 0.0044 °C
i = 2 (K + & Br – )

K f
= 0.52°C/m
m = 0.0042 m
{Now, answer the question} Boiling point = 100.0044°C
Example. “Determine the boiling point depression for ethanol, K b
= 1.22 K/m, when 0.50 g of
FeCl
3
is dissolved in 1.0 L of the solvent. The density of ethanol is 0.79 g/ml ” mass = density•volume = 0.79kg/L•1.0L = 0.79 kg of ethanol
m = mol/kg = 0.50 g • 1 mol FeCl
3
• 1 = mol/kg = 3.9X10
–4
162 g FeCl
3
0.79 kg ethanol
Variables | Formula | Work (filled in formula) |Answer & Units
∆T f
= x | ∆T f
= i•K f
•m | x = 4•1.22 K/m• 3.9X10
–4 m | x = 0.0019 K
i = 4 (Fe 3+ & 3Cl – )
K f
= 1.22°C/m {Although, K = 273 + °C, ∆T in K = ∆T in °C }: ∆T = 0.0019 K
m = 3.9X10
–4 m
Experimental Methods of Purifying Substances.
Purifying substances to create pure solvents or pure solutes is an important process both in the lab and in industry. The crude oil that comes from the ground becomes the gasoline and diesel oil that goes into cars, the tar on the road, the kerosene in lamps and many other products from plastics to petroleum jelly. To create these products the crude oil is put through a “cracking process.” This method of separating the different products that come from oil relies on the different boiling points of the different substances. http://commons.wikimedia.org/wiki/File:Crude_Oi l_Distillation.png
In the laboratory setting, different liquid substances formed in an experiment can be separated using the different boiling points of the different compounds. This method is called distillation. In the apparatus at the right, the mixture of liquids is placed in the round flask below the thermometer. Using a heat source, the temperature is increased until the solution begins to boil. The glass above the boiling solution heats up from the hot gases rising from the boiling solution, which keeps the substance hot enough to remain a vapor. The heat is adjusted slowly and carefully so only the lowest boiling liquid is able to reach the condensing column (or condenser). The purity of the liquid is checked by using the thermometer, which shows the boiling point of the liquid at its tip. If the temperature of the thermometer matches the boiling point of the pure substance than the distillate is collected in a flask at the end of the

condensing column until the boiling point increases. Several pure liquids can be collected in this way (some liquid between boiling points is thrown away to make sure only a pure liquid is collected).
Distillation and chromotography, which is discussed next, use differences in the physical properties of substances to separate a mixture. Any method that relied on chemical methods of separation would destroy some substance in the mixture; which is not helpful in producing pure substances. http://commons.wikimedia.org/wiki/File:Simple_distillat ion_apparatus.png
Laboratory distillation set-up
1: Heat source
2: Still pot
Mixture
3: Still head
4: Thermometer/Boiling point temperature
5: Condenser
6: Cooling water in
7: Cooling water out
8: Distillate/receiving flask
Pure Liquid
9: Vacuum/gas inlet
10: Still receiver
11: Heat control
12: Stirrer speed control
13: Stirrer/heat plate
14: Heating (Oil/sand) bath
15: Stirrer bar/anti-bumping granules
16: Cooling bath.
Another method of separating substances widely used in industry and in the lab is called chromatography. Chromatography is used in industry to make pure compounds for medicine and research. It is used to explore new types of molecules found in plants and animals (see the chromatography of spinach to the right), and it can be used for forensic identification of substances in cosmetics, soils, dyes, etc. DNA profiling uses a similar technique.
Chromatography comes in many forms: thin layer chromatography
(TLC), column chromatography, paper chromatography, gas chromatography
(GC), high pressure liquid chromatography (HPLC), and many others, but they all have a mobile phase that moves through a stationary phase. The mobile phase is a solvent,

which is liquid or gas and polar or nonpolar, that moves through the stationary phase pushing and pulling the different substances in the mixture.
The substances must travel through the entire stationary phase to the end where they are collected. Different types of monitoring like electrical conductivity and color show when the different substances reached the end of the stationary column.
Chromatography uses the physical properties of solubility and adsorption to separate the components of a mixture. The mobile phase dissolves the substances in the mixture. Some substances dissolve well and some dissolve poorly—often because polar substances dissolve in polar solvents, nonpolar substances dissolve nonpolar substances, and polar and nonpolar substances do not dissolve each other. Meanwhile the stationary layer will have surfaces that attract the different substances. If a substance in the mixture is attracted to the stationary layer—again because of polar and nonpolar interactions or because of hydrogen bonding—then it will move slowly with the solvent. If the substance in the mixture does not adsorb to the stationary phase then the substance is more likely to flow with the solvent. Together these two physical properties of being attracted to the mobile phase or to the http://commons.wikimedia.org/wiki/File:Ch romatography_of_chlorophyll_results.jpg
stationary phase give each substance a different rate of moving through the stationary phase.
A mixture of substances is placed on top of the stationary phase. The mobile phase (a solvent) passes through the stationary phase carrying the components of the mixture with it.

The components separate because they dissolve well or poorly in the mobile phase and because they have different attractions for the stationary phase. Pure substances are collected at the end of
Summary the stationary phase. http://commons.wikimedia.org/wiki/File:Ch romatography_column.PNG
Solutions are a common component of chemical reactions and many substances are solutions like air, salt water, flavored drinks, alloyed metals, and carbonated beverages. A solution is a homogeneous mixture with a solvent that is the compound or element doing the dissolving and is commonly the major component of the mixture and solutes, which are the compounds or elements that are dissolved in the mixture.
At a molecular level, a solution is created when the molecules, atoms, or ions of the solute are surrounded by solvent molecules and distributed widely and randomly throughout the volume of the container. To increase the rate at which the solute is broken up into individual particles, one can increase the temperature, increase the surface area of the solute by creating smaller pieces, or agitating the mixture by stirring, swirling, or shaking the container.
Furthermore, increasing pressure increases the rate of dissolving of gases.
Concentration is the amount of substance in a certain volume. In chemistry there are many units of concentration: parts per million—ppm—percent by mass or volume, mass per volume, moles of solute per liter of solution—Molarity, M—and moles of solute per kg of solvent— molality, m. Problems involving parts per million (or billion or trillion) and percent concentrations can be solved using ratio calculations. Molarity and molality problems are best solved using dimensional analysis to keep track of the units.
Molarity problems are important in experimental methods called titrations. These problems involve a mole ratio to change substances in a stoichiometry calculation. Molality is used in freezing point depression and boiling point elevation problems.
Freezing point depression and boiling point elevations are colligative properties caused by adding solute to a system at equilibrium at the freezing point or boiling point of a pure solvent. The added solute disrupts the equilibrium between the two states of matter so that only one state exists. By lowering freezing point or raising the boiling point temperatures, the equilibrium is reestablished at a new freezing or new boiling point. The change in freezing and boiling point is calculated using the equation: ∆T = i•K•m. The Van’t Hoff factor, i, equals the number of particles in one chemical formula formed when the substance is dissolved. For molecules, usually, only one particle is in each chemical formula so i = 1, but for ionic compounds each formula unit has multiple ions; e.g. NaCl has i = 2 and Al
2
(SO
4
)
2
has i = 5.
Mixtures can be separated into pure substances by the techniques of distillation and chromatography. Distillation uses differences in boiling points to separate the components of a mixture. Chromatography uses the dissolving of substances in a mobile phase that travels through a stationary phase, and the attraction of the components of the mixture for the material in the stationary phase. Both processes can separate more that one substance in a

mixture, but both use difference physical properties of the components of a mixture, because using chemical properties will destroy the needed compound or element.

Solutions
6. Solutions are homogenous mixtures of two or more substances. As a basis for understanding this concept: a.Students know the definitions of solute and solvent. b.Students know how to describe the dissolving process at the molecular level by using the concept of random molecular motion. c. Students know temperature, pressure, and surface area affect the dissolving process. d.Students know how to calculate the concentration of a solute in terms of grams per liter, molarity, parts per million, and percent composition. e. * Students know the relationship between the molality of a solute in a solution and the solution's depressed freezing point or elevated boiling point. f. * Students know how molecules in a solution are separated or purified by the methods of chromatography and distillation.
Starred standards are non-tested standards on the California Standards Test
Contributed by Kenneth Pringle
Edited by Kathleen Duhl
Formatted and Wiki Contribution by Christine Mytko